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.