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, $\overline{E}≠\overline{E ^\circ}$.$~\square$

Complex Computations and Inequalities (1.13-15)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercises 13-15:

13. For complex $x,y$ show$$||x|-|y||≤|x-y|$$ 14. If $z$ is complex and $|z|=1$, compute$$|1+z|^2+|1-z|^2$$ 15. Under what conditions does equality hold in the Schwarz inequality? I.e.,$$|\sum a_j\overline{b_j}|^2 = \sum |a_j|^2 \sum |b_j|^2$$for complex $a_j,b_j$.

Proof: (13) Assume $|x|≥|y|$. Then we must show $|x|-|y| > |x-y|$ is impossible. Multiply both sides by (positive) $|x|+|y|$ to obtain $|x|^2-|y|^2 > |x-y|(|x|+|y|) ≥ |x-y||x+y| = |x^2-y^2|$ so $|x^2| > |x^2-y^2|+|y^2| ≥ |x^2|$, a contradiction. The case is parallel when $|y|≥|x|$.

(14) Then $z=a+bi$ such that $a^2+b^2=1$. We see $|1-z|=|z-1|$ and now $|z+1|^2+|z-1|^2=(a+1)^2+b^2+(a-1)^2+b^2=2(a^2+b^2)+2=4$.

(15) Let $v_a,v_b∈\mathbb{C}^k$ be the the vectors of the $a_j$ and $b_j$. Let $A = \sum |a_j|^2$,$B = \sum |b_j|^2$, and $C = \sum a_j\overline{b_j}$. We claim $|C|^2=AB$ if and only if $v_b$ is associate to $v_a$ (i.e. $v_b=cv_a$ for some $c∈\mathbb{C}$) or at least one of $v_a$ or $v_b$ is zero. ($⇒$) Assume $|C|^2=AB$ and $v_a,v_b \neq 0$ so $A,B > 0$. As we have seen, $∑|Ba_j-Cb_j|^2 = B(AB-|C|^2)$. When $|C|^2=AB$ we then thus have $a_j=\dfrac{C}{B}b_j$ for all $j$, and thus $v_b = \dfrac{C}{B}v_a$. ($⇐$) When either one of $v_a$ or $v_b$ is zero the equality clearly holds, so assume $v_a,v_b \neq 0$ and $v_b = cv_a$ for some complex $c$. We can thus manipulate$$B=\sum |b_j|^2 = \sum |ca_j|^2 = |c|^2\sum |a_j|^2 = |c|^2A$$and$$\overline{c}A=\overline{c}\sum a_j\overline{a_j} = \sum a_j\overline{b_j} = C$$so for all $j$ we have$$Ba_j = |c|^2Aa_j = \dfrac{|c|^2a_j\overline{a_j}A}{\overline{a_j}} = \dfrac{|ca_j|^2}{\overline{a_j}}A = b_j \dfrac{\overline{b_j}}{\overline{a_j}} A = b_j \overline{c} A = Cb_j$$so that $∑ |Ba_j - Cb_j|^2 = B(|C|^2 - AB) = 0$. Since $v_b \neq 0$ and thus $B > 0$, we have $|C|^2 = AB$.$~\square$

Real Logs (1.7)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercise 7:

Fix $b > 1, y > 0$ and prove that there is a unique real $x$ such that $b^x = y$ by completing the following outline. Say $x = \text{log}_b y$.
(a) For natural $n$, $b^n - 1 ≥ n(b-1)$.
(b) Hence $b - 1 ≥ n(b^{1/n}-1)$.
(c) If $t > 1$ and $n > (b-1)/(t-1)$ then $b^{1/n} < t$.
(d) If $w$ is such that $b^w < y$ then $b^{w+1/n} < y$ for sufficiently large $n$.
(e) If $b^w > y$ then $b^{w-1/n} > y$ for sufficiently large $n$.
(f) Let $A = \{w~|~w < y\}$ and show $b^{\text{sup }A} = y$.
(g) Show $x$ is unique.

Proof: (a) We show $b^n ≥ n(b-1)+1$. This is clear when $n=1$, so by induction$$b^n = bb^{n-1} ≥ b((n-1)(b-1)+1)) = (n-1)b(b-1)+b$$and we must show $(n-1)b(b-1)+b ≥ n(b-1)+1$. Collecting terms on the left and manipulating we must thus show $(n-1)(b-1)^2 ≥ 0$, which is clear as all these terms are nonnegative.
(b) This is clear from the previous, as $b^{1/n} > 1$ by noting $c^n ≤ 1$ for $c ≤ 1$ by induction.
(c) Assume $b^{1/n} ≥ t$. Then $b - 1 < n(t-1) ≤ n(b^{1/n}-1)$, a contradiction by (b).
(d) We have $1 < \dfrac{y}{b^{w}}$. Observe $b^{1/n} ≤ \dfrac{b-1}{n}+1$ by (b) so we may choose $n$ such that $\dfrac{b-1}{n}+1 < \dfrac{y}{b^{w}}$ by conditioning $\dfrac{b-1}{n} < \dfrac{y}{b^w}-1 > 0$ and now $n > \dfrac{b-1}{\dfrac{y}{b^w}-1}$ so that $b^{w+1/n} < y$.
(e) As before, since we showed $b^{1/n}$ tends toward 1 we can choose $n$ such that $b^{1/n} < \dfrac{b^w}{y}$ to fulfill.
(f) By (a) we have $b^w$ is divergent so $A$ has $x = \text{sup }A$. Suppose $b^x < y$; then by (d) we have some $b^{x+1/n} < y$ so $x$ is not an upper bound. Suppose $b^x > y$; then by (e) choose $n$ for $b^{x-1/n} > y$ and $x$ is not a minimal upper bound. Therefore $b^x = y$.
(g) If $x' \neq x$ then $b^x - b^{x'} = b^{x'}(b^{x-x'}-1) = 0$ is a contradiction.$~\square$