Solution 1 to Problems Collection Proofs Problem 2

Please do not let the title confuse you. It is extremely out of date.

The problem is extremely difficult, and for that reason the solution is put on a separate page for conveniency.

For your reference, the problem is also available.

Problem

Show that the series

$\sum_{n=2}^\infty \frac{1}{n^m (\log{n})^p}$

where p and m are real numbers converge if $m>1$ or $m=1$ but $p>1$ and diverge otherwise.

Solution

Before we even get started, let's state a few definitions first.

Definition 1: A set $X$ , whose elements we shall call points, is said to be a metric space if with any two points $p$ and $q$ of $X$ there is associated a real number $d(p,q)$ , called the distance from $p$ to $q$ , such that

$(a) d(p,q)>0$ if $p \ne q; d(p,p)=0;$

$(b) d(p,q)=d(q,p);$

$(c) d(p,q)\le d(p,r)+d(r,q)$ , for any $r \in X$ .

Any function with these properties is called a distance function or a metric

Definition 2: Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$ .

(a) A neighborhood of a point $p$ is a set $N_r (p)$ consisting of all points such that $d(p,q)<r$ . The number $r$ is said to be the radius of $N_r (p)$ .

(b) A point $p$ is a limit point of a set $E$ if every neighborhood of $p$ contains a point $q \ne p$ such that $q \in E$ .

(c) If $p \in E$ and $p$ is not a limit point of $E$ , then $p$ is called a isolated point of $E$ .

(d) $E$ is closed if every limit point of $E$ is a point of $E$ .

(e) A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \subseteq E$

(f) $E$ is open if every point of $E$ is a interior point of $E$ .

(g) The complement of $E$ (denoted by $E^c$ ) is the set of points $p$ such that $p \not\in E$ but $p \in X$ .

(h) $E$ is bounded if there is a ream number $M$ and a point $q \in X$ such that $d(p,q)<M$ for all $p \in E$ .

Definition 3: Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$ .

(a) By a open cover of a set $E$ in $X$ we mean a collection ${G_{\alpha}}$ of open subsets of $X$ such that $E \subseteq \cup _{\alpha} G_{\alpha}$ .

(b) A subset $K$ of $X$ is said to be compact if every open cover of $K$ contain a finite subcover.

Definition 4: Let $X$ be a metric space and $E \subseteq X$ . Let $E'$ denote the set of limit points of $E$ . The closure of $E$ is said to be $\bar{E} =E \cup E'$ .

Lemma 1: Closed subsets of compact sets are compact.

Proof: Suppose $F \subseteq K \subseteq X$ , $F$ is closed and $K$ is compact. Let $\{ V_{\alpha} \}$ be a open cover of $F$ . We wish to show that $\{ V_{\alpha} \}$ have no finite subcover of $F$ . Let $F^c$ be adjoined to $\{ V_{\alpha} \}$ . Then we obtain an open cover $\Omega$ of $K$ . (Note that complements of closed sets are open since if $E$ is closed, all the limit points of $E$ is a point of $E$ so if $p \in E'$ , $p\in E$ . Every neighborhood of $p$ contain a point of $E$ that is not $p$ itself so no neighborhood of $p$ is completely in $E^c$ so $p$ is not a interior point of $E^c$ , so $E^c$ is open. Thus $F^c$ is open.) Since $K$ is compact, there is a finite subcollection $\phi$ of $\Omega$ that cover $K$ , and hence $F$ . If $F^c \subseteq \phi$ , we can exclude it from $\phi$ and can thus still have a open cover of $F$ . The obtained open cover is thus a finite subcollection of $\{ V_{\alpha} \}$ that cover $F$ , so $F$ is compact. $\square$

Definition 5: A subset $E$ of $R^k$ is said to be a k-cell if $E$ is the set of $x$ such that $E= \{ x=(x_1,\dots , x_k)| a_1 \le x_1 \le b_1, \dots , a_k \le x_k \le b_k \}$ , where the $a_i$ 's and $b_i$ 's are real numbers.

Lemma 2: Every k-cell is compact.

Proof: Let $I$ be a k-cell, consisting of the points $x$ such that $x=(x_1,\dots , x_k), a_1 \le x_1 \le b_1, \dots , a_k \le x_k \le b_k$ . Put

$\delta = \{ \sum_{i=1} ^{k} (b_i-a_i)^2 \} ^{\frac{1}{2}}$

Thus $x \in I,y \in I$ would imply $|x-y| \le \delta$ .

Suppose, for the sake of contradiction, that there exist an open cover $\{ G_{\alpha} \}$ of $I$ which contain no finite subcover. Put $c_i=\frac{a_i+b_i}{2}$ . The intervals $[a_j,c_j]$ and $[c_j,b_j]$ then determine $2^k$ k-cells $Q_i$ whose union is $I$ . Thus at least one of these sets, call it $I_1$ , cannot be covered by any finite subcollection of $\{ G_{\alpha} \}$ (otherwise $I$ could be so covered). We divide $I_1$ like we did $I$ , and obtain $I_2$ as we did $I_1$ . As the process continue we thus obtain a sequence of $I_n$ 's such that

(a) $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ ;

(b) $I_n$ is not covered by any subcollection of $\{ G_{\alpha} \}$ ;

(c) If $x \in I_n, y \in I_n$ , then $|x-y| \le 2^{-n} \delta$ .

We claim that there is a point $x*$ in every $I_n$ . In fact, we can even prove the stronger statement

"Let $k$ be a positive interger. If $\{ I_n \}$ is a sequence of k-cells such that $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ , then $\cap _{1} ^{\infty} I_n \ne {\o}$ ."

To prove that, it suffices to prove it with $k=1$ . The rest follows easily from induction. The $k=1$ case can be rephrased as follows:

"If $\{ I_n \}$ is a sequence of intervals such that $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ , then $\cap _{1} ^{\infty} I_n \ne {\o}$

To prove this, let $I_n=[a_n,b_n]$ . Then let $E$ denote the set of all the $a_i$ 's. Obviously $E$ is non-empty and is bounded above (by $b_1$ , for one). Then the least upper bound of $E$ exist, call it $x$ . If $m$ and $n$ are any positive integers, then

$a_n \le a_{m+n} \le b_{m+n} \le b_m$

so $x \le b_m$ for each $m$ . Since $a_m \le x$ , obviously $x \in \cap _{1} ^{\infty} I_n$ .

The result follow.

Thus there is a point $x*$ in every $I_n$ . Now, since $\{ G_{\alpha} \}$ covers $I$ it follows that $x* \in G_{\alpha}$ for some $\alpha$ . Since $G_{\alpha}$ is open there is a $r>0$ such that $|y-x*|<r$ implies $y \in G_{\alpha}$ . If n is so large that $2^{-n} \delta <r$ then (c) implies that $I_n \subseteq G_{\alpha}$ , which contradicts (b). A contradiction is obtained and the result must follow. $\square$

Corollary 1 (Heine-Borel Theorem): Suppose $E \subseteq R^k$ . Then the following two statements are equivalent:

(a) $E$ is closed and bounded

(b) $E$ is compact

Proof: It suffices to show that $(a) \Longleftrightarrow (b)$ . We shall only prove that $(a) \implies (b)$ since we will only use that part. However, a proof of the converse is given in a remark after the solution.

Since $E$ is closed and bounded, $E$ is a closed subset of a k-cell, a compact set. Thus $E$ is compact by Lemma 1. $\square$

Thus, in $R^k$ , every single bounded set have a compact closure.

Definition 6: Let $E$ be a subset of a metric space $X$ , and let $S$ be the set such that

$S= \{ d(p,q):|p,q \in E \}$ .

Then the smallest upper bound of $S$ is called the diameter of $E$ and is often written as $\text{diam} (E)$ .

Lemma 3: (a) Let $\bar{E}$ be the closure of $E$ in a metric space $X$ . Then

$\text{diam} (\bar{E})=\text{diam} (E)$

(b) If $K_n$ is a sequence of compact sets in $X$ such that $K_n \supseteq K_{n+1}, (n=1,2,3, \dots)$ and if

$\lim_{n \to \infty} \text{diam} (K_n)=0$

, then $\cap _{n=1} ^{\infty} K_n$ consist of exactly one point.

Proof: (a) Since $E \subseteq \bar{E}$ , it is obvious that

$\text{diam} (\bar{E})\supseteq \text{diam} (E)$

Thus it suffice to show that

$\text{diam} (\bar{E})\subseteq \text{diam} (E)$

To do so, choose $p,q \in \bar{E}$ . Then for each $\epsilon >0$ there exist points $p'$ and $q'$ such that $d(p,p')< \epsilon$ and that $d(q,q')<\epsilon$ . Hence,

$d(p,q) \le d(p,p')+d(p',q')+d(q',q) < 2\epsilon +d(p',q') \le 2\epsilon+ \text{diam} (E)$

It follows that

$\text{diam} (\bar{E}) \le 2 \epsilon + \text{diam} (E)$

. Since $\epsilon$ is ambitarily chosen, (a) is proved. $\square$

(b) Put $K=\cap _{n=1} ^{\infty} K_n$ . It suffice to show that $K$ has at most 1 point and that it exist. The former is easy. If $K$ have 1 or more points, $\text{diam} (K) >0$ which contradict the fact that $\lim_{n \to \infty} \text{diam} (K_n)=0$ .

It remains to show that $K$ is non-empty.

Sublemma 3.1: If $\{ K_{\alpha} \}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{ K_{\alpha} \}$ is non-empty, then $\cap K_{\alpha}$ is non-empty.

Proof: Fix a member $K_1$ of $\{ K_{\alpha} \}$ and let $G_{\alpha}= {K_{\alpha}}^{c}$ for each $\alpha$ . Suppose, for the sake of contradiction that no point of $K_1$ belong to every single member of $\{ K_{\alpha} \}$ . Note that every $G_{\alpha}$ is open by the parenthetical remark in the proof of lemma 1. Thus the sets $G_{\alpha}$ form a open cover of $K_1$ ; however, since $K_1$ is compact there exist finitely many indices $\alpha_{1}, \alpha_{2}, \alpha_{3}, \dots, \alpha_{n}$ such that

$K_1 \subseteq G_{\alpha_{1}} \cup G_{\alpha_{2}} \cup \dots \cup G_{\alpha_{n}}$

But this would mean

$K_1 \cap K_{\alpha_{1}} \cap K_{\alpha_{2}} \cap \dots \cap K_{\alpha_{n}}$

, contradictory to the hypothesis.

A contradiction is obtained so we are done. $\square$ .

The result follow from the sublemma. $\blacksquare$

Now, we show state another important lemma that will use the above lemma (lemma 3). Part (a) and (c) of the following lemma is called the Cauchy Criterion.

Before starting the proof, however, we have one more definition to make.

Definition 7: A sequence $\{ p_n \}$ of a metric space $X$ is said to be a Cauchy Sequence if for every $\epsilon >0$ there exist a integer $N$ such that $d(p_n,p_m)< \epsilon$ if $n,m \ge N$ .