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.$~\square$