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• Introduction
• Extended Real Numbers
• Metric Spaces

Introduction

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Extended Real Numbers

[definition] [deftitle]Definition %counter%[/deftitle]

We define $\conj{\RR} \coloneqq \RR \cup \{\pm\infty\}$ to be the extended real number system with the additional property that $-\infty < x < \infty$ for every $x \in \RR$. We also define:

• $x \pm \infty = \pm \infty$ for every $x \in \RR$.
• $x \cdot (\pm \infty) = \sgn(x)\cdot\pm\infty$ for every $x \in \RR$.
• $\infty + \infty = \infty$.
• $-\infty - \infty = -\infty$.
• $0 \cdot (\pm \infty) = 0$. [/definition]

We use the extended real numbers freely in interval notation. We will also allow endpoints to be $\pm\infty$.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Every subset $A$ of $\conj{\RR}$ has a supremum and infimum, denoted $\sup A$ and $\inf A$. That is, $\conj{\RR}$ is complete. [/theorem]

[proof] If $A$ is bounded above/below, we invoke the completeness of $\RR$. Otherwise, $\sup A = \infty$ and/or $\inf A = -\infty$ as necessary. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Every sequence $\{x_n\}$ in $\conj{\RR}$ has a limit superior and limit inferior written

[/theorem]

[proof] Note that $\sup_{n\ge k} x_n \ge \sup_{n \ge k+1} x_n$ for each $k$. So we can consider the set

Since $\conj{\RR}$ is complete, this set has infimum $x$ which is equal to $\limsup x_n$. A similar argument holds for the limit inferior. [/proof]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The sequence $\{x_n\}$ converges in $\RR$ if and only if

Furthermore, $\lim x_n = L$. [/theorem]

[proof] ($\implies$) Let $\eps > 0$. Then there is some $N$ for which

So then

This implies that

Thus, the claim follows.

($\impliedby$) Noting that

for every $n$, it follows by the squeeze theorem that

so $\lim x_n = L$. [/proof]

Metric Spaces

[definition] [deftitle]Definition %counter%[/deftitle]

A metric space $(X, d)$ is a (nonempty) set $X$ and a function $d:X \times X \to [0, \infty)$ such that:

• $d(x, y) = 0$ if and only if $x = y$.
• $d(x, y) = d(y, x)$ for every $x, y\in X$.
• $d(x, y) \le d(x, z) + d(z, y)$ for every $x, y, z\in X$. [/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $(X, d)$ be a metric space, $x\in X$, and $r > 0$. The open ball of radius $r$ centered at $x$ is

[/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $X$ be a metric space. We say that $U \subseteq X$ is open provided that for every $x \in X$, there is some open call $B(r, x)\subseteq U$. We say that a set is closed if its complement is open. [/definition]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Open balls are open. [/theorem]

[proof] Let $y \in B(r, x)$. Then triangle inequality implies that $B(r - d(x, y), y) \subseteq B(r, x)$. [/proof]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

$X$ and $\emptyset$ are clopen. [/theorem]

[proof] Obvious. [/proof]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

The union of any family of open subsets is open and the intersection of any finite family of open subsets is open. [/theorem]

[proof] Obvious. [/proof]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $X$ be a metric space and $E\subseteq X$. Then the interior $E^o$ of $E$ is the largest open subset of $X$ contained in $E$. Equivalently, the interior of $E$ is

[/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $X$ be a metric space and $E\subseteq X$. Then the closure $\overline{E}$ of $E$ is the smallest closed subset of $X$ containing $E$. Equivalently,

[/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $X$ be a metric space and $E\subseteq X$. We say that $E$ is dense in $X$ if $\overline{E} = E$ and nowhere dense if $E^o = \emptyset$. We say that $X$ is separable if has a countable dense subset. [/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $(X, d)$ be a metric space. A sequence $\{x_n\}$ converges to $x\in X$ provided that for every $\eps > 0$, there is some $N$ for which $d(x_n, x) < \eps$ whenever $n \ge N$. [/definition]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

let $X$ be a metric space, $E\subseteq X$, and $x \in X$. The following are equivalent:

1. $x\in \overline{E}$.
2. $B(r, x) \cap E \neq\emptyset$ for all $r > 0$.
3. There is a sequence $\{x_n\}$ in $E$ that converges to $x$. [/theorem]

[proof] (1) $\implies$ (2). Let $x \in \overline{E}$ and assume for contradiction that there is some $r > 0$ for which $B(r, x) \cap E$ is empty. Then $B(r, x)^c$ is a closed subset of $X$ containing $E$ but not $x$. Since $\overline{E}$ is the intersection of all closed subsets of $X$ containins $x$, it follows that $x \notin \overline{E}$ which is a contradiction.

(2) $\implies$ (3). Let $r_n = \frac{1}{n}$ for $n \in \NN$. By assumption, for each $n$, there exists some $x_n \in B_(r_n, x)$ and so this gives us a sequence $\{x_n\}$ which we claim converges to $x$. Let $\eps > 0$ be given. Then if $n > 1/\eps$, it follows that $\eps > 1/n$ and we know that $d(x_n, x) < \eps$ as desired. Thus, $x_n \to x$.

(3) $\implies$ (1). By definition, $E\subseteq \overline{E}$ and so $\{x_n\}$ is a sequence in the closed set $\overline{E}$. As $x_n \to x$, it follows that $x \in \overline{E}$ and we are done. [/proof]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. We say that $f:X \to Y$ is continuous at $a \in X$ provided that for every $\eps > 0$, there exists some $\delta > 0$ such that

We say that $f$ is continuous if it continuous at all points in its domain. In addition, if $\delta$ can always be picked indepedent of $x$, we say that $f$ is uniformly continuous. [/definition]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. Then $f:X \to Y$ is continuous if and only if the preimage of every open subset of $Y$ is an open subset of $X$. [/theorem]

[proof] ($\implies$) An equivalent way of formulating continuity of $f$ at $a$ is to say that for every $\eps > 0$, there exists some $\delta > 0$ such that $B(\delta, a) \subseteq f^{-1}(B(\eps, a))$. Now suppose that $U\subseteq Y$ is open and consider the preimage $V \coloneqq f^{-1}(U)$ under $U$. For any given $a \in V$, we know that the openness of $U$ implies that there is some $B(\eps_a, f(a)) \subseteq U$. As $f$ is continuous, there exists some $\delta > 0$ such that $B(\delta_a, a) \subseteq f^{-1}(B(\eps_a, f(a)))$. Since $f(B(\delta_a, a)) \subseteq B(\eps_a, f(a)) \subseteq U$, it follows that $B(\delta_a, a) \subseteq V$. So $V$ is necessarily open in $X$.

($\impliedby$) Let $a \in X$ and pick out some $\eps > 0$. Then we know that $V \coloneqq f^{-1}(B(\eps, f(a)))$ is open. The openness of $V$ implies that there is some $B(\delta, a) \subseteq V$ and so we see that $B(\delta, a) \subseteq f^{-1}(B(\eps, f(a)))$ which is precisely the statement that $f$ is continuous. [/proof]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $(X, d)$ be a metric space. We say that the sequence $\{x_n\}$ in $X$ is Cauchy provided that for every $\eps > 0$, there exists $N > 0$ so that $d(x_n, x_m) < \eps$ whenever $n, m \ge N$. If every Cauchy sequence in $E\subseteq X$ converges to a limit in $E$, then we say that $E$ is complete. [/definition]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

A closed subset of a complete metric space is complete, and a complete subset of an arbitrary metric space is closed. [/theorem]

[proof] For the first claim, suppose $E \subseteq X$ is closed and $X$ is complete. If $\{x_n\}$ is a Cauchy sequence in $E$, then the completeness of $X$ implies that $x_n \to x\in X$ and the closedness of $E$ implies that $x \in E$.

For the second claim, suppose $E\subseteq X$ is complete. Then if $\{x_n\}$ is a sequence in $E$ that converges to $x\in X$, we know that the sequence is Cauchy as for any $\eps > 0$, there exists some $N > 0$ such that

Thus,

The completeness of $E$ then implies that $x_n \to x’ \in E$ and it is easy to see that $x’ = x$ as there is some $N’$

and so $d(x, x’) = 0$. Thus, $x \in E$ and we are done. [/proof]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $(X, d)$ be a metric space. We define the distance from a point $x$ to a set $E\subseteq X$ as

Similarly, if $F \subseteq X$, then the distance from $E$ to $F$ is

We also define the diameter of $E$ to be

We say that $E$ is bounded given that $\diam E < \infty$. We say that $E$ is totally bounded if it can be covered by finitely many balls of radius $\eps$ FOR EVERY $\eps > 0$. [/definition]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Every totally bounded set is bounded. [/theorem]

[proof] Let $E \subseteq X$ be totally bounded. Then there is some covering of $E$ by finitely many open balls $\set{B(\eps, x_i)}_{i=1}^n$ where each $x_i \in E$. Then

\(d(x, y) \le 2\eps + \max\set{d(x_i, x_j) \mid i, j \in [n]}.\) [/proof]

[purplebox] [purpletitle]Remark %counter%[/purpletitle]

Bounded does NOT imply totally bounded. Consider $\RR$ with the discrete metric $d$. Obviously $\RR$ is bounded since the diameter of $\RR$ is just $1$. It is NOT totally bounded, however, because if we pick $\eps = 1/2$, then $B(x, \eps) = \set{x}$. [/purplebox]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

If $E$ is totally bounded, then $\overline{E}$ is as well. [/theorem]

[proof] Let $\eps > 0$ and suppose that $E \subseteq \bigcup_{i=1}^n B(\eps, x_i)$. For any $x \in \overline{E}$, we know that $B(\eps, x) \cap E \neq \emptyset$. Thus, there is some $B(\eps, x_i)$ such that $B(\eps,x)\cap B(\eps, x_i)$. Let $z$ be an element of this intersection. Then

Thus, $x \in B(2\eps, x_i)$ so it follows that $E \subseteq \bigcup_{i=1}^n B(2\eps, x_i)$. [/proof]

[theorem] [thmtitle]Theorem %counter% (Heine-Borel and Bolzano-Weierstrass)[/thmtitle]

Let $E$ be a subset of the metric space $X$. Then the following are equivalent:

1. $E$ is complete and totally bounded.
2. Every sequence in $E$ has a convergent subsequence.
3. $E$ is compact. [/theorem]

[proof] (1) $\implies$ (2). Let $\{x_n\}$ be a sequence in $E$. Granted that this sequence has a Cauchy subsequence, the completness of $E$ implies that said subsequence would converge in $E$. Thus, we show that such a subsequence exists.

Define $\eps_n = 1/n$. The total boundedness of $E$ implies that there is some $\eps_1$-ball $B_1$ that contains infinitely many terms of $\{x_n\}$. Applying the same argument to $E \cap B_1$, there is some $\eps_2$-ball $B_2$ that contains infinitely many terms of $\{x_n\}$. We continue this process inductively. Now we pick $x_{n_k} \in B_k$ for each $k \in \NN$. We claim that the newly obtained sequence $\{x_{n_k}\}$ is Cauchy. If $\eps > 0$ is given, then we can consider $k, j > N$ where $N > \frac{1}{\eps}$ (which implies $\eps > \eps_N$). Since $x_{n_k}, x_{n_j} \in B_N$, it follows that

Thus, the subsequence is Cauchy and we are done.

(2) $\implies$ (3). Skipped.

(3) $\implies$ (1). Total boundedness is immediate by the compactness of $E$. For completeness, we pick out some Cauchy sequence $\{x_n\}$ in $E$ and use the totally boundedness to show $\{x_n\}$ converges. For the sake of contradiction, assume that for all $x \in E$ there exists $\eps > 0$ such for every $N > 0$,

In particular, we may choose $x = x_m$ where $m \ge N$. Then it follows that $d(x_n, x_m) \ge \eps$ for every $n, m \ge N$, contradiction. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Every closed and bounded subset of $\RR^n$ is compact. [/theorem]

[proof] Let $E$ be a closed and bounded subset of $\RR^n$. We claim that every sequence in $E$ has a convergent subsequence. Recall that sequences in $\RR^n$ converge if and only if each component sequence converges, so it suffices to show that each component sequence has a convergent subsequence.

(Sketch of argument) Let $\{x_n\}$ be a sequence in a closed and bounded subset of $\RR$. If $x_n$ is eventually monotonically increasing, then we know that $x_n$ converges to the surpremum of the sequence. Similarly, if $x_n$ is eventually monotonically decreasing, then $x_n$ converges to the infimum of the sequence. If $x_n$ is not eventually monotonic, then it has infinitely many “valleys” or infinitely many “peaks”. Picking out these valleys or infinitely many “peaks” gives a convergent subsequence as the sequence is contained in a bounded subset. The closedness implies that the limit is actually in the set. [/proof]