## Real Analysis Review Day 9: Metric Spaces (Part II)

A sequence $\{x_n\}$ in a metric space $(X, \rho)$ is called Cauchy if $\rho(x_n, x_m)\to 0$ as $n, m\to\infty$. A subset $E$ of $X$ is called complete if every Cauchy sequence in $E$ converges and its limit is in $E$. For example, $\mathbb R^n$ is complete, while $\mathbb Q^n$ is not.

Proposition. A closed subset of a complete metric space is complete, and a complete subset of an arbitrary metric space is closed.

Proof. If $X$ is complete, $E\subset X$ is closed, and $\{x_n\}$ is a Cauchy sequence in $E$, $\{x_n\}$ has a limit in $X$. So $x\in\bar{E}=E$. If $E\subset X$ is complete and $x\in\bar{E}$, then there is a sequence $\{x_n\}$ in $E$ converging to $x$. $\{x_n\}$ is Cauchy, so its limit lies in $E$; thus $E=\bar E$. $\Box$

In a metric space $(X, \rho)$ we can define the distance from a point to a set and the distance between two sets. Namely, if $x\in X$ and $E, F\subset X$,

$\rho(x, E)=\inf\{\rho(x, y): y\in E\}$,

$\rho(E, F)=\inf\{\rho(x, y): x\in E, y\in F\}=\inf\{\rho(x, F): x\in E\}$.

Note that $\rho(x, E)=0$ iff $x\in\bar E$. We also define the diameter of $E\subset X$ to be

$diam E=\sup\{\rho(x, y): x, y\in E\}$.

$E$ is called bounded if $diam E<\infty$.

If $E\subset X$ and $\{V_\alpha\}_{\alpha\in A}$ is a family of sets such that $E \subset \cup_{\alpha\in A}V_\alpha$, $\{V_\alpha\}_\alpha\in A$ is called a cover of $E$, and $E$ is said to be covered by the $V_\alpha$‘s. $E$ is called totally bounded if for every $\epsilon>0$, $E$ can be covered by finitely many balls of radius $\epsilon$. Every totally bounded set is bounded, for if $x, y\in\cup_1^n B(\epsilon, z_j)$, say $x\in B(\epsilon, z_1)$ and $y\in B(\epsilon, z_2)$, then

$\rho(x, y)\le\rho(x, z_1)+\rho(z_1, z_2)+\rho(z_2, y)\le 2\epsilon+\max\{\rho(z_j, z_k): 1\le j, k\le n \}$.

If $E$ is totally bounded, so is $\bar E$, for it is easily seen that if $E\subset \cup_1^n B(\epsilon, z_j)$, then $\bar E\subset \cup_1^n B(2\epsilon, z_j)$.

Theorem. If $E$ is a subset of the metric space $(X, \rho)$, the following are equivalent:

1) $E$ is complete and totally bounded.

2) (The Bolzano-Weierstrass Property) Every sequence in $E$ has a subsequence that converges to a point in $E$.

3) (The Heine-Borel Property) If $\{V_\alpha\}_{\alpha\in A}$ is a cover of $E$ by open sets, there is a finite set $F\subset A$ such that $\{V_\alpha\}_{\alpha\in A}$ covers $E$.

Proof. 1) $\Rightarrow$ 2): Suppose that 1) holds and $\{x_n\}$ is a sequence in $E$. $E$ can be covered by finitely many balls of radius $2^{-1}$, and at least one of them must contain $x_n$ for finitely many many $n$: say, $x_n\in B_1$ for $n\in N_1$. $E\cap B_1$ can be covered by finitely many balls of radius $2^{-2}$, and at least one of them must contain $x_n$ for infinitely many $n\in N_1$: say $x_n\in B_2$ for $n\in N_2$. Continuing inductively, we obtain a sequence $x_n\in B_j$ for $n\in N_j$. Pick $n_1\in N_1$, $n_2\in N_2$, … such that $n_1. Then $\{x_{n_j}\}$ is a Cauchy sequence, for $\rho(x_{n_j}, x_{n_k})<2^{1-j}$ if $k>j$, and since $E$ is complete, it has a limit in $E$.

2) $\Rightarrow$ 1): If $E$ is not complete, there is a Cauchy sequence $\{x_n\}$ in $E$ with no limit in $E$. No subsequence of $\{x_n\}$ can converge in $E$, for otherwise the whole sequence would converge to the same limit. On the other hand, if $E$ is not totally bounded, let $\epsilon>0$ be such that $E$ cannot be covered by finitely many balls of radius $\epsilon$. Choose $x_n\in E$ inductively as follows. Begin with any $x_1\in E$ and having chosen $x_1, ..., x_n$, pick $x_{n+1}\in E\setminus \cup_1^n B(\epsilon, x_j)$. Then $\rho(x_n, x_m)>\epsilon$ for all $n, m$, so $\{x_n\}$ has no convergent subsequence.

1) + 2) $\Rightarrow$ 3): If suffices to show that if 2) holds and $\{V_\alpha\}_{\alpha\in A}$ is a cover of $E$ by open sets, $\exists\epsilon>0$ such that every ball of radius $\epsilon$ that intersects $E$ is contained in some $V_\alpha$, for $E$ can be covered by finitely many such balls by 1). Suppose to the contrary that for each $n\in\mathbb N$, $\exists$ a ball $B_n$ of radius $2^{-n}$ such that $B_n\cap E\not=\emptyset$ and $B_n$ is contained in no $V_\alpha$. Pick $x_n\in B_n\cap E$; by passing to a subsequence we may assume that $\{x_n\}$ converges to some $x\in E$. We have $x\in V_\alpha$ for some $\alpha$, and since $V_\alpha$ is open, $\exists\epsilon>0$ such that $B(\epsilon, x)\subset V_\alpha$. But if $n$ is large enough so that $\rho(x_n, x)<\epsilon/3$ and $2^{-n}<\epsilon/3$, then $B_n\subset B(\epsilon, x)\subset V_\alpha$, contradicting the assumption on $B_n$.

3) $\Rightarrow$ 2): If $\{x_n\}$ is a sequence in $E$ with no convergent subsequence, for each $x\in E$ there is a ball $B_x$ centered at $x$ that contains $x_n$ for only finitely many $n$. Then $\{B_x\}_{x\in E}$ is a cover of $E$ by open sets with no finite sub-cover. $\Box$

A set $E$ can possesses the properties 1)-3) is called compact. Every compact set is closed and bounded, and the converse is false.

Proposition. Every closed and bounded subset of $\mathbb R^n$ is compact.

Proof. Since closed subsets of $\mathbb R^n$ are complete, it suffices to show that bounded subsets of $\mathbb R^n$ are totally bounded. Since every bounded set is contained in some cube

$Q=[-R, R]^n=\{x\in\mathbb R^n: \max(|x_1|,..., |x_n|),

it is enough to show that $Q$ is totally bounded. Given $\epsilon>0$, pick an integer $k>R\sqrt{n}/\epsilon$, and express $Q$ as the union of $k^n$ congruent subcubes by dividing the interval $[-R, R]$ into $k$ equal pieces. The side length of these subcubes is $2R/k$ and hence their diameter is $\sqrt{n}(2R/k)<2\epsilon$, so they are contained in the balls of radius $\epsilon$ about their centers. $\Box$

Two metrics $\rho_1$ and $\rho_2$ on a set $X$ are called equivalent if

$C\rho_1\le\rho_2\le C'\rho_1$ for some $C, C'>0$.

Equivalent metrics define the same open, closed and compact sets, the same convergent and Cauchy sequence, and the same continuous and uniformly continuous mappings. Consequently, most results concerning metric spaces depend not on the particular metric chosen but only on its equivalent class.

“As you simplify your life, the laws of the universe will be simpler; solitude will not be solitude, poverty will not be poverty, nor weakness, weakness.” ~ Henry David Thoreau

References:

[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 15-16.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.

## Real Analysis Review Day 8: Metric Spaces (Part I)

Definition. A metric on a set $X$ is a function: $\rho: X\times X\to[0, \infty)$ such that

1) $\rho(x, y)=0$ iff $x=y$;

2) $\rho(x, y)=\rho(y, x)$ for all $x, y\in X$;

3) $\rho(x, x)\le\rho(x, y)+\rho(y, z)$ for all $x, y, z\in X$.

A set equipped with a metric is called a metric space.

Examples. 1)  The Euclidean distance $\rho(x, y)=|x-y|$ is a metric on $\mathbb R^n$.

2) $\rho_1(x, y)=\int_0^1|f(x)-g(x)|dx$ and $\rho_\infty(f, g)=\sup_{0\le x\le 1}|f(x)-g(x)|$ are metrics on the space of continuous functions on $[0, 1]$.

3) If $\rho$ is a metric on $X$ and $A\subset X$, then $\rho|(A\times A)$ is a metric on $A$.

4) If $(X_1, \rho_1)$ and $(X_2, \rho_2)$ are metric spaces, the product metric $\rho$ on $X_1\times X_2$ is given by

$\rho((x_1, x_2),(y_1, y_2))=\max(\rho_1(x_1, y_1), \rho_2(x_2, y_2))$.

5) Some other metrics used on $X_1\times X_2$:

$\rho_1(x_1, y_1)+\rho_2(x_2, y_2)$, $[\rho_1(x_1, y_1)^2+\rho_2(x_2, y_2)^2]^{1/2}$.

Definition. Let $(X, \rho)$ be a metric space. If $x\in X$ and $r>0$, the (open) ball of radius $r$ about $x$ is $B(r, x)=\{y\in X: \rho(x, y).

A set $E\subset X$ is open if for every $x\in E$, $\exists r>0$ such that $B(r, x)\subset E$, and closed if the complement is open.

Remark. $X$ and $\emptyset$ are both open and closed. The union of any family of open sets is open, and the intersection of any family of closed sets is closed. The intersection (resp. union) of any finite family of open (resp. closed) sets is open (resp. closed). Indeed, if $U_1, U_2, ..., U_n$ are open and $x\in\cap_1^n U_j$, for each $j$, $\exists r_j>0$ such that $B(r_j, x)\subset U_j$, and then $B(r, x)\subset\cap_1^n U_j$ where $r=\min(r_1, ..., r_n)$, so $\cap_1^n U_j$ is open.

Definition. If $E\subset X$, then the interior of $E$, denoted by $E$ the union of all open sets $U\subset E$ is the largest open set contained in $E$, which is called the interior of $E$ and is denoted by $E^0$. The intersection of all closed sets $F\supset E$ is the smallest closed set containing $E$; it is called the closure of $E$ and is denoted by $\bar{E}$. $E$ is said to be dense in $X$ if $\bar{E}=X$, and nowhere dense if $\bar{E}$ has empty interior. $X$ is called separable if it has a countable dense subset. (For example, $\mathbb Q^n$ is a countable dense subset of $\mathbb R^n$.) A sequence $\{x_n\}$ in $X$ converges to $x\in X$ (symbolically: $x_n\to x$ or $\lim x_n=x$) if $\lim_{n\to\infty}\rho(x_n, x)=0$.

Proposition. If $X$ is a metric space, $E\subset X$, and $x\in X$, the following are equivalent:

1) $x\in\bar{E}$.

2) $B(r, x)\cap E\not=\emptyset$ for all $r>0$.

3) There is a sequence $\{x_n\}$ in $E$ that converges to $x$.

Proof. If $B(r, x)\cap E=\emptyset$, then $B(r, x)^c$ is a closed set containing $E$ but not $x$, so $x\not\in\bar{E}$. Conversely, if $x\not\in\bar{E}$, since $(\bar{E})^c$ is open $\exists r>0$ such that $B(r, x)\subset(\bar E)^c\subset E^c$. Thus 1) is equivalent to 2). If 2) holds, for each $n\in\mathbb N$, $\exists x_n\in B(n^{-1}, x)\cap E$, so that $x_n\to x$. On the other hand, if $B(r, x)\cap E=\emptyset$, then $\rho(y, x)\ge r$ for all $y\in E$, so no sequence of $E$ can converge to $x$. Thus 2) is equivalent to 3). $\Box$

Remark. If $(X_1, \rho_1)$ and $(X_2, \rho_2)$ are metric spaces, a map $f: X_1\to X_2$ is called continuous at $x\in X_1$ if for every $\epsilon>0$, $\exists \delta>0$ such that $\rho_2(f(y), f(x))<\epsilon$ whenever $\rho_1(x, y)<\delta$. In other words, $f^{-1}(B(\epsilon, f(x)))\supset B(\delta, x)$. The map $f$ is called continuous if it is continuous at each $x\in X_1$ and uniformly continuous if the $\delta$ in the definition of continuity can be chosen independent of $x$.

Proposition. $f: X_1\to X_2$ is continuous iff $f^{-1}(U)$ is open in $X_1$ for every open $U\subset X_2$.

Proof. If the latter condition holds, then for every $x\in X_1$ and $\epsilon>0$, the set $f^{-1}(B(\epsilon, f(x)))$ is open and contains $x$, so it contains some ball about $x$; this means that $f$ is continuous at $x$. Conversely, suppose that $f$ is continuous and $U$ is open in $X_2$. For each $y\in U$, $\exists \epsilon_y>0$ such that $B(\epsilon_y, y)\subset U$, and for each $x\in f^{-1}(\{y\})$, $\exists\delta_x$ such that $B(\delta_x, x)\subset f^{-1}(B(\epsilon_y, y))\subset f^{-1}(U)$. Thus $f^{-1}(U)=\cup_{x\in f^{-1}}B(\delta_x, x)$ is open. $\Box$

“I understand now that the vulnerability I’ve always felt is the greatest strength a person can have. You can’t experience life without feeling life. What I’ve learned is that being vulnerable to somebody you love is not a weakness, it is strength.” ~ Elisabeth Shue

References:

[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 13-14.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.