Paracompactness and Perfect Maps (6.41.8)

James Munkres Topology, chapter 6.41, exercise 8:

Let $p:X→Y$ be a closed surjective continuous map such that $p^{-1}\{y\}$ is compact for each $y∈Y$. If $X$ is Hausdorff, show $X$ is paracompact if and only if $Y$ is.

Proof: ($⇐$) Let $Y$ be paracompact, and let $\mathcal{B}=\{U_α\}$ be an open cover of $X$. For each $y∈Y$, let $\mathcal{B}_y⊆\mathcal{B}$ be a finite subcover of $p^{-1}\{y\}$, and obtain open $V_y⊆X$ such that $p^{-1}\{y\}⊆V_y⊆∪\mathcal{B}_y$ and $p(V_y)⊆Y$ is a neighborhood of $y$. Then $\{p(V_y)\}$ is an open cover for $Y$, so let $\mathcal{A}$ be a locally finite open refinement covering $Y$. For each $y∈Y$, let $y∈A_y∈\mathcal{A}$. We claim $\mathcal{C}=\{p^{-1}(A_y)∩B~|~y∈Y, B∈\mathcal{B}_y\}$ is a locally finite open refinement of $\mathcal{B}$ covering $X$. It is clear $\mathcal{C}$ is an open refinement, and given $x∈X$, we have $x∈p^{-1}(y)∩B⊆p^{-1}(A_y)∩B$ for some $y∈Y$ and $B∈\mathcal{B}_y$, therefore $\mathcal{C}$ covers $X$. As well, let $U$ be a neighborhood of $p(x)$ intersecting $A_y$ for only finitely many $y∈Y$. Then $p^{-1}(U)$ is a neighborhood of $x$ intersecting $p^{-1}(A_y)$ for only finitely many $y∈Y$, hence intersecting $p^{-1}(A_y)∩B$ for only finitely many pairs $(y,B)$ when $B∈\mathcal{B}_y$. Therefore $\mathcal{C}$ is locally finite and $X$ is paracompact.

($⇒$) Let $X$ be paracompact, and let $\mathcal{B}$ be an open cover of $Y$. Then let $\mathcal{A}$ be a locally finite open refinement of $\{p^{-1}(B)~|~B∈\mathcal{B}\}$ covering $X$. We claim $\mathcal{C}=\{p(A)~|~A∈\mathcal{A}\}$ is a locally finite refinement of $\mathcal{B}$ covering $Y$, so that since $Y$ is regular by normality of $X$, $Y$ will be paracompact by Lemma 41.3. The only nontrivial quality to check is local finiteness; given $y∈Y$, for each $x∈p^{-1}(y)$ let $U$ be a neighborhood of $x$ intersecting only finitely many elements of $\mathcal{C}$. Since $p^{-1}(y)$ is compact, there exists a neighborhood about it intersecting only finitely many elements of $\mathcal{C}$, and furthermore a saturated sub-neighborhood $V_y$ of this one. Being saturated, $V_y∩A=ø$ implies $p(V_y)∩p(A)=ø$ for all $A∈\mathcal{A}$, so that $p(V_y)$ is a neighborhood of $y$ intersecting only finitely many elements of $\mathcal{C}$.$~\square$

Unions of Paracompact Spaces (6.41.7)

James Munkres Topology, chapter 6.41, exercise 7:

Let $X$ be a regular space. Show that (a) if $X$ is a finite union of closed paracompact subspaces, or (b) if $X$ is covered by the interiors of countably many closed paracompact subspaces, then $X$ is paracompact.

Proof: (a) Suppose $X=K_1∪K_2$ where $K_1,K_2⊆X$ are closed and paracompact. By induction the general case will follow from this one. Suppose $\{U_α\}$ is an open cover of $X$. Then $\{U_α∩K_1\}$ and $\{U_α∩K_2\}$ are open covers of $K_1$ and $K_2$ respectively, so let $\mathcal{B}_1 = \{B_β\}$ and $\mathcal{B}_2=\{B_γ\}$ be locally finite refinements of the two. We claim $\mathcal{C}=\mathcal{B}_1∪\mathcal{B}_2$ is a locally finite refinement of $\{U_α\}$ covering $X$. The refinement and covering conditions are evident, so we proceed to demonstrate local finiteness; assume $x∉K_1$. If $U$ is a neighborhood of $x$ such that $K_2∩U$ intersects only finitely many members of $\mathcal{B}_2$, then $U∩(X-K_1)$ is a neighborhood of $x$ intersecting among the same finite family from $\mathcal{B}_2$, and is disjoint from the members of $\mathcal{B}_1$. So we may assume $x∈K_1$, and similarly $x∈K_2$. As such let $U$ and $V$ be neighborhoods of $x$ such that $K_1∩U$ and $K_2∩V$ intersect only finitely many members of $\mathcal{B}_1$ and $\mathcal{B}_2$ respectively. If $U∩V$ intersects an element of $\mathcal{B}_1$, then since that element is within $K_1$ so too does $U∩V∩K_1⊆U∩K_1$, so that only finitely many elements of $\mathcal{B}_1$ are intersected. Similarly too for $\mathcal{B}_2$, and now $U∩V$ is a neighborhood of $X$ intersecting only finitely many elements of $\mathcal{B}$.

(b) Let $X=∪\text{int }K_i$ where each $K_i$ is paracompact. Again let $\{U_α\}$ be an open cover of $X$. For each $i$, let $\mathcal{B}_i$ be a locally finite refinement of $\{U_α∩K_i\}$ covering $K_i$, and further let $\mathcal{A}_i=\{B∩\text{int }K_i~|~B∈\mathcal{B}_i\}$. Then $\mathcal{A}_i$ is a locally finite open cover of $\text{int }K_i$ for each $i$, so that $\mathcal{A}=∪\mathcal{A}_i$ is a countably locally finite open refinement of $\{U_α\}$ covering $X$, so that $X$ is paracompact by Lemma 41.3.$~\square$

Product of a Compact and Paracompact Space (6.41.2a)

James Munkres Topology, chapter 6.41, exercise 2a:

Show that the product of a paracompact space $X$ and a compact space $Y$ is paracompact.

Proof: Let $\{U_α\}$ be an open cover of $X×Y$. For each $x∈X$, the space $\{x\}×Y≅Y$ is compact, so let it be covered by finitely many $U_{x_1},...,U_{x_n}$ with nontrivial intersection with $\{x\}×Y$, and let $W_x=∩π_1(U_{x_i})$. Then $\{W_x\}_{x∈X}$ is an open cover of $X$, so let $\mathcal{A}$ be a locally finite open refinement covering $X$. For each $A∈\mathcal{A}$, finitely many $U_{x_1},...,U_{x_n}$ cover $A×Y⊆W_x×Y$ (for some $x$), so let $C_A=\{(A×Y)∩U_{x_i}\}$. We claim $\mathcal{C}=∪_{A∈\mathcal{A}}C_A$ is a locally finite open refinement of $\{U_α\}$ covering $X×Y$.

Every element of each $C_A$ is contained in some $U_α$, so $\mathcal{C}$ is clearly a refinement. As well, given $z=x×y∈X×Y$, let $x∈A∈\mathcal{A}$ so that $z∈A×Y=∪C_A$ implying $z$ is contained in an element of $C_A$ and now $\mathcal{C}$ covers $X$. Finally, choose a neighborhood $U$ of $x$ intersecting only finitely many members $A_i∈\mathcal{A}$. Then $U×Y$ is a neighborhood of $z$ that can intersect only among the members of $C_{A_i}$, all of which are finite. Therefore $\mathcal{C}$ is locally finite and $X×Y$ is paracompact.$~\square$

Countable Local Finiteness (6.39.5-6)

James Munkres Topology, chapter 6.39, exercises 5-6:

5. If $X$ is second-countable, show a collection $\mathcal{A}$ of subsets of $X$ is countably locally finite if and only if it is countable.

6. Let $ℝ^ω$ have the uniform topology. Given $n$, let $\mathcal{B}_n$ be the collection of all subsets of the form $\prod A_i$ where $A_i=ℝ$ for $i≤n$ and $A_i$ equals $\{0\}$ or $\{1\}$ otherwise. Show $\mathcal{B}=∪\mathcal{B}_n$ is countably locally finite, but is neither countable nor locally finite.

Proof: (5) It suffices to consider when $\mathcal{A}$ is an uncountable, countably locally finite collection of subsets of $X$. Since countable unions of countable sets are countable, there exists an uncountable locally finite collection $\mathcal{B}$. Assume $ø∉\mathcal{B}$, and construct a choice function $f : \mathcal{B}→∪\mathcal{B}$ such that $f(B)∈B$ for each $B∈\mathcal{B}$. If $\{U_α\}$ is a countable basis for $X$ and $\mathcal{B}$ is locally finite, then the countable set $$\{V_n\}=\{U_α~|~f^{-1}(U_α)\text{ finite}\}$$ covers $X$. But now $$\mathcal{B}=f^{-1}(X)=f^{-1}(∪V_n)=∪f^{-1}(V_n)$$ is a countable union of finite sets, so $\mathcal{B}$ is countable, a contradiction.

(6) It's clear that $\mathcal{B}_n$ is uncountable for any $n$, so $\mathcal{B}$ is not countable. As well, $1^ω$ is contained in every subset of the form $\prod A_i$ where, for some $N$, $A_i=ℝ$ for every $i≤N$, and $A_i=\{1\}$ otherwise, so that $\mathcal{B}$ is not even point-finite, let alone locally finite. But the $1/2$-neighborhood of any element in $ℝ^ω$ intersects at most one element of $\mathcal{B}_n$, so $\mathcal{B}_n$ is evidently locally finite, hence $\mathcal{B}$ is countably locally finite.$~\square$

Stone-Cech Compactification of Discrete Spaces (5.38.7-8)

James Munkres Topology, chapter 5.38, exercises 7-8:

7. Let $X$ be a discrete space.
(a) Show that if $A⊆X$, then $\overline{A}∩\overline{X-A}=ø$ where their closures are taken in $β(X)$.
(b) Show that if $U⊆β(X)$ is open, then $\overline{U}$ is open.
(c) Show that $X$ is totally disconnected.

8. Show the cardinality of $β(ℕ)$ is at least as great as $I^I$ where $I=[0,1]$.

Proof: 7. (a) Define a function $f : X→ℝ$ by $f(A)=1$ and $f(X-A)=0$. Letting $F : β(X)→ℝ$ extend $f$, we see $F^{-1}(1)$ and $F^{-1}(0)$ are disjoint closed sets in $β(X)$ containing $A$ and $X-A$ respectively, so these latters' closures are disjoint.

(b) Note $\overline{U∩X}∪\overline{X-U∩X}$ is the whole space $β(X)$ since its complement is an open set not intersecting $X$, so by part (a) $\overline{U∩X}$ is open. Now evidently $\overline{U∩X}⊆\overline{U}$, but also $\overline{U}⊆\overline{U∩X}$ since if there exists $x∈\overline{U}$ with a neighborhood $V$ disjoint from $U∩X$, let $y∈V∩U$ and now $V∩U$ is a neighborhood of $y$ not intersecting $X$ hence $y∉\overline{X}$, a contradiction.

(c) Let $x,y∈β(X)$ be distinct. Since $β(X)$ is Hausdorff let $U$ be open such that $x∈U$ and $v∉\overline{U}$. Then by part (b) $\overline{U}∪β(X)-\overline{U}$ is a separation of $β(X)$ disconnecting $x$ from $y$.

8. As we've seen (cf. 5.31.16a), $ℝ^I≅(0,1)^I$ has a countable dense subset, so $\overline{(0,1)^I}=[0,1]^I$ does as well, call it $S$. Letting $f : ℕ→S⊆[0,1]^I$ be a surjection that is automatically continuous, we obtain a map of $β(ℕ)$ into $I^I$ containing $S$, and since images of compact sets are compact hence closed in a Hausdorff space, the map is a surjection and the claim is proven.$~\square$

Munkres Review Chapters 1-4

(1) The ordered square is connected. Proof: Linear continua are connected.

(2) $ℝ^ω$ in the uniform topology is disconnected. Proof: The sets of bounded and unbounded sequences in $ℝ^ω$ are both open in this metric, and form a separation.

(3) The ordered square is not path connected. Proof: Suppose there exists a path from $0×0$ to $1×1$. Since $I_0^2$ is a linear continuum, this implies the path is surjective and $I_0^2$ is an image of the separable $[0,1]$. But $I_0^2$ itself is not separable, for $r×(1/4,3/4)$ for $r∈I$ is a collection of uncountably many disjoint open subsets.

(4) $ℝ_K$ is not path connected. Proof: Suppose $f : [0,1]→ℝ_K$ is a path from $0$ to $1$. Then $f^{-1}(0)$ is a closed set not containing $1$, so $r = \text{sup }f^{-1}(0) < 1$. Since $[r,1]≅[0,1]$ we may assume $f(x) > 0$ for $x > 0$.

Now, let $a_n = \text{inf } f^{-1}(1/n)$ for $n∈ℕ^+$. We see $a_{n+1} < a_n$ by connectivity of continuous images, but also $a_n \nrightarrow 0$ in $[0,1]$ since $1/n \nrightarrow 0$ in $ℝ_K$. Hence let $a > 0$ be such that $a < a_n$ for all $n$. But then $f(a) > 0$ so $1/N < f(a)$ for some $N$, implying $a_N < a$ again by connectivity of continuous images, a contradiction.

(5) The ordered square is not locally path connected. Proof: The proof of (3) extends to show that the path components of $I_0^2$ are precisely $r×[0,1]$ for $r∈[0,1]$, which are not open.

(6) $ℝ_l$ is not locally compact at any of its points. Proof: Let $x∈ℝ_l$ and $U$ be a hood of $x$. Suppose $V$ is a hood of $x$ such that $\overline{V}⊆U$ is compact. Then $x∈[a,b)⊆V$ for some $a,b∈ℝ$, and since $[a,b)$ is also closed in $\overline{V}$ this implies $[a,b)$ is compact. However, $[a,b)$ is not compact even in $ℝ$.

(7) $ℝ^ω$ in the uniform topology is not locally compact. Proof: Suppose $C$ is a compact subset of $ℝ^ω$ containing a hood of $i=(0,0,...)$. Then $B[i,ε]=[-ε,ε]^ω$ is compact for some $ε∈(0,1)$. But $\{e_n\}_{n∈ℕ^+}$ (when $e_n$ is the point with zeros in every coordinate except the $n\text{th}$ in which it is $ε$) is an infinite subset containing no limit point (as $d(e_n,e_m) = ε$ for every $n≠m$), so $[-ε,ε]^ω$ cannot even be limit point compact.

(8) $ℝ^ω$ in the uniform topology is not second countable, separable, or Lindelof. Proof: We see each pair of distinct $x,y∈\{0,1\}^ω$ are of distance $1$ in $ℝ^ω$, so that $ℝ^ω$ cannot be separable and hence not second countable. As well, $\{B(x,3/4)~|~x∈\{0,1\}^ω\}$ is an uncountable open cover of the closed subset $[0,1]^ω$, yet there are not even any proper subcovers, so $ℝ^ω$ cannot be Lindelof.

(9) $ℝ^I$ is not locally metrizable. Proof: Suppose some basis element $U=\prod_{i∈I} U_i$ of $ℝ^I$ were metrizable. Then since $U_i=ℝ$ for all but finitely many $i∈I$, and since $I-F$ for any finite subset $F⊆I$ is still uncountably infinite, we see $ℝ^I$ can be imbedded in $U$. But $ℝ^I$ itself is not metrizable, as it is not normal.

(10) $ℝ^I$ is not Lindelof. Proof: Regular Lindelof spaces are normal, which $ℝ^I$ is not.

Countability Axioms and Topological Groups (4.30.18)

James Munkres Topology, chapter 4.30, exercise 18:

Show in a first-countable topological group $G$ that second countability, separability, and the Lindelof condition are equivalent.

Proof: Let $\{B_n\}$ be a countable basis about $e$. We may presume $B_n⊇B_{n+1}$ for all $n$, and if necessary by setting $B_n'=B_n∩B_n^{-1}$ we may presume $B_n^{-1}=B_n$.

Suppose $G$ has a countable dense subset $D$, and that $U$ is a hood about $g$. Then since $g×e∈m^{-1}(U)$ and $\{gB_n\}$ is a basis about $g$, there exists $n$ such that $gB_n×B_n⊆m^{-1}(U)$. Choose $d∈D∩gB_n$ and we see $d=gb$ for some $b∈B_n$, hence $g=db^{-1}∈dB_n$, so $dB_n$ is a hood about $g$. Furthermore, given $c∈B_n$ we see $dc=gbc∈m(gB_n×B_n)⊆U$, so that $dB_n⊆U$. Therefore $\{dB_n~|~d∈D,n∈ℕ\}$ is a countable basis for $G$.

Suppose $G$ is Lindelof. Then for each $n∈ℕ$, there is a countable subset $D_n⊆G$ such that $∪_{d∈D_n}dB_n=G$. We show $D=∪D_n$ is dense in $G$; let $U⊆G$ be a hood about $g$. Then $gB_n⊆U$ for some $n$, and we also see $g∈∪_{d∈D_n}dB_n$ so write $g=db$ for some $d∈D_n⊆D$ and $b∈B_n$. But now $d=gb^{-1}∈gB_n⊆U$ so $d∈U$ and $D$ is a countable dense subset of $G$.$~\square$

Separability of Large Euclidean Product Spaces (3.30.16)

James Munkres Topology, chapter 3.30, exercise 16:

(a) Show that the product space $ℝ^I$ contains a countable dense subset.
(b) Show that if $|J| > |\mathcal{P}(ℕ)|$, then $ℝ^J$ does not contain a countable dense subset.

Proof: (a) For every oddly finite sequence $q_1,...,q_{2n+1}$ of rationals such that $0 ≤ q_2 < q_4 < ... < q_{2n} ≤ 1$, let there be an associated function $I → ℝ$ that takes value $q_{2k-1}$ on $(q_{2k-2},q_{2k})$ for each $k = 1, ..., n$ (where $q_0=0$), takes value $q_{2n+1}$ on $(q_{2n},1]$, and is zero on each of $q_2,...,q_2n$. We see the collection of such functions is countable, and we claim it forms a countable dense subset of $ℝ^I$: For a given basis element $∏_{i∈I}U_i$ of open sets $U_i⊆ℝ$ such that $U_i \not = ℝ$ for only finitely many $i∈I$, let $i_1 < ... < i_n$ be exactly all such that $U_{i_j} \not = ℝ$. We may choose rationals $q_2,...,q_{2(n-1)}$ such that $i_1 < q_2 < i_2 < q_4 < ... < q_{2(n-1)} < i_n$ (unless $n=1$ where one chooses $i_1 < q_2$ and say $q_3=0$) and for each $q_1,q_3,...,q_{2n-1}$ choose rationals such that $q_{2k-1}∈U_{i_k}$ for each $k=1,...,n$. Then the associated function on $q_1,...,q_{2n-1}$ takes a value within $U_i$ on each $i$, and hence is contained in the basis element.

(b) Suppose $D$ is a countable subset of $ℝ^J$. Fix some nonempty interval $(a,b)$, and choose some disjoint nonempty interval $(c,d)$. Then since the function $f : J → \mathcal{D}$ given by $f(α) = D ∩ π_α^{-1}(a,b)$ cannot be injective, we find some distinct $α,β∈ℝ$ such that for each $d∈D$ we have $d(α)∈(a,b)$ iff $d(β)∈(a,b)$. But now the basis element $∏U_j$ where $U_α=(a,b)$ and $U_β=(c,d)$ and $U_j=ℝ$ otherwise cannot contain a point of $D$.$~\square$

G-Delta Sets and First Countability (3.30.1)

James Munkres Topology, chapter 3.30, exercise 1:

(a) A $G_δ$ set in a space $X$ is a countable intersection of open sets of $X$. Show that in a first-countable $T_1$ space, every one-point set is a $G_δ$ set.
(b) There is a familiar space wherein every one-point set is a $G_δ$ set, which is nevertheless not first countable. What is it?

Proof: (a) Let $x∈X$, and let $\{B_i\}$ be a countable basis at $x$. Suppose $y∈∩B_i$ while $y \neq x$. Then since $X$ is $T_1$, let $U$ be a neighborhood of $x$ not containing $y$. We see $U$ is open and contains $x$, yet contains no element of the basis $\{B_i\}$ seeing as $y \not \in U$, a contradiction.

(b) Let $ℝ^\omega$ be under the box topology. Then given $x∈ℝ^\omega$, when we let $U_n=∏(x_n-1/n,x_n+1/n)$ we see $∩U_n=\{x\}$, so that every one-point set in $ℝ^\omega$ is $G_δ$. To show that $ℝ^\omega$ is not first countable, suppose $\{B_n\}$ is a countable basis at any particular point $x$. Then for each $n∈ℕ$, we may choose an interval $(a_n,b_n)$ such that $x_n∈(a_n,b_n) \subset π_n(B_n)$. Hence, let $U=∏(a_n,b_n)$; we see $x∈U$ yet $B_n \not \subseteq U$ for all $n$ since $π_n(B_n) \not \subseteq π_n(U)$, so that $\{B_n\}$ is not a countable basis at $x$, a contradiction.$~\square$

Fixed Point Analysis and Partial Differential Equations

Suppose $f : ℝ^2→ℝ$ is continous and globally Lipschitz in its second coordinate, i.e. there exists some $m∈ℝ$ such that for all $x,y_1,y_2∈ℝ$ $$|f(x,y_1)-f(x,y_2)|≤m·|y_1-y_2|$$ Further suppose $h > 0$ such that for the aforementioned $m$ we observe $mh < 1$. Then when $c[0,h]$ is the space of continuous functions $[0,h]→ℝ$ under the maximum absolute difference metric $d$, and $y_0∈ℝ$ is fixed, show that the operator $$T : c[0,h]→c[0,h]$$ $$T(φ)(x)=y_0+∫_0^x f(t,φ(t)) \mathrm{d}t$$ is well defined and has a unique fixed point.

Proof: Note that by the fundamental theorem of calculus, $T(φ)$ is continuous on $[0,h]$ so that $T$ is indeed well defined. Since $c[0,h]$ is a complete metric space, by Banach's theorem it suffices to show that $T$ is a contracting map to ensure existence and uniqueness of a fixed point. Observe the inequalities given $φ_1,φ_2∈c[0,h]$: $$d(T(φ_1),T(φ_2)) = \max_{x∈[0,h]} |(y_0+∫_0^x \! f(t,φ_1(t))~\mathrm{d}t) - (y_0+∫_0^x \! f(t,φ_2(t))~\mathrm{d}t)| =$$ $$\max_{x∈[0,h]} |∫_0^x \! f(t,φ_1(t)) - f(t,φ_2(t))~\mathrm{d}t| ≤ \max_{x∈[0,h]} ∫_0^x \! |f(t,φ_1(t)) - f(t,φ_2(t))|~\mathrm{d}t≤$$ $$\max_{x∈[0,h]} m·∫_0^x \! |φ_1(t)-φ_2(t)|~\mathrm{d}t ≤ mh·\max_{x∈[0,h]} |φ_1(x)-φ_2(x)| = mh·d(φ_1,φ_2)$$ so that $T$ is a contracting map of factor no larger than $mh < 1$.

Banach Fixed Point Theorem

(Banach) Let $X$ be a nonempty complete metric space, and let $T : X→X$ be a contracting map, i.e. $d(T(x),T(y)) ≤ θ·d(x,y)$ for all $x,y∈X$ for some fixed $θ < 1$. Then $T$ has a unique fixed point.

Proof: (Banach) Uniqueness of fixed points is clear, since if $T$ fixes $x$ and $y$, then $θ·d(x,y) ≥ d(T(x),T(y))=d(x,y)$ and $θ ≥ 1$ unless $d(x,y)=0$ and $x=y$. To exhibit the fixed point, let $x_0∈X$ be any point, and define $x_{n+1}=T(x_n)$. Then when $α=d(x_0,x_1)$ it follows by induction that $d(x_n,x_{n+1}) ≤ θ^nα$. As well, by the triangle inequality it follows that $d(x_n,x_m) ≤ \sum_{k=n}^{m-1} θ^kα$ when $n≤m$. Since $θ < 1$ this forms part of a convergent geometric series, showing $(x_n)$ forms a Cauchy sequence in $X$. Let $x$ be the point of convergence of this sequence. $T$ is continuous since contraction maps are generally continuous, so $T(x)$ is the point of convergence of the sequence $(T(x_n))=(x_{n+1})$, showing $T(x)=x$.

Function Spaces from Compact Sets into Complete Metric Spaces

Let $X$ be a compact space, and let $(Y,d)$ be a complete metric space. We show that the set $F$ of continuous functions from $X$ to $Y$ under the metric $$h(f,g)=\max_{x∈X}~d(f(x),g(x))~~~~~\text{for }f,g∈F$$ induces a complete metric space.

Note that when $f,g : X→Y$ are continuous, then since $d : Y×Y→ℝ$ is continuous, we observe $d∘(f×g) : X→ℝ$ is continuous and due to compactness of $X$ indeed attains a maximum value, so that the function $h : F×F→ℝ$ is well defined. To verify that it is a metric, we see $h(f,g)=0$ iff $d(f(x),g(x))=0$ iff $f(x)=g(x)$ for all $x∈X$, i.e. $f=g$. As well, $$h(f,g)=\max_{x∈X}~d(f(x),g(x))=\max_{x∈X}~d(g(x),f(x))=h(g,f)$$ Finally, let $a,b,c∈F$. Then $$h(a,c) = \max_{x∈X}~d(a(x),c(x)) ≤ \max_{x∈X}~[d(a(x),b(x))+d(b(x),c(x))] ≤$$$$\max_{x∈X}~[d(a(x),b(x))]+\max_{x∈X}~[d(b(x),c(x))] = h(a,b)+h(b,c)$$ so that $h(a,c) ≤ h(a,b)+h(b,c)$ and $(F,h)$ is a metric space.

Now we show completeness. Let $(f_n)$ be a Cauchy sequence in $F$. We show that for each $x∈X$, the sequence $(f_n(x))$ is Cauchy in $Y$. To wit, given $ε > 0$, choose $N$ such that $n,m > N$ implies $h(f_n,f_m) < ε$. Then $d(f_n(x),f_m(x)) ≤ \max_{x∈X}~d(f_n(x),f_m(x)) = h(f_n,f_m) < ε$. Thus since $Y$ is complete write $f_n(x)→f(x)$ for each $x∈X$. It now suffices to show $f : X→Y$ is continuous and $f_n→f$. We shall prove the latter convergence is uniform and the former will follow by the uniform limit theorem.

Let $ε > 0$ be given. Since $(f_n)$ is Cauchy, let $N$ be such that $n,m > N$ imply $h(f_n,f_m) < ε/2$. Then $d(f_n(x),f_m(x)) < ε/2$ for all $x∈X$. Now let $n > N$ be given. Since $f_n(x)→f(x)$ choose $m > N$ such that $d(f_m(x),f(x)) < ε/2$. Then $d(f_n(x),f(x)) ≤ d(f_n(x),f_m(x))+d(f_m(x),f(x)) < ε$ so the convergence is uniform.

Local Compactness and Products of Quotient Maps (3.29.11)

James Munkres Topology, chapter 3.29, exercise 11:

(a) Let $Z$ be a locally compact Hausdorff space, and let $p : X→Y$ be a quotient map. Show that the map $$π=p×1 : X×Z→Y×Z$$ is a quotient map.

(b) Let $p : A→B$ and $q : C→D$ be quotient maps, and let $B$ and $C$ be locally compact Hausdorff. Show $p×q$ is a quotient map.

Proof: (a) It is clear that $π$ is continuous, so it suffices to show $A⊆Y×Z$ is open when $π^{-1}(A)⊆X×Z$ is open. For any given $y×z∈A$, when $x×z$ is such that $π(x×z)=y×z$ (i.e. $x∈p^{-1}(y)$) we find a neighborhood $U$ of $x$ saturated with respect to $p$ and a neighborhood $V$ of $z$ such that $U×V⊆π^{-1}(A)$, so that $π(U×V)=p(U)×V$ will be a neighborhood of $y×z$ contained in $A$. In this fashion, let $U_1×V$ be a basis element in $π^{-1}(A)$ containing $x×z$ since $π^{-1}(A)$ is open. Since $Z$ is locally compact Hausdorff, we may assume $V$ is such that $\overline{V}$ is compact and $U×\overline{V}⊆π^{-1}(A)$.

Given open $U_i$ such that $x×z∈U_i×\overline{V}⊆π^{-1}(A)$, we construct $U_{i+1}$ such that $x×z∈U_{i+1}×\overline{V}⊆π^{-1}(A)$ and $p^{-1}(p(U_i))⊆U_{i+1}$, and we shall set $U=∪U_i$. To see this construction, note that for each $u∈p^{-1}(p(U_i))$ we see $u×\overline{V}$ is compact, so that by the tube lemma we may choose a neighborhood $V_u$ of $u$ such that $V_u×\overline{V}⊆π^{-1}(A)$. Letting $U_{i+1}=∪V_u$, we see each of the two desired properties are fulfilled.

We see $p^{-1}(p(U))=U$ by the second desired property of each $U_i$, so that $U$ is saturated. Since also $U×V⊆U×\overline{V}⊆π^{-1}(A)$ we have demonstrated the proper $U$ and $V$, so that $π$ is a quotient map.

(b) This is evident by considering the composition of quotient maps $p×q=(1×q)∘(p×1)$.$~\square$

Neighborhoods of Compact Metric Subspaces (3.27.2)

James Munkres Topology, chapter 3.27, exercise 2:

Let $X$ be a metric space with metric $d$, and let $A⊆X$ be nonempty.
(a) Show $d(x,A)=0$ if and only if $x∈\overline{A}$.
(b) Show that if $A$ is compact, $d(x,A)=d(x,a)$ for some $a∈A$.
(c) Define the $ε$-neighborhood of $A$ by $$U(A,ε)=\{x~|~d(x,A) < ε\}$$ Show that $U(A,ε)=∪_{a∈A}B_d(a,ε)$.
(d) Assume $A$ is compact; let $U$ be an open set containing $A$. Show that some $ε$-neighborhood of $A$ is contained in $U$.

Proof: (a) ($⇒$) This implies that for every $ε > 0$ there is a point $a∈A$ such that $d(x,a) < ε$, implying every neighborhood of $x$ intersects $A$ and $x∈\overline{A}$. ($⇐$) To obtain a point $a∈A$ of distance $δ < ε$ from $x$ for given $ε > 0$, take any element from $A∩B_d(x,ε)$.

(b) Let $α=d(x,A)$. Then $d : x×A→ℝ$ is continuous, so consider the nonempty closed sets $d^{-1}([α,α+1/n])$; the family has the finite intersection property, and so their intersection contains a point $a∈A$ which is seen to satisfy $d(x,a)=α$.

(c) ($⊆$) Suppose $x∈U(A,ε)$, so that $d(x,A) < z < ε$ for some $z$; choose $a∈A$ such that $d(x,a) < z$ and we see $x∈B_d(a,ε)$. ($⊇$) This is clear since $d(x,A) ≤ d(x,a)$ for all $a∈A$ by definition.

(d) Observe the closed $K=X-U$. Letting $ε=\text{inf }d(k,A)$ for $k∈K$ we see that if $ε > 0$, then $A⊆U(A,ε)⊆U$. So we may assume a sequence $k_n∈K$ such that $d(k_n,A)→0$ is decreasing. By (b), we may obtain another sequence $k_n×a_n$ such that $d(k_n,a_n)→0$ is also decreasing. Since $A$ is compact, let $a∈A$ be a limit point of $a_n$. Reusing notation, if $ε > 0$, we shall find $k∈K$ such that $d(a,k) < ε$, so that $a∈A∩K$, a contradiction. To wit, let $n$ be such that $d(a_n,k_n) < ε/2$, and let $m > n$ be such that $d(a,a_m) < ε/2$. Then by the triangle inequality $d(a,k_m) < ε$, and we are done.$~\square$

Partial Converse to the Uniform Limit Theorem (3.26.10)

James Munkres Topology, chapter 3.26, exercise 10:

(a) Let $f_n : X→ℝ$ be a sequence of continuous functions, such that $f_n(x)→f(x)$ for each $x∈X$. If $f$ is continuous, and if the sequence $(f_n)$ is monotone increasing, and if $X$ is compact, show that the convergence is uniform.

(b) Give examples to show that the converse fails to hold if either of the requirements of $X$ being compact or $f_n$ being monotone increasing is deleted.

Proof: (a) For each $n$, let $K_n=\{x∈X~|~|f_n(x)-f(x)|≥ε\}$. It is routine to show $K_n$ is closed, by showing its complement is open: Suppose $|f_n(x)-f(x)|<ε$. Then $f(x)∈(f_n(x)-ε,f_n(x)+ε)=V$, so let $U$ be a neighborhood of $f(x)$ contained in $V$, and we see $f^{-1}(U)⊆X$ is open such that $f(U)⊆V$ so that $U$ is a neighborhood of $x$ also contained in the complement of $K_n$. Now, since $(f_n)$ is monotone increasing, and hence $f(x)≥f_n(x)$ by convergence, we see $K_{n+1}⊆K_n$ for all $n$. If we assume each $K_n$ is nonempty, then finite intersections $∩K_i=K_{\text{max }i}$ are also nonempty, so since $X$ is compact, there exists $x∈X$ such that $x∈K_n$ for all $n$, i.e. $|f_n(x)-f(x)|≥ε$ for all $n$. However, this is impossible since $f_n(x)→f(x)$ for all $x$. Hence $K_N=ø$ for some $N$, hence $K_n=ø$ for all $n≥N$, that is, $|f_n(x)-f(x)|<ε$ for all $x$, for sufficiently large $n$.

(b) Suppose $X$ need not be compact; then the sequence of functions $(f_n)$ defined by $$f_n(x)= \{ \begin{array}{lr} 0 & : x ≤ n\\ n-x & : x > n \end{array}$$ is seen to be monotone increasing with $f_n(x)→0$ for all $x$, yet does not converge uniformly to the zero function.

Suppose $f_n$ need not be monotone increasing; then let $[0,1]=∪[a_n,b_n]$ be any countably infinite partition of $[0,1]$ into disjoint closed intervals where $a_n≠b_n$ for all $n$. Then when $c_n=\dfrac{a_n+b_n}{2}$, define a sequence of functions $(f_n)$ by $$f_n(x)= \{ \begin{array}{lr} 0 & : x ∉ [a_n,b_n]\\ \dfrac{x-a_n}{c_n-a_n} & : x ∈ [a_n,c_n]\\ \dfrac{x-b_n}{c_n-b_n} & : x ∈ [c_n,b_n] \end{array}$$ It is clear by the pasting lemma that each $f_n$ is a continuous function $[0,1]→[0,1]$. As well, it is clear $f_n(x)≠0$ for at most one value of $n$, so that $f_n(x)→0$ for each $x$. However, since $f_n(c_n)=1$ for each $n$, we do not have uniform convergence to the zero function.$~\square$

Generalization of the Tube Lemma (3.26.9)

James Munkres Topology, chapter 3.26, exercise 9:

Let $A⊆X$ and $B⊆Y$ be compact. Then show that for every open neighborhood $N⊆X×Y$ of $A×B$ we have $$A×B⊆U×V⊆N$$ for some open $U⊆X$ and $V⊆Y$.

Proof: We first show the partially generalized lemma when $A=\{a\}$ is a single point. Since $a×B≅B$ is compact, we obtain $$a×B⊆\bigcup(U_i×V_i)$$ when $U_i⊆X$ and $V_i⊆Y$ are open, for finitely many indices $i$. We may assume each $U_i$ is a neighborhood of $a$. We let $U=∩U_i$ and $V=∪V_i$ and claim they satisfy the lemma presented; this is evident since $U×V_i⊆U_i×V_i⊆N$ for all $i$ hence $U×V=∪(U×V_i)⊆N$, and also $a×b∈a×B$ implies $b∈V_i$ for some $i$ so $a×B⊆U×V$.

So by this partially generalized tube lemma, for each $a∈A$ let $U_a⊆X$ be a neighborhood of $a$ and $V_a⊆Y$ be open such that $U_a×V_a⊆N$. Since $A$ is compact, we may choose a finite subset $\mathcal{A}⊆A$ such that $A⊆∪_{α∈\mathcal{A}}U_α$. We let $U=∪U_α$ and $V=∩V_α$ and claim they satisfy the lemma presented. Since $U_α×V⊆U_α×V_α⊆N$ we see $U×V=∪(U_α×V)⊆N$. As well, for each $α∈\mathcal{A}$ we have $α×B⊆U_α×V_α$ so that $B⊆V_α$, implying $B⊆V$. And since the $U_α$ cover $A$ by construction we have $A×B⊆∪(U_α×V)⊆U×V$.$~\square$

Graph Criterion for Continuity into Compact Spaces (3.26.8)

James Munkres Topology, chapter 3.26, exercise 8:

Let $Y$ be a compact Hausdorff space. Then show $f : X→Y$ is continuous if and only if the graph of $f$ $$G_f = \{x×f(x)~|~x∈X\}$$ is closed in $X×Y$.

Proof: ($⇒$) Suppose $f$ is continuous. We show $(X×Y)-G_f$ is open, so let $x×y∉G_f$. Then $f(x)≠y$, so since $Y$ is Hausdorff choose a neighborhood $V$ of $y$ not containing $f(x)$. Now $Y-V$ is closed in $Y$, so since compact Hausdorff spaces such as $Y$ are regular, choose a neighborhood $W$ of $y$ so that in particular $\overline{W}⊆V$. Now, if $x∈\overline{f^{-1}(W)}$, then since $f$ is continuous we derive $$f(x)∈f(\overline{f^{-1}(W)})⊆\overline{W}⊆V$$ so $f(x)∈V$ despite the choice of $V$. Hence choose some neighborhood $U$ of $x$ such that $U∩f^{-1}(w)=ø$. We claim $U×W$ is a neighborhood of $x×y$ disjoint from $G_f$; assume otherwise that $z×f(z)∈U×W$ for some $z∈X$. Then $z∈U∩f^{-1}(W)$, a contradiction.

($⇐$) Suppose $G_f$ is closed. Then let $K⊆Y$ be closed. We see $$f^{-1}(K)=π_1(G_f∩[X×K])$$ since $f(x)∈K$ implies $x×f(x)∈G_f∩[X×K]$, and $x×k∈G_f$ for $k∈K$ implies $f(x)=k∈K$. Now since $Y$ is compact we see projection $π_1$ is a closed map, so $f^{-1}(K)⊆X$ is closed.$~\square$

Path-Connected 1-Manifold Unembeddable in R^n (3.24.12)

James Munkres Topology, chapter 3.24, exercise 12:

12. Recall $S_Ω$ denotes the minimal uncountable well-ordered set. Let $L$ denote the ordered set $S_Ω×[0,1)$ in the dictionary order with its smallest element deleted, known as the long line.

(a) Let $X$ be an ordered set, and let $a < b < c$ be elements of $X$. Show $[a,c)≅[0,1)$ as ordered sets iff $[a,b),[b,c)≅[0,1)$.
(b) Let $x_0 < x_1 < ...$ be a sequence of increasing elements of $X$, and suppose $b=\text{sup }\{x_n\}$. Show $[x_0,b)≅[0,1)$ iff $[x_i,x_{i+1})≅[0,1)$ for all $i$.
(c) Let $a_0$ denote the smallest element of $S_Ω$. For each $a∈S_Ω$ distinct from $a_0$, show $[a_0×0,a×0)$ in $S_Ω×[0,1)$ has order type $[0,1)$.
(d) Show $L$ is path connected.
(e) Show that every point in $L$ has a neighborhood homeomorphic with an open interval in $ℝ$.
(f) Show that $L$ cannot be embedded in $ℝ^n$ for any $n$.

Proof: First, note that $[0,1)≅[a,b)$ for all real $a < b$ by composing a translation with a positive scalar multiplication. As well, $$f : [0,1) → [0,∞)$$ $$f(x)=\dfrac{1}{1-x}-1$$ is bijective and order-preserving, so $[0,1)≅[0,∞)$.

(a) ($⇒$) If $[a,c)≅[0,1)$ by the order isomorphism $f$ then $[a,b)≅[0,f(b))≅[0,1)$ by the above. Similarly, $[b,c)≅[0,1)$. ($⇐$) If $[a,b)≅[0,1)$ and $[b,c)≅[0,1)$ by the isomorphisms $f_0,f_0$ respectively, then define $f : [a,c)→[0,2)$ by $f(x)=f_n(x)+n$ where $n=0$ if $x∈[a,b)$ and $n=1$ if $x∈[b,c)$. It is clear $f$ is surjective and order preserving, so that $f$ is an order isomorphism. Since $[0,2)≅[0,1)$ the claim follows.

(b) ($⇒$) As before, when $f : [x_0,b) → [0,1)$ is an isomorphism, we have $[x_i,x_{i+1})≅[f(x_i),f(x_{i+1}))≅[0,1)$. ($⇐$) For each $i∈ℕ$ let $f_i : [x_i,x_{i+1})→[0,1)$ be the isomorphism. Define $$f : [x_0,b)→[0,∞)$$ $$f(x)=f_n(x)+n$$ where $n$ is unique such that $x∈[x_n,x_{n+1})$. As before, it is clear that $f$ is surjective and order preserving, hence an isomorphism. Since $[0,1)≅[0,∞)$ the claim follows.

(c) Suppose $a∈S_Ω \setminus \{a_0\}$ is the minimal element violating the claim. Let $S=\{s∈S_Ω~|~s < a\}$. If $S$ is finite, it is clear there is a maximal element $m$ of $S$ that is the direct predecessor of $a$, so that $[a_0×0,m×0)≅[0,1)$ by minimal assumption, and $[m×0,a×0)=m×[0,1)≅[0,1)$, implying $[a_0×0,a×0)≅[0,1)$ by (a). So we may assume that $S$ is infinite and has no maximal element.

If $X$ is a countably infinite ordered set without a maximal element, define a function $g : ℕ→X$ constructed from a bijection $f : ℕ→X$ by $g(0)=f(0)$ and $g(n+1)=f(m)$ where $m∈ℕ$ is minimal such that $f(m) > g(n)$, which $m$ must exist since each element $g(n)∈X$ precedes an infinite number of elements if $X$ has no maximal element. Then $g$ shall be called a climb of $X$, and has the property that $g(0) < g(1) < ...$ is an increasing sequence, and if $X$ is embedded in $Y$ with the least upper bound property and $X$ is bounded above, then $\text{sup }X=\text{sup }\{g(n)\}_{n∈ℕ}$ (following from the fact that $g(n) ≥ f(n)$ by induction).

So let $g$ be a climb of $S$. Then it is clear that $g(0)×0 < g(1)×0 < ...$ is an increasing sequence minimally upper bounded by $a×0$, and the claim follows from (b) with an application of (a) to show each $[g(n)×0,g(n+1)×0)≅[a_0×0,g(n+1)×0)≅[0,1)$.

(d) This shall follow from part (e) when we show each two points $p,q$ of $L$ contains a neighborhood $U$ homeomorphic to $ℝ$. For if $f : ℝ→U$ is a homeomorphism, then $f : ℝ→L$ is also continuous, and we may let $r_1=f^{-1}(p)$ and $r_2=f^{-1}(q)$, and when $g : [0,1]→ℝ$ is a path from $r_1$ to $r_2$, we see $f∘g : [0,1] → L$ is a path from $p$ to $q$.

(e) As promised in (d), we shall show further that each two points $p,q∈L$ contains a neighborhood $U$ homeomorphic to $ℝ$. Assume $p < q$. Since $S_Ω$ evidently contains no maximal element, we may assume $p < q < α×0$ for some $α∈S_Ω$. Since $[a_0×0,α×0)$ in $S_Ω×[0,1)$ has order type $[0,1)$, we see $(-∞,α×0)$ in $L$ has order type $(0,1)$. Since surjective, order-preserving maps are homeomorphisms between sets endowed with the order topology, we see $(-∞,α×0)≅(0,1)$ as topological spaces. Since $(0,1)≅(-1/2,1/2)$ by translation, and $(-1/2,1/2)≅(-1,1)$ by positive scalar multiplication, it is routine to check that $f : (-1,1)→ℝ$ defined by the individual homeomorphisms on its closed halves $[0,1)≅[0,∞)$ and $(-1,0]≅(-∞,0]$ constructed similarly as above is a homeomorphism, so that $(0,1)≅ℝ$ and the claim is verified.

(f) Note that $ℝ^n$ has the countable basis $$\bigcup_{q∈ℚ^n}\bigcup_{m∈ℕ^+}B(q,1/m)$$ so that if $L$ was homeomorphic to a subspace of $ℝ^n$, it too would possess a countable basis. Also, note that for any two bases $\mathcal{A},\mathcal{B}$ of a topology, we may construct a new basis $\mathcal{C}$ consisting of only the basis elements of $\mathcal{A}$ that are contained within some element of $\mathcal{B}$. This is to show that if there is a countable basis $\mathcal{A}$ of $L$, then we may let $\mathcal{B}$ be the open intervals (without infinity) of $L$. For each $C∈\mathcal{C}$ let $C⊆(α_C×r_c,β_C×q_c)$, and for each $α∈S_Ω$ let $S_α$ denote its slice in $S_Ω$. Now when $π_1$ is projection to $S_Ω$ we note $$\bigcup_{C∈\mathcal{C}}π_1(C)⊆\bigcup_{C∈\mathcal{C}}π_1(α_C×r_c,β_C×q_c)⊆\bigcup_{C∈\mathcal{C}} S_{β_C} ⊂ S_Ω$$ since countable unions of countable sets are countable, and $S_Ω$ is uncountable. Hence not every point of the form $α×0$ is contained in a basis element, a contradiction.$~\square$

Quotient Topology Examination (2.22.4,6)

James Munkres Topology, chapter 22, exercises 4,6:

4(a) Define an equivalence relation on the plane $X=ℝ^2$ given by $$x_0×y_0 \sim x_1×y_1 \text{ if } x_0+y_0^2 = x_1+y_1^2$$ Let $X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it?

(b) Repeat (a) for the equivalence relation $$x_0×y_0 \sim x_1×y_1 \text{ if } x_0^2+y_0^2=x_1^2+y_1^2$$ 6. Let $ℝ_K$ denote the topology of $ℝ$ granted by the basis of the usual intervals $(a,b)$ and also the intervals $(a,b)-K$ where $K=\{1/n~|~n∈ℤ^+\}$. Let $Y$ be the quotient space induced by collapsing $K$ to a point, and let $p:ℝ_K→Y$ be the quotient map.

(a) Show $ℝ_k$ is $T_1$ non-Hausdorff.
(b) Show $p×p:ℝ_K^2→Y^2$ is not a quotient map.

Proof: (4)(a,b) The two equivalence relations are those induced by the functions $f(x,y)=x+y^2$ onto $ℝ$ and $g(x,y)=x^2+y^2$ onto $[0,∞)$, respectively. As these are both surjective, continuous maps, it suffices by Corollary 22.3(a) to prove that these functions are quotient maps. It will do to verify that they are open; as such, we observe their actions on the basis elements $$f((a,b)×(c,d))= \{ \begin{array}{lr} (a,b+d^2) & : c ≤ 0\\ (a+c^2,b+d^2) & : c > 0 \end{array}$$ To note that $g$ is open, note that in $ℝ$ and hence in $[0,∞)$ the addition of two open intervals is open, and that the squaring map on $(a,b)$ is $[0,\text{max}\{a^2,b^2\})$ if $a < 0 < b$, and $(a^2,b^2)$ if $0 < a < b$ and $(b^2,a^2)$ if $a < b < 0$.

(6)(a) We note that since each point in $ℝ_K$ constitutes a closed set, and $K$ is closed (every point not contained in $K$ has a neighborhood $(a,b)-K$ not intersecting $K$), that consequently every point in $Y$ constitutes a closed set, and thus is $T_1$. Now, assume open neighborhoods $U_0,U_K$ of $0,K∈Y$ respectively. Then $p^{-1}(U_0)$ is a neighborhood in $ℝ_K$ of $0$ not intersecting $K$, so that $p^{-1}(U_0)=(a,b)-K$ for some $a < 0 < b$. Let $n$ be such that $1/n < b$; then since $p^{-1}(U_K)$ is a neighborhood of $K$, we may assume $1/n∈(c,d)$ for some $0 < c < 1/n < d$; choose some $e∈(c,1/n)-K$ so that $f(e)∈U_0∩U_K$ and $U_0$ and $U_K$ are not disjoint.

(b) We first show that the diagonal $D=\{x×y~|~x=y\}⊆Y^2$ is not closed, particularly that $0×K$ is a limit point. To this end, assume a basis neighborhood $U_0×U_K$ of $0×K$ not intersecting $D$; this would imply disjoint neighborhoods $U_0,U_K$ about $0,K$ respectively, which is impossible by the above. However, if $S$ is the (closed) diagonal of $ℝ_K^2$, then $(p×p)^{-1}(D)=S∪(K×K)$ is closed in $ℝ_K^2$ so that $p×p$ isn't a quotient map.$~\square$

Matching Preclusion of (n,k)-Star Graphs

$$S_{n,1}$$
Consider $G=S_{n,1}$ ($n \geq 4$). Let $F=F_V ∪ F_E$ be any optimal strong matching preclusion set of vertices of and edges respectively. Now, $G-F_V≅K_{n-|F_V|}$ since itself $G≅K_n$, so it suffices to classify optimal matching preclusion of $K_{n-|F_V|}$ generally. By X, for any even number $r$, $\text{mp}(K_r)=r-1$ and all optimal solutions are trivial except when $r=4$ and there is the particular nontrivial preclusion formed by removing the edge set of a three-cycle. When $r$ is odd, we can observe the $r$ edge-distinct almost perfect matchings on $K_r$ formed by choosing a vertex, projecting a diameter from it, and pairing up points across the diameter. In such a situation we have proven $\text{mp}(K_r) ≥ r$. Thus $\text{smp}(G)=n-1$ and optimal solutions occur with trivial deletions and the deletions of $n-4$ vertices together with the edge set of a three-cycle.

$$S_{n,2}$$
Consider $G=S_{n,2}$ ($n \geq 9$). For each $i=1,...,n$, let $H_i ≅ K_{n-1}$ be the region of $G$ whose nodes are those with $i$ in the last position, and let $F_i=H_i∩F$. We shall call $H_i$ a component of $G$, and to prevent confusion there will be no mention of the traditionally established meaning of this term. Edges between components will be referred to as cross edges; note there is exactly one cross edge incident to any given vertex. Without loss of generality assume $|F_1| \geq |F_i|$ for all $i$. We tackle the problem based on the size of $F_1$. But first, a lemma which we shall use later:

Lemma 1: $S_{n,2}$ and $S_{n,2}-H_i$ for any component $H_i$ are "weakly" Hamiltonian connected for up to and including $n-7$ faults, that is, there remains a Hamilton path between any two vertices in distinct components. Proof: The latter graph shall prove to be more restrictive, so we focus our attention there. Note that by Y, each $K_{n-1}$ component may sustain $(n-1)-4$ faults and remain Hamiltonian connected. As well, after faulting, contracting every component down to a single point yields a $K_{n-1}$ graph with at most $n-6$ faults, so that it too is Hamiltonian connected, sustaining up two $2$ more faults. Now, on the original faulted graph let $a,b$ be vertices in distinct components. Pick a Hamilton path connecting the components of $a$ and $b$ after faulting the component contraction on up to two edges corresponding to cross edges from $a$ and $b$. Use this path to construct a Hamilton path between $a$ and $b$, traveling through each component and using Hamiltonian connectivity in each to start and end at the appropriate vertices for cross edging while hitting every vertex along the way.$~\square$

$|F_1|=n-1$

We perform casework on the number of vertices deleted.

Case 1: If $|F_V|=n-1$, the rest of the graph is Hamiltonian. Therefore, there exists either a perfect matching or almost perfect matching.

Case 2: If $0 < |F_V| < n-1$ then $|F_V|≠n-2,n-3$ (as otherwise $|F_E|≤0,1$ respectively and $|F_E|+|F_V| < n-1$). Now, we claim there are at most three isolated vertices; within $H_1-F_1$ let $s$ denote the number of isolated vertices, $r$ the remaining vertices, and $d$ the deleted vertices. Note that $|F_E| ≥ {s \choose 2} + rs$ to account for isolated-isolated edges and isolated-remaining edges respectively, and $|F_V| ≥ d$ to account for deleted vertices, hence ${s \choose 2} + rs + d ≤ n-1$. However, $rs ≥ r$ and ${s \choose 2} > s$ assuming $s > 3$, implying $d+{s \choose 2} + rs > d + s + r = |H_1| = n-1$, a contradiction unless $s≤3$. As well, $s≠2$ since the one vertex's isolation requires $n-2$ object deletions, necessitating $r=1$, but this would mean the "remaining" vertex is itself isolated. This also shows $s=3$ implies $H_1-F_1$ is merely three isolated nodes.

Now, assume $s=3$; let $x,y,z$ be the unique vertices left in $H_1$. Then $x,y,z$ have sole cross neighbors $a,b,c$ respectively distinct components of $G$. Evidently $G-H_1-a-b-c$ still contains a Hamilton path since none of the cross edges between the remaining $K_{n-2}$ and $K_{n-1}$ components are faulted by removing $a,b,c$ and each component remains Hamiltonian connected. Thus use this path and the edges $(x,a),(y,b),(z,c)$ to obtain a desired matching.

Assume $s=1$; undelete a vertex $t$ (and along with it its neighboring edges). Now in $H_1-(F-t)$ there are no isolated vertices and only the previously isolated vertex (call it $v$) has degree one, so by our analysis of $S_{n,1}$ we may observe a perfect or almost perfect matching $M$ of $H_1-(F-t)$. If $M$ is perfect, necessarily $v$ is matched with $t$; let $(v,r)$ be a cross edge. Now $G-H_1-r$ has a Hamilton path as discussed above, so from all this obtain a desired matching on $G$ after redeleting $t$. If $M$ is almost perfect, perform the same trick as above while also considering the cross edge of the unmatched vertex.

Assume $s=0$; the same argument as above holds when we undelete a vertex, consider the induced maximal submatching, eject up to two cross edges, and use a Hamilton path on the rest.

$|F_1|=n-2$

Case 1: If $H_1-F_1$ contains a perfect matching $M$ then a Hamilton path can be applied to the rest (there will be but one fault in $G-H_1$, so e.g. start in the faulted or a faulted-cross-edge-incident component and continue using the Hamiltonian connectivity of the components), and if $M$ is almost perfect, the result is routine; if necessary choose a different unmatched vertex to ensure it retains a cross edge neighbor $r$ (possible if $F$ is nontrivial), match it with this neighbor, and find a Hamilton path on the rest.

Case 2: If there is not a perfect or almost perfect matching, then there are two possibilities: a trivial deletion, or the deletion of $n-5$ vertices and the edge set of a three-cycle. In the event of a trivial deletion, we may still obtain a matching missing only the isolated $v$ and a remaining vertex $s$, since the remaining vertices form a complete subgraph. If $F$ is nontrivial we can if necessary rechoose $s$ so that $s,v$ have cross neighbors $q,r$ respectively. In the event of the second possibility let $s$ and $v$ be two degree-one vertices with accessible cross neighbors $q,r$ (and then match the other two vertices in $H_1-F_1$). In both events finish the desired matching by constructing a Hamilton path on the rest, which is possible as all the other components remain Hamiltonian connected (see $|F_1|=n-4$ for an examination of a more difficult construction).

$|F_1|=n-3$

We have a perfect or almost perfect matching in $H_1-F_1$.
Case 1: If there's a perfect matching, the rest of the graph contains a Hamilton path, as every other component is still Hamiltonian connected, so use this to construct a desired matching.

Case 2: If there's an almost perfect matching, then let $m$ be the missed vertex. Note that $H_1-F_1$ does not isolate any vertices, and furthermore at most one vertex has degree one (observe otherwise degree-one vertices $u,v$; originally $d(u)+d(v)=2n-4$, so $n-3$ deletions must contribute to removing at least $2n-6=2(n-3)$ from their combined degree. Since faulting a vertex removes $2$ and faulting an edge distinct from $(u,v)$ removes only $1$, we either have the deletion of every vertex besides $u$ and $v$ [but now there isn't an almost perfect matching], or $H_1-F_1$ is $K_3$ missing an edge [if $u,v$ have faulted cross edges this is a trivial "fishtail" preclusion of $G$ which allows a maximal matching missing two vertices, and if not we may proceed with the main argument]). Hence we may find at least three almost perfect matchings on $H_1-F_1$ each missing distinct vertices, so we may assume $m$ has an unfaulted cross edge $(m,m')$. Now the argument is a simpler case of the one we shall handle for $|F_1|=n-4$ now, where we find a Hamilton path on the rest of $G-H_1-F'-m'$.

$|F_1|=n-4$

We have a perfect or almost perfect matching in $H_1-F_1$. Let $F'=F-F_1$.
Case 1: If there's a perfect matching, the rest of the graph contains a Hamilton path, as every other component is still Hamiltonian connected, so use this to construct a desired matching.

Case 2: If there's an almost perfect matching, then since $H_1$ contains a Hamilton cycle by Q necessarily of odd length ≥3, we may assume this matching misses $m$ which has an intact cross edge $(m,m')$ (unless it is the semitrivial disconnection of $H_1-F_1≅K_3$ from the rest of the graph); include the cross edge $(m,m')$, and observe that the remaining components $H_i≅K_{n-1}$ have at most $4$ faults each, so that by Y they remain Hamiltonian connected. As well, the component contraction of $G-H_1-F$ yields a $K_{n-1}$ with at most $3$ faults, so it retains a Hamilton path. Use these facts to construct a Hamilton path on $G-H_1-F'-m'$ and obtain a desired matching.

$|F_1|≤n-5$

The component contraction of $G-F'$ contains a perfect or almost perfect matching as it is itself a $K_n$ ($n > 4$) with at most $n-1$ edge faults and no isolated vertices unless all $n-1$ cross edges of a component are removed and $F$ is trivial, and at most one vertex of degree one since there are not enough faults otherwise. Add one edge back in possibly to leave every vertex with at least two unfaulted edges to apply Z to find (1) there is a Hamilton path on this component contraction after we re-remove the edge. As well, (2) each $H_i≅K_{n-1}$ component remains Hamiltonian connected as mentioned above. Use these two facts to construct a Hamilton path on $G-F$ and obtain a desired matching.

$$S_{n,3}$$

Consider $G=S_{n,3}$ ($n \geq 11$). For each $i=1,...,n$, let $H_i ≅ S_{n-1,2}$ be the region of $G$ whose nodes are those with $i$ in the last position, and let $F_i=H_i∩F$. We shall call $H_i$ an outer component of $G$, and we shall refer to individual components of $H_i$ as inner components of $G$. Similarly, we shall use the terms inner and outer cross edges; note each point of $G$ lies in a component in which it projects exactly one inner cross edge, and outside of which it projects one outer cross edge. We shall handle cases based on the size of $F_1$. But first, a lemma:

Lemma 2: $S_{n,3}-H_i$ has a Hamilton path for up to and including $n-8$ faults. Proof: Note that by the first lemma each component remains weakly Hamilton connected, and also that the (outer) component contraction on the whole faulted graph yields a $K_{n-1}$ with no faults (there are $n-2$ outer cross edges between any two outer components), so it retains a Hamilton path. Use these facts to construct a Hamilton path by starting in a given outer component, selecting a point in a distinct inner component to which to run a Hamilton path in accordance with the Hamilton path on the outer component contraction (this is possible for each component as at most $n-7$ of the $n-2$ cross edges between components will be unavailable; $n-8$ from faulting and $1$ from requiring it be from a distinct inner component), and continuing in this fashion throughout each outer component.$~\square$

$|F_1|=n-1$

Note $F$ must contain at least two vertices, otherwise the outer cross edges form a desired matching. Thus, let $a,b∈F$ be vertices and consider the perfect or almost perfect matching on $H_1-(F_1-a-b)$; we may assume $(a,b)$ is not part of this matching otherwise $G-F$ has a desired matching, and similarly neither $a$ nor $b$ are missed by this matching, so let $(a,c)$ and $(b,d)$ be edges in the matching. Consider the outer cross edges $e,f$ of $c,d$ respectively; we may find a perfect or almost perfect matching on $G-H_1-e-f$ by lemma 2 since $n≥10$, and perfectly match the rest.

$|F_1|=n-2$

Case 1: We have a perfect or almost perfect matching on $H_1-F_1$. With a perfect matching, we can perfectly match every other unfaulted outer component and maximally match the remaining one to obtain a desired matching on $G-F$. If there's an almost perfect matching missing $m$, then if $F$ is not trivial we may possibly rechoose $m$ so that it has an available outer cross edge $m''$. Add the matching on $H_1-F_1-m''$ plus $(m,m'')$ to a Hamilton path on the rest by lemma 2 to obtain a desired matching on $G-F$.

Case 2: We don't have a perfect or almost perfect matching. By our work on $S_{n,2}$, we may break this down into three subcases based on the semitrivial preclusions.

Trivial: The vertex $t$ is isolated from the odd number of vertices left, a maximal matching missing $t$ and $v$. If $F$ is nontrivial we may assume an outer cross edge $(t,t'')$ and possibly rechoose $v$ to assume its own outer cross edge $(v,v'')$. By the lemma finish a desired matching on $G-F$ with a Hamilton path on the rest since $n≥11$. Component Disconnection: Perform the same trick on the missed vertices inside and outside the disconnected component. Fishtail: Choose a matching missing one of the "fins" that has an unfaulted outer cross edge, and now perform the same trick again.

$|F_1|=n-3$

If $H_1-F_1$ has a perfect matching, by lemma 2 we may use a path to find a desirable matching on $G-F$ since $n≥10$. If there is an almost perfect matching, then we cannot have two vertices of degree one in $H_1-F_1$, since each has $n-2$ neighbors originally, and by considering (inner) cross edges they do not have the same set of neighbors. Since there are no isolated vertices, this implies (as in the case of $|F_1|=n-3$ for $S_{n,2}$) three distinct matchings missing distinct vertices, so that we may assume an almost perfect matching missing $m$ with an intact outer cross edge $(m,m'')$. As usual apply lemma 2 to finish a matching with a path on the rest.

$|F_1|≤n-4$

We turn our focus momentarily to prove a fact about $S_{n,2}$, that is, that it contains a Hamiltonian cycle for up to and including $n-3$ deletions from $K$ (for $n≥6$). We break it down by case on the size of $K_1 := K∩H_1$. If $|K_1|=n-3$, then $H_1-F_1$ contains a Hamiltonian path, so use its endpoints and the Hamiltonian connectivity of the other components to complete the cycle. If $|K_1|=n-4$, then $H_1-F_1$ contains a Hamiltonian cycle by Q (this result isn't making any sense!); use it to find a Hamiltonian path as before on $H_1$ with endpoints of unfaulted cross edges, and again use Hamiltonian connectivity of the other components to complete the cycle. If $|K_1|≤n-5$ then every component is Hamiltonian connected, and the component contraction of $S_{n,2}-K$ yields a $K_n$ with at most $n-3$ edge faults, and so contains a Hamiltonian cycle; use these to construct the cycle.

Now, we return to $G-F$. Note that each faulted outer component $H_i-F_i$ now contains a Hamiltonian cycle by the above. Each pair of outer components share $n-2$ outer cross edges, so we divide into two cases based on whether all $n-2$ between a pair have been faulted.

Case 1: Every pair of faulted outer components retains an outer cross edge between them. Then pair up odd-length Hamiltonian cycles within each outer component, leaving one out if necessary. Using an outer cross edge between each pair, we may perfectly match each pair of odd-length Hamiltonian cycle components and all the even-length ones individually, giving a desired matching on $G-F$.

Case 2: The outer cross edges between $H_i$ and $H_j$ have all been faulted, and necessarily every other pair retains a common cross edge. The process is as easy as above unless $H_i-F_i$ and $H_j-F_j$ are the unique odd-length Hamiltonian cycles. However, in this case $n-2$ of the $n-1$ deletions are within $H_i$, $H_j$ or the outer cross edges between them; this implies there is at most one other component with a fault. Choose an unfaulted component $H_k$, send two outer cross edges into it from $H_i$ and $H_j$ (preferably into distinct inner components), and use Hamiltonian connectivity on $H_k$ to take care of $H_i$, $H_j$, and $H_k$ and obtain a desired matching on $G-F$.

$$S_{n,k}$$

We shall prove by induction on $k$ that $S_{n,k}$ is super strongly matched when $n-8≥k≥3$. We have established the base case when $k=3$. Now, observe general $G=S_{n,k}$. We carry familiar terminology into this case.

$|F_1|=n-1$

Case 1: $H_1-F_1$ retains a perfect or almost perfect matching. In either case we may perfectly match the rest as each component is even, to obtain a desired matching on $G-F$.

Case 2: $H_1-F_1$ does not contain a perfect or almost perfect matching. Similar as to $S_{n,3}$ when $|F_1|=n-1$ we may undelete two vertices $a,b$ to find they take part in a matching with $c,d$ respectively, which have cross edge neighbors $e,f$ respectively. If $e,f$ are in the same outer component then perfectly match it, the rest, and the maximal matching in $H_1-(F_1-c-d)$ to obtain a desired matching, and if they are in different outer components, send an outer cross edge between the components to the same effect.

$|F_1|=n-2$

Case 1: $H_1-F_1$ has a perfect or almost perfect matching. The argument is verbally identical to the parallel case in $S_{n,3}$, though we argue the end by considering outer cross edges as above rather than by Hamilton path.

Case 2: We don't have a perfect or almost perfect matching. By induction this must only be a trivial deletion isolating $m$ and resulting in a matching missing $m$ and $t$. If $F$ itself is not trivial, we may use the outer cross edge $(m,m'')$ and possibly rechoose $t$ to also keep its outer cross edge $(t,t'')$. Add these to the matching and if necessary cross edge between two components to obtain a desired matching on $G-F$.

$|F_1|=n-3$

$H_1-F_1$ contains a perfect or almost perfect matching. It is familiar enough to clean up in the former case, and in the latter case it is simple enough to prove no two vertices in $H_1$ can have the same set of neighbors, hence $H_1-F_1$ leaves at most one vertex with degree one and we may use three missed-vertex-distinct almost perfect matchings to select one with a missed vertex having an unfaulted outer cross edge. The cleanup is again routine.

$|F_1|≤n-4$

When $n > 7$ and $k≥4$ we have at least $(n-2)(n-3)=n^2-5n+6 > n-1$ outer cross edges between any pair of components, so for each pair of faulted components with only an odd number of vertices, we may connect them by an outer cross edge and have few enough faults to perfectly match what remains within that pair. This gives rise to a desired matching on $G-F$ and finishes the inductive step.

Analytic Characterizations of Continuous Functions of the Lower Limit Topology (2.18.8)

James Munkres Topology, chapter 2.18, exercise 8:

8. (a) Suppose $f:ℝ→ℝ$ is "continuous from the right," that is, $$\lim_{x→a^+}f(x)=f(a)$$ for all $a∈ℝ$. Show that $f$ is continuous when considered as a function from $ℝ_l$ to $ℝ$.
(b) What sort of functions $f:ℝ→ℝ$ are continuous when considered as functions from $ℝ$ to $ℝ_l$? As maps from $ℝ_l$ to $ℝ_l$?

Proof: (a) Let $(a,b)⊆ℝ$. We show that for each $x∈f^{-1}(a,b)$ we have $x∈[x,c)$ for some $c$ so that $f^{-1}(a,b)$ is open. Assume not; then for each $y > x$ we have $f[x,y)⊈(a,b)$. Let $y_1,y_2,...$ be an infinite sequence of terms approaching $x$ from the right such that $f(y_i)∉(a,b)$ for all $i$. But now by hypothesis $f(x)$ is a limit point of $f(y_i)$, implying $x$ is a limit point of the complement of the open $(a,b)$ so that by its closure $f(x)∉(a,b)$, contradiction.

(b) Say $p∈ℝ$ is a local minimum of $f$ if there exists $a,b∈ℝ$ such that $p∈(a,b)$ and $q∈(a,b)⇒f(p)≤f(q)$. We show that $f:ℝ→ℝ_l$ is continuous at $p$ iff $f:ℝ→ℝ$ is continuous and $p$ is a local minimum of $f$, and continue to prove $f$ is continuous iff $f$ is constant. ($⇒$) Let $f:ℝ→ℝ_l$ be continuous at $p$. Then $f'=i∘f:ℝ→ℝ$ is continuous at $p$ seeing as it is the composition of two functions continuous at $p$, the inclusion $i:ℝ_l→ℝ$ being continuous since $ℝ_l$ is finer than $ℝ$. Now, let $(a,b)$ be an open neighborhood of $p$ within $f^{-1}[f(p),∞)$; we have $q∈(a,b)⇒f(q)∈[f(p),f(p)+1)⇒f(p)≤f(q)$ so that $p$ is a local minimum. ($⇐$) Let $f(p)∈[c,d)$. Choose a neighborhood $U$ of $p$ on which $p$ is minimum, and a neighborhood $V$ of $p$ for which $f(V)⊆(-∞,d)$, so that $U∩V$ is a neighborhood of $p$ for which $f(U∩V)⊆[f(p),d)⊆[c,d)$, so $f:ℝ→ℝ_l$ is continuous at $p$.

Suppose $f:ℝ→ℝ_l$ is continuous, so that it is continuous at each point $p$, so that every point of $ℝ$ is a local minimum of $f$ and $f:ℝ→ℝ$ is continuous. Suppose $f(x)≠f(y)$ for some $x < y$, and assume $f(x) > f(y)$ as the argument will be symmetric; let $c=\text{sup }\{z~|~t∈[x,t]⇒f(t)≥f(x)\}$. Evidently if $f(c)≥f(x)$ then $c$ is not a local minimum since for each neighborhood of $c$ we have some element $d > c$ within that neighborhood such that by the supremum definition of $c$ we see $f(d) < f(x) ≤ f(c)$. But if $f(c) < f(x)$ by some margin $ε$ we observe each neighborhood of $c$ contains some $d < c$ such that $f(d)≥f(x)$ and hence $f(c) < f(d)$ by a margin at least as large as $ε$ so that $f:ℝ→ℝ$ cannot be continuous at $c$.

Similarly, we can show that a map $f:ℝ_l→ℝ_l$ is continuous at a point $p$ iff $f:ℝ→ℝ$ is continuous at $p$ and $f$ is locally increasing at $p$, that is, there is $ε > 0$ such that $f$ is increasing on $[p,p+ε)$. By a similar method we find that the continuous functions are precisely the increasing functions.$~\square$

Conditions for Comparability of Product Topologies (2.16.5)

James Munkres Topology, chapter 2.16, exercise 5:

Let $X$ and $X'$ denote the same set under the topologies $J$ and $J'$ respectively; let $Y$ and $Y'$ denote the same set under the topologies $T$ and $T'$ respectively.

(a) Show that if $J⊆J'$ and $T⊆T'$ then the topology of $X'×Y'$ is finer than the topology of $X×Y$.
(b) Does the converse of (a) hold?
(c) What can you say if $J⊆J'$ and $T \subset T'$?

Proof: (a) The general basis element for $X×Y$ is $U×V$ when $U⊆X$ and $V⊆Y$ are open. Since the topologies of $X'$ and $Y'$ are finer than $X$ and $Y$ respectively, $U⊆X'$ and $V⊆Y$ are open and $U×V⊆X'×Y'$ is open. Hence the generated topology of $X×Y$ is contained in that of $X'×Y'$.

(b) Suppose $U⊆X$ and $V⊆Y$ are open, with $x∈U$ and $v∈V$. Then $(u,v)∈U×V⊆X×Y$ is contained in an open set, so since the topology of $X'×Y'$ is finer than that of $X×Y$ we have $(u,v)$ is contained in a basis element of $X'×Y'$, so write $(u,v)∈U'×V'⊆U×V$. Hence $u∈U'⊆U$ and $v∈V'⊆V$ showing $U$ and $V$ are also open in $X$ and $Y$ respectively.

(c) Suppose $X$ is nonempty. Then if the topology of $Y'$ is strictly finer than that of $Y$, let $V$ be open in $Y'$ and not in $Y$. As such, choose $v∈V$ such that there is no open $W⊆Y$ such that $v∈W⊆V$. Choose some $x∈X$. Then $(x,v)$ is contained in the open set $X×V⊆X'×Y'$. Now there cannot be a basis element $U×W⊆X×V⊆X×Y$ with open $U⊆X$ and $W⊆V⊆Y$ containing $(x,v)$ by specific choice of $v$. Hence the topology of $X'×Y'$ is strictly finer than that of $X×Y$.$~\square$

Namesake Property of Completions (3.24(c))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 24(c):

Let $X$ be a metric space, and let $X^*$ be the metric space of equivalence classes of Cauchy sequences under the relation $\{p_n\} \text{~} \{q_n\}$ if $\lim_{n→∞} d(p_n,q_n)=0$ and under the distance metric $Δ(P,Q)=\lim_{n→∞} d(p_n,q_n)$ for any Cauchy sequences $\{p_n\}$, $\{q_n\}$ in $P,Q$ respectively. Prove $X^*$ is complete.

Proof: Lemma: Let $s$ be a Cauchy sequence in some metric space. Say $s$ is a quick Cauchy sequence if $n,m≤N$ implies $d(s(n),s(m)) < \dfrac{1}{N}$. Then a subsequence of $s$ is a quick Cauchy sequence, so particularly the class $S$ of $s$ under the above relation contains a representative that is a quick Cauchy sequence. Proof: Since $s$ is Cauchy, for all $n∈ℕ^+$ let $N_n$ be such that $a,b≥N_n$ implies $d(s(a),s(b)) < \dfrac{1}{n}$. Let $s'$ be the subsequence such that $s'(n)=s(N_n)$, and it's seen that $n,m≥N$ implies $d(s'(n),s'(m))=d(s(N_n),s(N_m)) < \text{max}(\dfrac{1}{n},\dfrac{1}{m}) ≤ \dfrac{1}{N}$.

Now, let $\{S_n\}$ be a Cauchy sequence in $X^*$, with $s_n$ a quick Cauchy sequence of $S_n$ for all $n$. Let $s$ be the sequence with $s(n)=s_n(n)$ for all $n$.

We prove $s$ is Cauchy: Let $ε > 0$. Let $α∈ℕ$ be such that $\dfrac{2}{α} < ε$, and let $N'$ be such that $n,m ≥ N'$ implies $Δ(S_n,S_m) = \lim_{z→∞} d(s_n(z),s_m(z)) < ε - \dfrac{2}{α}$. Then if $n,m ≥ \text{max}(α,N')$ we see $$d(s(n),s(m)) = d(s_n(n),s_m(m))$$$$≤ d(s_n(n),s_n(z))+d(s_n(z),s_m(z))+d(s_m(z),s_m(m)) < \dfrac{1}{α}+ε-\dfrac{2}{α}+\dfrac{1}{α} = ε$$ for $z$ sufficiently large.

We prove $S$, the class of $s$, is the point of convergence of $\{S_n\}$: Let $ε > 0$. Let $α$ be such that $\dfrac{1}{α} < ε$, and let $N'$ be such that $n,m≥N'$ implies $d(s(n),s(m)) < ε-\dfrac{1}{α}$ since $s$ is Cauchy, then choose $N ≥ \text{max}(α,N')$. Given $n≥N$, observe $Δ(S,S_n) = \lim_{z→∞} d(s(z),s_n(z))$. For $z≥N$ we bound $$d(s(z),s_n(z))=d(s_z(z),s_n(z))$$$$≤d(s_z(z),s_n(n))+d(s_n(n),s_n(z)) < ε-\dfrac{1}{α}+\dfrac{1}{α} = ε$$ so $Δ(S,S_n)≤ε$ when $n≥N$, and we are done.$~\square$

Convergence of a Related Series (3.7)

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 7:

Let $a_n≥0$. Show that the convergence of $∑a_n$ implies the convergence of $$\sum \dfrac{\sqrt{a_n}}{n}$$
Proof:

Lemma: If $a_n≥a_{n+1}≥0$ and $∑a_n$ converges, then $∑\sqrt{a_{2^n}}$ converges. Proof: By theorem 3.25, $∑2^na_{2^n}$ converges, so by the root test $\text{lim sup }\sqrt[n]{2^na_{2^n}}=2\text{lim sup }\sqrt[n]{a_{2^n}}≤1$ so $\text{lim sup }\sqrt[n]{a_{2^n}}≤1/2$ implying $∑\sqrt{a_{2^n}}$ converges by the root test by $\text{lim sup }\sqrt[n]{\sqrt{a_{2^n}}}≤\sqrt{1/2} < 1$ (since one may show without advancing in theory that $\text{lim sup }\sqrt{x_n}≤\sqrt{\text{lim sup }x_n}$ for any positive sequence $x_n$).

Now, every nonempty subset of a convergent series' terms $\{a_n\}$ has an absolute greatest element, since by the epsilon-delta method there are only finitely many elements absolutely greater than any chosen term. Since $∑a_n=∑|a_n|$, use this to inductively construct a rearrangement $∑b_n$ such that $b_n≥b_{n+1}$ without losing convergence.

If we prove $∑\dfrac{\sqrt{b_n}}{n}$ converges, then though $∑\dfrac{\sqrt{a_n}}{n}$ is not generally a rearrangement, we may observe the effect of transpositions on partial sums to show the former bounds the latter, demonstrating the latter's convergence. So by 3.25, the former is equivalent to $∑\sqrt{b_{2^n}}$ converging, which is established by the lemma.$~\square$

Behavior of a Complex-Valued Series (3.6(d))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 6(d):

Investigate the behavior of $∑a_n$ when $a_n=\dfrac{1}{1+z^n}$ for complex values of $z$.

Proof: Assume $|z| > 1$. Then $$||\dfrac{a_{n+1}}{a_n}|-\dfrac{1}{|z|}|=||\dfrac{\dfrac{1}{1+z^{n+1}}}{\dfrac{1}{1+z^n}}|-|\dfrac{1}{z}||=||\dfrac{1+z^n}{1+z^{n+1}}|-|\dfrac{1}{z}||$$$$≤|\dfrac{1+z^n}{1+z^{n+1}}-\dfrac{1}{z}|=|\dfrac{z-1}{z+z^{n+2}}|=|\dfrac{z-1}{z}|·\dfrac{1}{|1+z^n|}$$$$≤|\dfrac{z-1}{z}|·\dfrac{1}{|z|^n-1}→0$$ as $n→∞$, so $\text{lim }|\dfrac{a_{n+1}}{a_n}|=\dfrac{1}{|z|} < 1$ and $∑a_n$ converges by the ratio test. Assume $|z| ≤ 1$. Then $|\dfrac{1}{1+z^n}|≥\dfrac{1}{1+|z|^n}≥\dfrac{1}{2}$ and $∑a_n$ diverges seeing as $a_n$ doesn't converge to $0$.$~\square$

Rearrangement of a Different Convergence (Example 3.53)

Walter Rudin Principles of Mathematical Analysis, chapter 3, example 53:

Consider the convergent series $$1-1/2+1/3-1/4+...$$ and one of its rearrangements $$1+1/3-1/2+1/5+1/7-1/4+1/9+1/11-1/6+...$$ As we have seen, the rearrangement $s_n$ if it converges does so at a point larger than the point of convergence for the original series. Prove $s_n$ converges.

Proof: For any sequence $a_n$, let $f~:~ℕ→ℕ$ be such that $a_{f(n)}$ and $a_n-a_{f(n)}$ are convergent sequences. Then we see $(\text{lim }a_n-a_{f(n)})+(\text{lim }a_{f(n)})=\text{lim }a_n$ so that $a_n$ converges. As it applies to this problem, we will show the subsequence $s_{3n}$ converges, and since $\text{lim }s_n-s_{f(n)}=0$ where $f(n)$ rounds $n$ to the next highest multiple of $3$, $s_n$ converges (technically $s_{f(n)}$ has different terms from $s_{3n}$, but $\text{lim }s_{f(n)}=\text{lim }s_{3n}$ since the former merely has finite copies of each of the latter's terms).

It suffices to compare the sum's terms $a_n$ to $\dfrac{4}{n^2}$, which forms a convergent series: $$|a_n|=a_n=\dfrac{1}{4n-3}+\dfrac{1}{4n-1}-\dfrac{1}{2n}=\dfrac{8n-3}{2n(4n-3)(4n-1)}$$$$≤\dfrac{4}{(4n-3)(4n-1)}≤\dfrac{4}{n^2}~~\square$$

Connected Base for Euclidean Space (2.29)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercise 29:

Prove that every open set in $ℝ^1$ is the union of an at most countable collection of disjoint segments.

Proof: We shall prove something stronger—namely, that every open set in $ℝ^k$ is the union of an at most countable collection of disjoint open connected spaces, which by Theorem 2.47 implies the proposition for $k=1$. This case could in fact be strengthened to path-connected spaces but is beyond the development of current theory.

Lemma 1 (Subspace Nonconnnectedness): Let $E=A∪B$ for disjoint, nonempty, separated $A,B$. Every subset of $E$ containing a point of $A$ and a point of $B$ is nonconnected. Proof: Let $F⊆E$ be such that $C=F∩A$ and $D=F∩B$ are nonempty. Now, note that for spaces $X,Y$ that$$\overline{X∩Y}=(X∩Y)∪(X∩Y)'⊆(X∩Y)∪(X'∩Y')$$$$=(X∪X')∩(Y∪Y')∩(X∪Y')∩(Y∪X')⊆\overline{X}∩\overline{Y}$$So observe $C∪D=(A∪B)∩F=F$ and $\overline{C}∩D=\overline{A∩F}∩D⊆\overline{A}∩\overline{F}∩D⊆\overline{A}∩B$ is empty, and similarly $C∩\overline{D}$ is empty.

Lemma 2 (Upper Bounds of Chains of Connected Spaces): Let $\{E_i\}$ be a collection of connected spaces totally ordered by inclusion. Then $E=∪E_i$ is connected. Proof: Assume $E=A∪B$ for disjoint, nonempty, separated $A,B$. Then some indexed summand $E_α$ must contain a point of $A$, and similarly $E_β$ for $B$. Then by hypothesis either $E_α⊆E_β$ or $E_β⊆E_α$, in either case one of them being nonconnected by Lemma 1, a contradiction.

Lemma 3 (Union of Nondisjoint Connected Spaces): Let $E$ and $F$ be connected spaces with nonempty intersection. Then $E∪F$ is connected. Proof: Assume $E∪F=A∪B$ for disjoint, nonempty, separated $A,B$. We may assume some $x∈E∩F$ is contained in $A$. As well, not all of $E$ and all of $F$ can be contained in $A$ since $B$ is nonempty, hence either $E$ or $F$ must not be connected by Lemma 1.

Now, let $E⊆ℝ^k$ be an open set. Define a relation $\text{~}$ on $E$ by $a \text{~} b$ if there exists a connected subspace of $E$ containing both $a$ and $b$. This is reflexive since $E$ is an open set admitting a neighborhood of any point, and neighborhoods are convex hence connected (cf. Exercise 2.21 and the example below Definition 2.17); clearly symmetrical; and transitive by Lemma 3.

Choose a set of representatives in $E$ inherited from this equivalence relation, and for any representative observe the collection $C$ of connected subspaces of $E$ containing the representative; by Lemma 2 and Zorn's Lemma there exists a maximal connected subspace under inclusion. This maximal subspace must in fact contain every point in the equivalence class, since we might otherwise construct a larger connected subspace by Lemma 3. Hence $E$ is a union of disjoint open connected spaces, and it remains only to prove that there are a countable number of them. Generally, we will prove that there cannot be a disjoint collection of uncountably many nonempty open sets in $ℝ^k$.

Observe any collection of disjoint nonempty open sets in Euclidean space. Take a point from each set, and choose a neighborhood of each so that they are disjoint. However, as we've seen, within each of these neighborhoods we may find a member of the countable base for $ℝ^k$ (cf. Exercises 2.22-23), implying the collection is at most countable.$~\square$

Limit Points and Housekeeping (2.6-9)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercises 6-9:

6. Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E'=\overline{E}'$. Generally, is it true that $E'=E''$?
7. Let $A_n$ be subsets of a metric space for $n∈ℕ$.
(a) If $B_n=∪_{i=0}^n A_i$ prove that $\overline{B_n}=∪_{i=0}^n \overline{A_i}$.
(b) If $B=∪A_n$ prove that $\overline{B}⊇∪\overline{A_n}$.
Show that this inclusion can be proper.
8. Is every point of every open set $E⊆ℝ^2$ a limit point of $E$? Answer the same question for closed sets in $ℝ^2$.
9. Let $E ^\circ$ denote the set of all interior points of a set $E$, called the interior of $E$.
(a) Prove that $E ^\circ$ is always open.
(b) Prove that $E$ is open iff $E ^\circ = E$.
(c) If $G ⊆ E$ and $G$ is open, show $G⊆E ^\circ$.
(d) Prove ${E^\circ}^c = \overline{E^c}$.
(e) Does $E ^\circ = \overline{E}^\circ$?
(f) Does $\overline{E}=\overline{E ^\circ}$?

Proof: Lemma 1: Let $∪A_i$ be a union of subsets of a metric space. Then $(∪A_i)'⊇∪A_i'$, and there is equality when the union is finite. Proof: The $⊇$ direction is clear since limit points of a subset are limit points of the whole set, so it remains to show $⊆$ when the union is finite. To wit, let $p$ be a limit point of $∪A_i$. For each $i$, let $r_i=\text{inf }\{r∈ℝ~|~A_i∩N_r(p)-\{p\}≠∅\}$. If $r_i > 0$ for all $i$ then since the indices are finite choose some $r$ such that $0 < r < \text{min}_i~r_i$ and we would have $N_r(p)∩(∪A_i)=\{p\}$ and $p$ is not a limit point, a contradiction. Hence $r_i=0$ for some $i$, therefore $p$ is a limit point of $A_i$.

6. We now show $E''⊆E'$ so that $E'$ is closed. We have in fact already seen this, since a limit of limit points is a limit of the original points, since they lie arbitrarily close to these limit points. Now, by the lemma we see $\overline{E}'=(E∪E')'=E'∪E''=E'$. Finally, let $E=(0,1)$. We see $E'=\{0,1\}$ and $E''=∅$ so that generally $E'$ need not equal $E''$.

7. (a) We see$$\overline{B_n}=B_n∪B_n'=(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i)'=$$$$(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i')=∪_{i=0}^n (A_i∪A_i')=∪_{i=0}^n \overline{A_i}$$with a $⊇$ containment between the appropriately modified terms three and four if the union is infinite. When $A_i=\{1/(i+1)\}$ for all $i$, we see $\overline{B}$ contains $0$, whereas $∪\overline{A_i}=∪A_i$ does not.

8. $ℝ^2$ contains no isolated points, and every point of an open set is interior, and since interior and non-isolated implies limit point, every point of an open set in $ℝ^2$ is a limit point. Clearly, $\{0\}$ is a closed set for which $0$ is not a limit point.

9. Lemma 2: $E^\circ = \overline{E^c}^c$. Proof: ($⊆$) Suppose $p∈E^\circ$. Then $p∈E$ so $p∉E^c$ and also $p∉{E^c}'$, hence $p∉\overline{E^c}$ so $p∈\overline{E^c}^c$. ($⊇$) Suppose $p∈\overline{E^c}^c$, so $p∉\overline{E^c}=E^c∪{E^c}'$ implying $p∈E$ and $p$ is not a limit point of $E^c$, i.e. $p$ is an interior point of $E$ and is contained in $E$.

(a) This follows from the lemma since $\overline{E^c}$ is closed.

(b) This will follow from (c).

(c) Assume $G⊆E$ is open and $G⊈E^\circ$. Then $G'=G∪E^\circ$ is an open set contained in $E$ with inclusions $E^\circ = \overline{E^c}^c ⊂ G' ⊆ E$. We then observe $E^c ⊆ G'^c ⊂ \overline{E^c}$ are inclusions with $G'^c$ closed containing $E$, yet $\overline{E^c}$ is supposed to be the smallest closed set containing $E^c$, a contradiction.

(d) This was shown in the lemma.

(e) Let $E=ℝ-\{0\}⊆ℝ^1$. Then $E^\circ=E$ yet $\overline{E}^\circ=ℝ^\circ=ℝ$. Hence, generally, $E ^\circ ≠ \overline{E}^\circ$.

(f) Let $E=\{0\}$. Then $\overline{E}=E$ and $\overline{E^\circ}=\overline{∅}=∅$. Hence, generally, $\overline{E}≠\overline{E ^\circ}$.$~\square$

Infinitude of Primitive Solutions for x^3+y^3=7z^3 (15.1.45)

Let $V=\mathcal{Z}(x^3+y^3+7z^3)⊂ℂ^3$. Then $\mathcal{I}(V)=(x^3+y^3+7z^3)⊂ℂ[x,y,z]$ (cf. 15.3.24).
(a) Show that $$Φ(x)=x(y^3-7z^3),~~~~~Φ(y)=y(7z^3-x^3),~~~~~Φ(z)=z(x^3-y^3)$$ defines a $ℂ$-algebra homomorphism from $k[V]$ to itself.
(b) Let $φ~:~V→V$ be the morphism corresponding to $Φ$. Observe that $(-2,1,1)∈V$ and compute $φ(-2,1,1)$.
(c) Prove that there are infinitely many points $(a,b,c)∈V$ with $a,b,c∈ℤ$ and the greatest common divisor of $a,b,c$ is $1$.

Proof: (a) It sufficies to show $Φ(x^3+y^3+7z^3)∈(x^3+y^3+7z^3)$, and indeed general polynomial division will reveal $x^3+y^3+7z^3~|~Φ(x^3+y^3+7z^3)$.

(b) We see $φ$ is the morphism defined by $φ(a,b,c)=(a(b^3-7c^3),b(7c^3-a^3),c(a^3-b^3))$, so $φ(-2,1,1)=(12,15,-9)$.

(c) We may assume $a,b,c$ for $(a,b,c)∈V$ are pairwise relatively prime when $a,b,c$ have a trivial greatest common factor; to see this, assume two elements are divisible by a prime $p$. By reducing $x^3+y^3+7z^3=0$ modulo $p^3$, we see that the other element too must be divisible by $p$, impossible if they have trivial greatest common factor.

Let $ψ(a,b,c)=\dfrac{φ(a,b,c)}{d}$ when $d$ is the greatest common factor of the three coordinates of $φ(a,b,c)$. We shall show the first coordinate of $(a,b,c)$ is smaller than or equal in absolute value than that of the first coordinate of $Ψ(a,b,c)$ as well as are the second coordinates when $c≠0$, but that they cannot both be equalities, so $Ψ^i(-2,1,1)≠Ψ^j(-2,1,1)$ for nonnegative $i≠j$.

Assume an arbitrary prime $p~|~d$ when $d$ is the greatest common divisor of the coordinates of $φ(a,b,c)=(a(b^3-7c^3),b(7c^3-a^3),c(a^3-b^3))$ when $(a,b,c)∈V$. We have $p~|~a$ or $p~|~(b^3-7c^3)$; assume the former. Then since $(a,b)=1$ we have $p~|~(7c^3-a^3)$. Reduce modulo $p$ to observe either $p=7$ or $p~|~c$, only the former being possible. But now reduce $a^3+b^3+7c^3=0$ modulo $p=7$ to still get $p~|~b$. Hence $p\not |~a$. Since $p$ was arbitrary this implies $d~|~(b^3-7c^3)$, so $a~|~a'$ when $a'$ is the first coordinate of $Ψ(a,b,c)$. This shows $|a|≤|a'|$, and by very similar reasoning $|b|≤|b'|$.

The only possible case for $|a|=|a'|$ and $|b|=|b'|$ implies $|d|=|b^3-7c^3|=|7c^3-a^3|$. But either $b^3-7c^3=7c^3-a^3$ or $b^3-7c^3=a^3-7c^3$ together with the hypothesis $a^3+b^3+7c^3=0$ implies $c=0$, a contradiction. Therefore the absolute inequality is strict and the proposition is established.$~\square$

The Fitting Ideal of Level 0 is Independent of Choice of Generators (15.1.36-37)

Let $M$ be a finitely generated module over $R$.

36. (a) Show that for all $p≥n$, the Fitting ideal of $M$ is also the ideal in $R$ generated by all the $n×n$ minors of all $p×n$ matrices.
(b) Let $A$ be a fixed $p×n$ matrix as in (a) and let $A'$ be a $p×n$ matrix obtained from $A$ by any elementary row or column operation. Show that the ideal in $R$ generated by all the $n×n$ minors of $A$ is the same as the ideal in $R$ generated by all the $n×n$ minors of $A'$.

37. Suppose $m_1,...,m_n$ and $m_1',...,m_{n'}'$ are two sets of generators for $M$. Let $\mathcal{F}$ be the Fitting ideal calculated with respect to $m_1,...,m_n$ and let $\mathcal{F}'$ be the Fitting ideal calculated with respect to $m_1,...,m_n,m_1',...,m_{n'}'$.

(a) Show that $m_s'=a_{s',1}m_1+...+a_{s',n}m_n$ for some $a_{s',i}∈R$ for all $1 ≤ s ≤ n'$, so that $(-a_{s',1},...,-a_{s',n},0,...,0,1,0,...,0)$ is a relation among $m_1,...,m_n,m_1',...,m_{n'}'$.
(b) If $A$ is an $n×n$ matrix whose rows are the coefficients of relations among $m_1,...,m_n$, show that $\text{det }A=\text{det }A'$ where $A'$ is an $(n+n')×(n+n')$ matrix whose rows are the coefficients of relations among $m_1,...,m_n,m_1',...,m_{n'}'$. Deduce $\mathcal{F}⊆\mathcal{F}'$.
(c) Prove $\mathcal{F}'⊆\mathcal{F}$ and conclude $\mathcal{F}=\mathcal{F}'$.
(d) Deduce from (c) that the Fitting ideal is independent of choice of generators for $M$.

Proof: (36)(a) $⊆$ is clear by simply fitting every $n×n$ matrix of relations into a $p×n$ relations matrix whose minor will be included in the latter ideal, and $⊇$ is also clear since every such minor is the determinant of an $n×n$ relations matrix.

(b) Let $a$ be an $n×n$ minor, the determinant of an $n×n$ matrix $N$. Then let $a'$ be the determinant of the matrix $N'$, which is identical to $N$ save for one of its rows or columns being replaced with another part of a row or column in the $p×n$ matrix. Then since determinants are $R$-bilinear on rows and columns, we have $a+r·a'$ is the $n×n$ minor under the column or row operation on $p×n$ by adding $r$ times the specified row or column to another affecting $N$. Since $N'$ is undisturbed, we can retrieve $a$ the original minor in the new ideal, and since this minor was arbitrary, this shows that the two ideals are equal.

(37)(a) Self-explanatory.

(b) Let $A'$ be the block matrix $$\begin{bmatrix} A & 0 \\ B & I \end{bmatrix}$$ where $I$ is the $n' × n'$ identity, and $B$ is the $n' × n$ matrix converting $n × 1$ vectors in terms of $m_i$ into $n' × 1$ vectors in terms of $m_i'$. Then $\text{det }A = \text{det }A'$ where $A'$ is a matrix of relations among $m_1,...,m_n,m_1',...,m_{n'}'$ as desired.

(c) Let $A$ be an $(n+n')×(n+n')$ relations matrix, and observe that its determinant ideal is contained in the $(n+n')×(n+n')$ minor-generated ideal of the $(n+2n')×(n+n')$ block matrix $A'$, where the top $(n+n')×(n+n')$ block is $A$ and the bottom $n'×(n+n')$ block is $[B~~I]$ as above. By 36(b) we may perform row operations on $A'$ to remain $A''$ a block matrix $$\begin{bmatrix} C & 0 \\ B & I \end{bmatrix}$$ where $C$ is an $(n+n')×n$ relations matrix in terms of $m_1,...,m_n$. The nontrivial minors of this matrix are those $n×n$ minors of $C$, which by 36(a) are all contained in $\mathcal{F}$. Since $\mathcal{F}'$ is the smallest ideal containing all these determinant ideals, we have $\mathcal{F}'⊆\mathcal{F}$ and $\mathcal{F}=\mathcal{F}'$.

(d) When $\mathcal{F}''$ is the Fitting ideal calculated with respect to $m_1',...,m_{n'}'$, we have $\mathcal{F}=\mathcal{F}'=\mathcal{F}''$.$~\square$

Product of Algebraic Sets and its Coordinate Ring (15.1.28)

Prove that if $V$ and $W$ are algebraic sets, then so is $V×W$ and $k[V×W]≅k[V] ⊗_k k[W]$.

Proof: If $V=\mathcal{Z}(I')$ then also $V=\mathcal{Z}(I)$ where $I=\mathcal{I}(\mathcal{Z}(I'))$, with the additional helpful property that $\mathcal{I}(V)=I$. We shall assume the same for $W=\mathcal{Z}(J)$. Lemma: Let $V⊆\mathbb{A}^n$, $W⊆\mathbb{A}^m$ be algebraic sets. Then $\mathcal{I}(V×W)=\mathcal{I}(V)+\mathcal{I}(W)$. Proof: Clearly $⊇$ holds. Now assume $f∈\mathcal{I}(V×W)$ yet $f∉\mathcal{I}(V)+\mathcal{I}(W)$. (Following proof of lemma borrowed from MSE) We can write any polynomial in $k[x_1,...,x_n,y_1,...,y_m]/(\mathcal{I}(V)+\mathcal{I}(W))$ as $$\sum_{i=1}^r \overline{f_ig_i}$$ where $f_i∈k[x_1,...,x_n]$ and $g_i∈k[y_1,...,y_m]$ for all $i$. Let $f$ as above be minimal such that one of its reduction sums has the least summand terms $r$ of the form above of all the polynomials of its hypothesis. Now, assume $f_i(a)=0$ for all $a∈V$ and all $i$; then since $\mathcal{I}(V)=I$ we have $f_i∈I$, hence $f∈\mathcal{I}(V)+(\mathcal{I}(V)+\mathcal{I}(W))=\mathcal{I}(V)+\mathcal{I}(W)$, a contradiction. Therefore $f_k(a)≠0$ for some $k$ and $a∈V$, where we may reorder $f_1=f_k$. Now by the hypothesis for $f$, for all $b∈W$ we have $f(a)(b)=(\sum f_i(a)g_i(b))+t(a)(b)=\sum f_i(a)g_i(b)=0$ (for some negligible $t∈\mathcal{I}(V)+\mathcal{I}(W)$) hence $\sum f_i(a)g_i=p∈\mathcal{I}(W)$. Since $f_1(a)≠0$ write $g_1=\dfrac{p-f_2(a)g_2-...-f_r(a)g_r}{f_1(a)}$ to write $\overline{f}$ in strictly fewer summands, a contradiction.

Proceed with the exercise. To see that $V×W$ is an algebraic set, observe $V×W=\mathcal{Z}(\mathcal{I}(V)+\mathcal{I}(W))$, with $⊆$ being evident and $⊇$ resulting from $V=\mathcal{Z}(\mathcal{I}(V))$ and $W=\mathcal{Z}(\mathcal{I}(W))$. Now, define a $k$-bilinear map $k[V]×k[W]→k[V×W]$ by $(f,g)↦fg$ (well-defined since $\mathcal{I}(V)+\mathcal{I}(W)⊆\mathcal{I}(V×W)$) to induce a surjective group homomorphism $Φ~:~k[V] ⊗_k k[W]→k[V×W]$. This is also a ring homomorphism. To prove bijection, it suffices to check that the ring homomorphism $Φ'~:~k[x_1,...x_n,y_1,...,y_m]→k[V] ⊗_k k[W]$ given by $Φ'(x_i)=x_i ⊗ 1$ and $Φ'(y_i)=1 ⊗ y_i$ factors through $\mathcal{I}(V)+\mathcal{I}(W)$, as then it would be a two-sided inverse to $Φ$ when induced on $k[V×W]$. Indeed, when $f∈\mathcal{I}(V)$ or $g∈\mathcal{I}(W)$ is a generator of $\mathcal{I}(V×W)=\mathcal{I}(V)+\mathcal{I}(W)$, we see $Φ'(f)=f⊗1=0⊗1=0$ and $Φ'(g)=1⊗g=0$.$~\square$

Problem 8 Lemma (Existence of a Complete Sequence)

Lemma: Let $W$ be a tournament directed graph on $2n+1$ nodes where every node projects an arrow to exactly $n$ other nodes. Then there exists on $W$ a sequence $p_1→p_2→...→p_{2n+1}→p_{2n+2}=p_1$ such that $p_i \neq p_j$ for all $i \neq j$ for $1 \leq i,j \leq 2n+1$.

Proof: The results in my attached exercise on Vandermonde determinants imply that there must exist some sequence $p_1→...→p_m→p_{m+1}=p_1$ involving distinct nodes, so let us observe such a sequence where $m$ is assumed to be maximal of all possible sequences of that form. Assume $m \neq 2n+1$. Denote by $P$ the set of nodes in the sequence and set $Q=W \setminus P$. For all $w \in W$ denote by $\text{img }w$ the set of nodes toward which $w$ projects an arrow.

First, a sublemma: Let $q \in Q$ such that $p_i→q→p_j$ for some $i,j$. Then it follows that we must have some $k$ for $p_k→q→p_{k+1}$ (else all the nodes could be shown to point toward $q$ iff $p_i$ does, a contradiction for $p_j$). Then $p_1→...→p_k→q→p_{k+1}→...→p_m→p_{m+1}=p_1$ is a strictly larger sequence, a contradiction by the maximality of $m$. Therefore for all $q \in Q$ we have $P \subseteq \text{img }q$ or $P \cap \text{img }q = \emptyset$.

Assume $m > n$. Then since $1 \leq |Q| \leq n$ we may choose some $q \in Q$, where necessarily $)\text{img }q) \cap P \neq \emptyset$, therefore $P \subseteq \text{img }q$, but now $n < |P| \leq |\text{img }q| = n$, a contradiction.

Therefore $m \leq n$. Then by the sublemma $p_i→q \iff p_1→q$ for $q \in Q$. Since $|P \setminus \{p_1\}| < n$ we must have $p_1 → q$ for some $q \in Q$. Now $(\text{img }p_1) \setminus P = (\text{img }P) \setminus P$ and since $|Q| \geq n+1$ we observe the nonempty $S = (\text{img }P) \setminus P$ and $T = W \setminus (S \cup P)$. Note $P$, $S$, and $T$ partition $W$. Now, $|S| \leq n$ and $(\text{img }S) \cap P = \emptyset$, hence for all $s \in S$ there exists $t \in T$ such that $s→t$. Choose such $s$ and $t$ and observe that $p_1→s→t→p_2→...→p_m→p_{m+1}=p_1$ is a strictly larger sequence once more.$\square$

The motivation behind this lemma was to justify the privilege of imagining every such tournament directed graph as a polygon of nodes, with each successive vertex pointing toward the next. However, the utility of this was never realized.

Computation of Algebraic Sets and Coordinate Rings (15.1.24)

Let $k$ be an infinite field, and let $V=\mathcal{Z}(xy-z)$. Prove that $V$ is isomorphic to $\mathbb{A}^2$ and provide an explicit isomorphism $φ$ from $V$ to $\mathbb{A}^2$ and associated $k$-algebra isomorphism $Φ$ from $k[V]$ to $k[\mathbb{A}^2]$ and their inverses. Is $V=\mathcal{Z}(xy-z^2)$ isomorphic to $\mathbb{A}^2$?

Proof: We see $V$ is the set of all points of the form $(x,y,xy)$, so there is an isomorphism $φ~:~V→W$ given by $φ(x,y,xy)=(x,y)$ with inverse $ψ~:~W→V$ given by $ψ(x,y)=(x,y,xy)$. We see $(xy-z)⊆\mathcal{I}(V)$ naturally and by reducing modulo $(xy-z)$ any $f∈\mathcal{I}(V)$ to a polynomial in $k[x,y]$ we see the remainder must be $0$ as it would annihilate all of $\mathbb{A}^2$, hence $\mathcal{I}(V)⊆(xy-z)$. Thus this induces the isomorphism $Φ~:~k[W]→k[V]$ where $k[V]=k[x,y,z]/(xy-z)$ and $k[W]=k[x,y]$ given by $Φ(f(x,y))=\overline{f(x,y)}$ with inverse $Ψ~:~k[V]→k[W]$ given by $Ψ(\overline{f(x,y,z)})=f(x,y)$.

For the second part, we shall show $k[V]≇k[W]$ where $W=\mathbb{A}^2$ so that $V≇W$ as algebraic sets. First, we have $(xy-z^2)⊆\mathcal{I}(V)$. Assume $\mathcal{I}(V)⊈(xy-z^2)$ and let $f∈\mathcal{I}(V) \setminus (xy-z^2)$ be minimal with respect to the lexicographic ordering $z > y > x$, so $f=z·g_1(x,y)+g_2(x,y)$.

Let $k^2$ be the subset of nonzero squares in $k$, necessarily infinite since each square has only a finite number of square roots, and for each nonzero $b∈k$ let $k^2/b = \{v/b~|~v∈k^2\}$. Note $V$ is the set of points of the form $(x,y,\sqrt{xy})$ whenever $xy$ is square, so for each nonzero $b∈k$ we have an infinite number of $a∈k^2/b$ such that $$f(a,b,\sqrt{ab})=f(a,b,-\sqrt{ab})=$$$$(\sqrt{ab})g_1(a,b)+g_2(a,b)=(-\sqrt{ab})g_1(a,b)+g_2(a,b)=0$$ Since $\sqrt{ab}$ is nonzero, for each such $b$ we have $g_1(x,b)∈k[x]$ has an infinite number of zeros $x=a$, so $g_1(x,b)=0$ for these $b$. Since there are an infinite number of such $b$, we have $g_1(x,y)∈k(x)[y]$ has an infinite number of roots $y=b$ in $k⊆k(x)$, hence $g_1(x,y)=0$ and by the same argument we find $g_2(x,y)=0$. Hence $f=0$ and we have shown $\mathcal{I}(V)=(xy-z^2)$.

We see $k[W]=k[x,y]$ is a UFD (cf. 9.3.8), and we shall prove $k[V]≇k[W]$ by showing $k[V]=k[x,y,z]/(xy-z^2)$ is not a UFD. Assume otherwise and observe the element $x∈k[V]$, which we shall prove is prime; we may write any element in $k[V]$ uniquely as a polynomial in $k[x,y][z]$ of degree $≤1$, so assume $x=(g_1+z·g_2)(g_3+z·g_4)$. Then $x=g_1g_3+xy·g_2g_4+z(g_1g_4+g_2g_3)$ is the unique reduced form, so $g_1g_3+xy·g_2g_4=x$ and $g_1g_4+g_2g_3=0$. Taking the first relation modulo $x$ we may assume without loss that $x~|~g_1$, and now taking the second relation modulo $x$ necessarily either $x~|~g_2$ or $x~|~g_3$; the former implies $x~|~(g_1+z·g_2)$, and the latter that the relation $g_1g_3+xy·g_2g_4$ involves adding monomial terms of degree strictly larger than $x$, hence the expression does not equal $x$. Thus $x$ must be prime. But $xy=z^2$ would imply $x~|~z$, a contradiction by observing unique representation. Hence $k[V]$ is not a UFD and not isomorphic to $k[W]$, and so neither is $V$ isomorphic to $\mathbb{A}^2$.$~\square$

Problem 10

10. How many zeros are there at the end of the number 1000!=(1)(2)(3)...(999)(1000)? How many for 1001!, 1002!, 1003!, 1004!, and 1005!?

Proof: Let $n=2^a5^bm$ with $2,5~\not |~m$. Then $n$ has $\text{min}\{a,b\}$ zeros at the end, as $10^c~|~n⇔2^c,5^c~|~n⇔c≤\text{min}\{a,b\}$. By Legendre's theorem for the power of $p$ in $n!$$$v_p(n!)=\sum_{k=1} \lfloor n/p^k \rfloor$$we see we will have to pay attention only to the power of $5$ in computing the number of zeros at the end. Use Legendre's theorem for this application on $1000$ to obtain $249$ zeros, and the same goes for $1001!,...,1004!$ as they don't have any greater powers of $5$, and $1005=3·5·67$ contributes one extra zero to $1005!$ for a total of $250$.