Lecture 3. Metric Spaces

Many important phenomena do not take place in a flat space. Just think of taking a hike to your favorite mountain or of the relativistic universe Einstein gave us. While, eventually, we would like to be able to perform analysis on a nonlinear object, the first step is to understand simpler matters as, say, convergence in such nonlinear spaces. In $\mathbb{R}$ we measure the distance between two points $x$ and $y$ as the "space" in between them, that is, $d(x,y)=|y-x|$. This obviously requires access to addition. Think now of two points $A$ and $B$ on a donut.

How distant are they?

While not the only one, a "natural way" to go about defining a distance would be to take the length of the shortest path from $A$ to $B$. Something like what is depicited in the figure. Keeping this example in mind, tet us start with a set $X$ and see how a distance function $d$ should look like.

Where should the function be defined? What values should it take on?

Well, since it needs to deliver the distance between two points $x,y\in X$ and since we think of lengths as positive, we need to have $$ d:X\times X\to [0,\infty). $$ We also would like that there be no distance from a point to itself $$ d(x,x)=0,\: x\in X $$ since we can always stay put as a way to go from $x$ to $x$. We would, however, like to have that $$ d(x,y)>0\text{ if }x\neq y, $$ since we expect it to be some way to go between two any distinct points. Finally, our intuition above, also tells us to require that $$ d(x,z)\leq d(x,y)+d(y,z)\text{ for }x,y,z\in X. $$ This follows from the fact that we can obtain a path from $x$ to $z$ by concatenating the shortest paths from $x$ to $y$ and from $y$ to $z$. The latter might, however, not be the shortest, hence the inequality. These properties are precisely what will yield the definition of a metric. As an exercise, show by concrete example using the minimal path intuition that the triangle inequality above can hold with a strict inequality.

Metric Spaces

Definitions (Distance Functions and Metric Spaces)

(i) A distance function $d$ on a non-empty set $X$ is a function $d: X\times X\to [0, \infty)$ satisfying the following three properties
$\quad$ 1. (symmetry) $d(x, y)=d(y,x)$ for all $x, y\in X$.
$\quad$ 2. (positivity) $d(x,y)\geq 0$ for any $x, y\in X$ and $d(x,y)=0\iff x=y$.
$\quad$ 3. (triangle inequality) $d(x, z)\le d(x, y)+d(y,z)$ for all $x, y, z\in X$.
(ii) A metric space $(X, d)$ consists of a set $X$ and a distance function $d$ on $X$.

Question

Example

Example

It is easy to verify that, for any $x, y\in \mathbb{R}^n$, one has $d_2(x,y)=d_2(y,x)$ and that $d_2(x,y)=0$ if and only if $x=y$. As for the triangle inequality, it suffices to prove that $$ \|x+y\|_2\le \|x\|_2+\|y\|_2\iff \|x\|_2^2+2x\cdot y +\|y\|_2^2\le (\|x\|_2+\|y\|_2)^2 $$ since $d_2(x,y)=\|x-y\|_2$. This is the same as Schwarz's inequality $$ x\cdot y\le \|x\|_2\|y\|_2. $$

Question

Example

For any two points $x, y\in \mathbb{R}^n$, it is easy to see that $d_p(x,y)=d_p(y,x)$, that $d_p(x, y)\ge 0$ and that $d_p(x, y)=0$ if and only if $x=y$. We prove the triangle inequality for the case $p=1$. Since $d_p(x,y)=\|x-y\|_p$. It suffices to prove $$ \|x+y\|_p\le \|x\|_p+\|y\|_p,\quad x, y\in \mathbb{R}^n. $$ When $p=1$, it is easy to see that $$ \|x+y\|_1\le (|x_1|+\cdots+|x_n|+|y_1|+\cdots+|y_n|=\|x\|_1+\|y\|_1. $$

Example

For any two points $x, y\in \mathbb{R}^n$, it is easy to see that $d_\infty (x, y)=d_\infty(y,x)$, $d_\infty(x, y)\ge 0$ and that $d_\infty(x, y)=0$ if and only if $x=y$. Next we show that $d_\infty$ satisfies the triangle inequality, i.e. that $$ d_\infty(x,z)\le d_\infty(x,y)+d_\infty(y,z). $$ Without loss of generality, we may assume that $\|z-x\|_\infty=|z_1-x_1|$, so that \begin{equation*} \|z-x\|_\infty=|z_1-x_1|\le |z_1-y_1|+|y_1-x_1| \le\|z-y\|_\infty+\|y-x\|_\infty \end{equation*}

Exercise

Exercise

Question

Example

For any $x, y\in \mathbb{R}^n$, it is clear that $\tilde{d}(x,y)=\tilde{d}(y,x)$ and $\tilde{d}(x, y)=0$ if and only if $x=y$. For the proof of the triangle inequality, let $$ g(t)=\frac{t}{1+t},\quad t\in (0,\infty), $$ then $$ g'(t)={1\over (1+t)^2}>0,\quad t\in [0, \infty) $$ This implies that $g(t)$ is increasing function on $[0,\infty)$. Therefore, for any $x, y, z\in \mathbb{R}^n$, one has \begin{eqnarray*} \tilde{d}(x,z)&=&g(d(x,z)) \le g(d(x,y)+d(y, z))\\ &=& {d(x,y)+d(y, z)\over 1+d(x,y)+d(y, z)} \le {d(x,y)\over 1+d(x,y)}+{d(y, z)\over 1+d(y, z)} =\tilde{d}(x,y)+\tilde{d}(y,z). \end{eqnarray*} The proof is complete.

Example

For any $x, y\in X$, it is easy to see that $d_0(x, y)=d_0(y,x)$ and $d_0(x, y)=0 $ if and only if $x=y$. For $x,y, z\in X$, without loss of generality, one may assume that $x\ne z$, then either $x\ne y$ or $z\ne y$. Thus $$ d_0(x,z)=1\le d_0(x,y)+d_0(y,z). $$ By definition $(X, d_0)$ forms a metric space.

Topology of Metric Spaces

Definitions (Balls and Open Sets)

Let $(X, d)$ be a metric space. Then
(i) A ball centered at $x_0\in X$ with radius $r$ is defined by $$ B(x_0, r)=\{x\in X: d(x, x_0)< r\}. $$
(ii) Let $A$ be a set of $X$. A point $x_0\in A$ is said to be an interior point of $A$ if there is $r>0$ such that $B(x_0, r)\subset A$.
(iii) We say that $A$ is open if every point of $A$ is an interior point of $A$.

Proposition

Let $(X, d)$ be a metric space. Then
(i) The ball $B(x_0, r)$ is open for any $x_0\in X$ and $r>0$.
(ii) If $\overset{\circ}{A}$ denotes the subset of $A$ consisting of all its interior points, then $\overset{\circ}{A}$ is open.

(i) To prove that $B(x_0, r)$ is open, by definition, one needs to show that every point $y\in B(x_0, r)$ is an interior point of $B(x_0, r)$. Let $r(y)=r-d(y, x_0)>0$, we claim that $B(y, r(y))\subset B(x_0, r)$. In fact, for any $x\in B(y, r(y))$, by the triangle inequality, one has $$ d(x, x_0)\le d(x, y)+d(y, x_0) < r(y)+d(y,x_0)=r $$ This implies $x\in B(x_0, r)$ and $B(y, r(y))\subset B(x_0, r)$. The claim is proved and $y$ is an interior point. Therefore, $B(x_0, r)$ is open.
(ii) For any $x_0\in\overset{\circ}{A}$, then there is a ball $B(x_0, r)\subset A$. Notice that if $B(x_0, r)\subset A$ then $B(x_0, r)=B(x_0, r)_0\subset\overset{\circ}{A}$ (see next exercise). This implies $\overset{\circ}{A}$ is open.

Exercise

Complete Metric Spaces

Definition (Cauchy Sequences and Completeness)

Let $(X, d)$ be a metric space.
(i) We say that a sequence $\{x_n\}_{n=1}^\infty$ converges to a limit $x\in X$ iff $$ \lim_{n \to \infty} d(x_n,x) = 0. $$ (ii) We say that a sequence $\{x_n\}_{n=1}^\infty$ is a Cauchy sequence if, for any $\epsilon>0$, there exists $N$ such that if $m, n\ge N$ then $$ d(x_n, x_m)\leq\epsilon. $$ (iii) The metric space $(X, d)$ is called complete if every Cauchy sequence converges in $X$.

Example

(a) Let $\{x^m=(x_1^m,\cdots, x_n^m)\}_{m=1}^\infty$ be a Cauchy sequence in $(\mathbb{R}, d_2)$. Then for each $1\le j\le n$, $\{x_j^m\}_{m=1}^\infty$ is a Cauchy sequence in $\mathbb{R}$. There is a $x_j\in \mathbb{R}$ such that $\lim_{m\to\infty} x_j^m=x_j$. Let $x=(x_1,\cdots, x_n)$. Then $$ d_2(x^m, x)=\|x^m-x\|_2\le \sum_{j=1}^n |x^m_j-x_j|\to 0 \quad \text{ as } m\to \infty. $$
(b) Let $x_n=(1+\frac{1}{n})^n$. Then $$ \lim_{n\to \infty} x_n=e\in \mathbb{R} $$ Then $\{x_n\}_{n=1}^\infty$ is a Cauchy sequence in $(\mathbb{Q}^n, d_2)$, it does not have a limit in $(\mathbb{Q}^n, d_2)$. So, $(\mathbb{Q}^n, d_2)$ is not complete.
(c) $(\mathbb{Z},d_0)$ is a complete metric space. Since for any Cauchy sequence $\{x_n\}_{n=1}^\infty$, there is $N$ such that if $m>n\ge N$, one has $$ d_0(x_m, x_n)<1/2 $$ This implies that $x_m=x_n=X_N$ if $m,n\ge N$. Therefore, $x_n\to X_N$ as $n\to\infty$.

Definition (Contractive Sequences)

A sequence $\{x_n\}_{n=1}^\infty$ in $X$ is said to be contractive iff there is a constant $c\in [0, 1)$ such that $$ d(x_n, x_{n+1}) \le cd (x_{n-1}, x_n)\quad \text{ for }n=2,3,\dots. $$

Theorem

Every contractive sequence in a metric space $(X,d)$ is a Cauchy sequence.

Using the triangle inequality and the inequality defining contractivity we see that \begin{eqnarray*} d(x_m,x_n)&\le& d(x_m, x_{n+1}) + d(x_{n+1}, x_n)\\ &\le&\sum_{j=1}^{m-n} d(x_{n+j}, x_{n+j-1}) \le \sum_{j=1}^{m-n} c^j d(x_n, x_{n-1})\\ &\le & \sum_{j=1}^{m-n} c^j c^{n-2} d(x_2, x_1) =\dfrac{1-c^{m-n}}{1-c} c^{n-1} d(x_2, x_1) \le \frac{ c^{n-1}}{(1-c) } d(x_2, x_1). \end{eqnarray*} Observing that $$ \frac{c^{n-1}}{1-c} d(x_1, x_2) \to 0\text{ as }n \to \infty\, , $$ we can find $N\in \mathbb{N}$ such that $$ c^n / (1-c) \cdot d(x_2, x_1) < \epsilon\, , $$ so that $d(x_m, x_n)\leq\epsilon$, whenever $m\geq n \ge N$. This proves the theorem.

Corollary

In a complete metric space $(X, d)$, every contractive sequence has a limit in $X$.

Definition (Contractive Maps)

Let $(X, d)$ be a metric space. A map $f: X\to X$ is said to be a contractive map if there is $c\in (0,1)$ such that $$ d\bigl(f(x), f(y)\bigr)\le c\, d(x, y),\quad x, y\in X. $$

Theorem (Banach Fixed Point Theorem)

In a complete metric space $(X, d)$, every contractive map $f:X\to X$ has a unique fixed point in $X$.

The uniqueness is obvious. To prove the existence. Let $x_0\in X$ be arbitrary and define $$ x_n=f(x_{n-1})\quad \text{ for } \: n=1,2,3,\dots. $$ Then $\{x\}_{n=1}^\infty$ is a contractive sequence and therefore has a limit $x_\infty\in X$. Since $f$ is continous, one can easily see that $x_\infty=f(x_\infty)$.

Remark

Let $f(x) = x + \frac{1}{1+e^x}$ and $X=\mathbb{R}$. Then $$ f'(x) = 1 -\frac{e^x}{(1+e^x)^2} \in (0,1). $$ and so $$ |f(x) - f(y)| = |f'(\xi)||x-y| =\Big |1 - \frac{e^\xi}{(1+e^\xi)^2} \Big||x-y| < |x-y|. $$ But $f(x)=x$ has no solution in $\mathbb{R}$.

Proposition

If $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ are Cauchy sequences in a metric $(X,d)$, then $\bigl(d(x_n, y_n)\bigr)_{n\in \mathbb{N} }$ is a convergent sequence in $\mathbb{R}$.

Observe that \begin{align*} d(x_n, y_n) - d(x_m, y_m)& \le d(x_n, x_m) + d(x_m, y_n) - d(x_m, y_m)\\ & \le d(x_n, x_m) + d(x_m, y_m) +d(y_m, y_n) - d(x_m, y_m) \\ & = d(x_n, x_m) + d(y_m, y_n) \end{align*} for all $m,n\in \mathbb{N} $. By interchanging the roles of $x_n$ and $y_m$ it follows that $$ |d(x_m, y_m) - d(x_n, y_n) |\le d(x_m, x_n) + d(y_m, y_n) $$ By assumption there is $N\in \mathbb{N} $ such that $$ d(x_m, x_n)\leq\epsilon/2\text{ and }d(y_m, y_n)\leq\epsilon/2\text{ if }m,n\geq N. $$ Then $$ |d(x_n, y_n) - d(x_m, y_m) | < \epsilon/2 + \epsilon/2 = \epsilon\text{ when }m,n \ge N. $$ Therefore, $\bigl(d(x_n, y_n)\bigr)_{n\in \mathbb{N} }$ is a Cauchy sequence in $\mathbb{R}$ and by completeness, it possesses a limit.

Union and intersection of sets

Proposition

(i) Let $(A_\alpha)_{\alpha \in I}$ be a family (finite or infinite) of set in $X$. Then $$ \left ( \bigcup_{\alpha \in I} A_\alpha \right )^c = \bigcap_{\alpha \in I} A_\alpha^c $$
(ii) Let $(A_\alpha)_{\alpha \in I}$ be a family of open sets in a metric space $(X, d)$. Then $\cup_{\alpha\in I} A_\alpha $ is open.
(iii) If $A_1,\cdots, A_m$ are open in a metric space $(X, d)$, then $\cap_{j=1}^m A_j$ is also open

(i) It suffices to notice that \begin{eqnarray*} x \in \left ( \bigcup_{\alpha \in I} A_\alpha \right ) ^c \iff x \not \in \bigcup_{\alpha \in I} A_\alpha\iff x \not \in A_\alpha \mbox{ for any } \alpha \in I\\ \iff x \in A_\alpha^c\mbox{ for all } \alpha \in I \iff x \in \bigcap_{\alpha\in I} A_\alpha^c\, . \end{eqnarray*} (ii) is obvious.
(iii) For any $x_0\in \cap_{j=1}^m A_j$, there are $r_j>0$ such that $B(x_0, r_j)\subset A_j$ for $1\le j\le m.$ Let $$ r_0=\min\{ r_j: 1\leq j\leq m \}. $$ Then $B(x_0, r_0)\subset \cap _{j=1}^m A_j$. Therefore, $\cap_{j=1}^m A_j$ is open.