Metric spaces

Examples.

Metrics in $\mathbb{R}$. Consider the set $\mathbb{R}$ of real numbers. The absolute difference function $|\cdot - \cdot|: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a metric on $\mathbb{R}$. Indeed for any real numbers $x, y, z$, we have $|x - y| \geq 0$, $|x - y| = 0$ if and only if $x = y$, $|x - y| = |y - x|$, and $|x - z| \leq |x - y| + |y - z|$.

Let $n \geq 1$. The set $\mathbb{R}^n$ is the set of all $n$-tuples of real numbers. For the sake of brevity, we shall denote an element of $\mathbb{R}^n$ by $x = (\xi_1, \xi_2, \ldots, \xi_n)$, where $\xi_i \in \mathbb{R}$ for $1 \leq i \leq n$. We define the Euclidean metric on $\mathbb{R}^n$ to be the map ${\ell^2}: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ given by $$ {\ell^2}(x, y) = \sqrt{\sum_{i = 1}^n (\xi_i - \eta_i)^2}. $$

In $\mathbb{R}^3$ and $\mathbb{R}^2$ one can visualize ${\ell^2}(x, y)$ as the distance between the points $x$ and $y$ in the Euclidean plane and space, respectively. The Euclidean metric is the most commonly used metric on $\mathbb{R}^n$ as it is the natural generalization of the absolute difference function on $\mathbb{R}$ and provides a natural notion of distance between points in $\mathbb{R}^n$.

The Manhattan metric (also known as the taxicab metric) on $\mathbb{R}^n$ is the map ${\ell^1}: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ given by $$ {\ell^1}(x, y) = \sum_{i = 1}^n |\xi_i - \eta_i| $$ for $x = (\xi_1, \xi_2, \ldots, \xi_n)$ and $y = (\eta_1, \eta_2, \ldots, \eta_n)$. The Manhattan metric is so-called because it is the distance a taxicab would have to travel in a rectangular grid of streets (similar to those found in Manhattan) to get from the point $x$ to the point $y$.

Finally, we define the Chebyshev distance on $\mathbb{R}^n$ as $$ {\ell^\infty}(x, y) = \max_{1 \leq i \leq n} |\xi_i - \eta_i|. $$

These three metrics are related by the inequalities $$ {\ell^\infty}(x, y) \leq {\ell^2}(x, y) \leq {\ell^1}(x, y) \leq n {\ell^\infty}(x, y). $$

Discrete metric. Let $X$ be an arbitrary set. The discrete metric on $X$ is the map $\delta: X \times X \to \mathbb{R}$ given by $$ \delta(x, y) = \begin{cases} 0 & \text{if } x = y \\ 1 & \text{if } x \neq y. \end{cases} $$

Example. In $\mathbb{R}^2$ with the $\ell^2$ metric, the open ball or radius $1$ centered at the origin is the open disc $$ B_1((0, 0)) = \{(x, y) \in \mathbb{R}^2: x^2 + y^2 \lt 1\}. $$ On the other hand, using the $\ell^1$ metric, the earlier-defined open ball is the set $$ B_1((0, 0)) = \{(x, y) \in \mathbb{R}^2: |x| + |y| \lt 1\}, $$ which geometrically is a square (diamond) with vertices $(\pm 1, 0)$ and $(0, \pm 1)$. Finally, using the standard discrete metric, the ball is reduced to the singleton $\{(0, 0)\}$.

Example. Let $X$ be an arbitrary set and consider the standard discrete metric $\delta$ in $X$. If $A$ is a subset pf $X$, then $\operatorname{int}(A) = A$ and $\operatorname{ext}(A) = X \setminus A$. $A$ has no boundary points.

Proof.

(2) Let $x \in \bigcup_{U \in \mathfrak{U}} U$. Then $x \in U$ for some $U' \in \mathfrak{U}$. Since $U'$ is open, there exists $\varepsilon > 0$ such that $B_{\varepsilon}(x) \subset U'$. But $U' \subset \bigcup_{U \in \mathfrak{U}} U$, so $B_{\varepsilon}(x) \subset \bigcup_{U \in \mathfrak{U}} U$.

Proof. Suppose $(a_n) \to a$ and $(a_n) \to b$. We want to show that $a = b$.

Let $\varepsilon := \frac{1}{3}\rho(a, b) > 0$. Now since $(a_n) \to a$, there exists $N_1 \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon$ for all $n \geq N_1$. Similarly, since $(a_n) \to b$, there exists $N_2 \in \mathbb{N}$ such that $\rho(a_n, b) \lt \varepsilon$ for all $n \geq N_2$.

Now choose $m := \max\{N_1, N_2\}$. Then for all $n \geq m$, we have $\rho(a_n, a) \lt \varepsilon$ and $\rho(a_n, b) \lt \varepsilon$. Using the triangle inequality, we have $$ \begin{align} \rho(a, b) & \leq \rho(a_n, a) + \rho(a_n, b) = \rho(a, a_n) + \rho(a_n, b) \\ & \lt \varepsilon + \varepsilon = \frac{2}{3}\rho(a, b) \end{align} $$ so that $\rho(a,b) \lt \frac{2}{3}\rho(a, b)$. This is a contradiction since $\rho(a, b) \gt 0$. Therefore $a = b$.

Example. We will show that the sequence $\displaystyle\left(\frac{n+1}{n}\right)$ in $\mathbb{R}$ with the usual metric converges to $1$.

Let $(a_n)$ be the sequence described above. Then if $(a_n) \to 1$, we need to show that for any $\varepsilon \gt 0$ and for sufficiently large $N$ we have $$ \left\lvert \frac{n+1}{n} - 1 \right\rvert = \left\lvert \frac{n+1-n}{n} \right\rvert = \left\lvert \frac{1}{n} \right\rvert \lt \varepsilon $$ and thus $1/n \lt \varepsilon$ and $n \gt 1/\varepsilon$. Therefore choosing $N \in \mathbb{N}$ such that $N \gt 1/\varepsilon$ will suffice.

We can proceed with the formal proof.

Let $\varepsilon \gt 0$. Choose $N \in \mathbb{N}$ such that $N \gt 1/\varepsilon$. Then for all $n \geq N$, we have $n \gt 1/\varepsilon$ and thus $1/n \lt \varepsilon$. Therefore $$ \left\lvert \frac{1}{n} \right\rvert \lt \varepsilon $$ and so $(a_n) \to 1$.

Example. We will show that the sequence $(a_n) = n$ does not converge in $\mathbb{R}$.

Tending towards contradiction, suppose that $(a_n)$ converges to some value $\nu$ and consider $\varepsilon = 1 > 0$. Then because $(a_n)$ converges, there must exist some natural number $N$ such that for any $n \in \mathbb{N}$ with $n \geq N$, we have $\rho(a_n, \nu) \lt \varepsilon$. In particular, we have $\rho(n, \nu) \lt 1$ which implies that $n \lt \nu + 1$, so that $\mathbb{N}$ is bounded above by $\nu + 1$. This is a contradiction and so $(a_n)$ does not converge.

Proof. To prove that $(a_n)$ is bounded, we need to find a point $r \in X$ and a real number $M \gt 0$ such that $\rho(a_n, r) \leq M$ for all $n \in \mathbb{N}$. Now because $(a_n)$ is convergent (to some point $a$, say), there exists $N \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon$ for all $n \geq N$. Fix $\varepsilon = 1$. Then in particular, we have $\rho(a_n, a) \lt 1$.

Define $$ M := \max\{\rho(a_1, a), \rho(a_2, a), \dots, \rho(a_{N-1}, a), 1\}. $$ Thus for all $n \geq N$, $\rho(a_n, a) \lt 1 \leq B$ (because $(a_n)$ is convergent) and for all $n \lt N$, $1 \leq M$ (by our definition of $M$). Therefore $\rho(a_n, a) \leq M$ for all $n \in \mathbb{N}$ and so $(a_n)$ is bounded.

Proof. Because $(a_n)$ converges to $a$, for every $\varepsilon \gt 0$ there exists $N \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon$ for all $n \geq N$. Now let $(a_{n_k})$ be a subsequence of $(a_n)$. Since the $n_k$’s are strictly increasing $n_k \geq k$[1] and thus for all $k \geq N$, we have $n_k \geq N$, so that $\rho(a_{n_k}, a) \lt \varepsilon$. Therefore $(a_{n_k})$ converges to $a$.

Proof.

(1) Let $\varepsilon > 0$. Then $\varepsilon / 2 > 0$, and since $(a_n) \to a$ and $(b_n) \to b$, there exists $N_1, N_2 \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon / 2$ for all $n \geq N_1$ and $\rho(b_n, b) \lt \varepsilon / 2$ for all $n \geq N_2$. Let $N := \max\{N_1, N_2\}$. Then for all $n \geq N$, we have $$ \begin{align} \rho((a_n + b_n), (a + b)) & \leq \rho(a_n, a) + \rho(b_n, b) \\ & \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align} $$ Therefore $(a_n + b_n) \to a + b$.

(2) To prove this, we want to show that for any $\varepsilon \gt 0$, there exists $N \in \mathbb{N}$ such that $\rho(ca_n, ca) \lt \varepsilon$ for all $n \geq N$. From $\rho(ca_n, ca) \lt \varepsilon$, we have $\|c\| \cdot \rho(a_n, a) \lt \varepsilon$ (where $\|c\|$ is the norm of $c$).

We now proceed with the actual proof. Let $\varepsilon \gt 0$. Then $\displaystyle\frac{\varepsilon}{\|c\|} \gt 0$. Since $(a_n) \to a$, there exists $N \in \mathbb{N}$ such that $\rho(a_n, a) \lt \displaystyle\frac{\varepsilon}{\|c\|}$ for all $n \geq N$. Therefore for all $n \geq N$, we have $$ \begin{align} \rho(ca_n, ca) & = \|c\| \cdot \rho(a_n, a) \\ & \lt \|c\| \cdot \frac{\varepsilon}{\|c\|} = \varepsilon. \end{align} $$ Therefore $(ca_n) \to ca$.

(3) First observe that $$ \begin{align} \rho(a_n b_n, ab) & = \rho(a_n b_n, a b_n + a b - a b_n) \\ & \leq \rho(a_n b_n, a b_n) + \rho(a b_n, a b) + \rho(a b, a b_n) \\ & =

Proof.

(1) We prove the contrapositive. Suppose $a \lt 0$. Let $\varepsilon \gt 0$ be such that $a + \varepsilon \lt 0$. Then since $(a_n) \to a$, there exists $N \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon$ for all $n \geq N$. Taking $n = N$, we have $\rho(a_N, a) \lt \varepsilon$ and thus $a_N \lt a + \varepsilon \lt 0$. Therefore $a_N \lt 0$ and so $a_n \not\geq 0$ for all $n \geq N$. Therefore $(a_n)$ does not converge to $a$.

(2) Consider $N = 1$, i.e., $a_n \leq b_n$ for all $n \in \mathbb{N}$. Let $(c_n)$ be the sequence defined by $c_n = b_n - a_n$. Then $c_n \geq 0$ for all $n \in \mathbb{N}$ and $c_n \to b - a$. By (1), $b - a \geq 0$ and thus $a \leq b$. If $N \gt 1$, then we apply the same argument to the $N$-tail of $(a_n)$ and $(b_n)$.

(3) We have $0 \leq a_n - \xi \leq \eta - \xi$ for all $n \in \mathbb{N}$ from the hypothesis, whence, defining the constant sequence $(x_n) := \eta - \xi$ and the sequence $(y_n) := a_n - \xi$, we have $(x_n) \to \eta - \xi$ and $(y_n) \to a - \xi$. By (1) and (2), we have $0 \leq a - \xi \leq \eta - \xi$ and thus $\xi \leq a \leq \eta$.

Proof.

(1) Let $(a_n) \to a$. Then for every $\varepsilon \gt 0$, there exists $N \in \mathbb{N}$ such that $\rho(a_n, a) \lt \varepsilon / 2$ for all $n \geq N$. Therefore for all $n, m \geq N$, we have $$ \rho(a_n, a_m) \leq \rho(a_n, a) + \rho(a, a_m) \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$ Therefore $(a_n)$ is Cauchy.

(2) Let $(a_n)$ be Cauchy. Then for every $\varepsilon \gt 0$, there exists $N \in \mathbb{N}$ such that $\rho(a_n, a_m) \lt \varepsilon$ for all $n, m \geq N$. In particular, for $\varepsilon = 1$, there exists $N \in \mathbb{N}$ such that $\rho(a_n, a_m) \lt 1$ for all $n, m \geq N$. Moreover, since $N \geq N$, we have $\rho(a_n, a_N) \lt 1$ for all $n \geq N$.

Define $$ M := \max\{\rho(a_1, a_N), \rho(a_2, a_N), \dots, \rho(a_{N-1}, a_N), 1\}. $$ Then for all $n \geq N$, we have $\rho(a_n, a_N) \lt 1 \leq M$ (by our choice of $N$ above) and for all $n \lt N$, $\rho(a_n, a_N) \leq M$ (by our definition of $M$). Therefore $\rho(a_n, a_N) \leq M$ for all $n \in \mathbb{N}$ and so $(a_n)$ is bounded.

(3) The necessary condition is trivial. If we suppose that $(a_n) \to a$, then by Theorem 3.5, every subsequence of $(a_n)$ converges to $a$. We prove sufficiency.

Suppose $(a_{n_k})$ is a subsequence of $(a_n)$ that converges to $a$. Let $\varepsilon > 0$. Then since $(a_n)$ is Cauchy, there exists $N \in \mathbb{N}$ such that for all $n, m \geq N$, we have $$\rho(a_n, a_m) \lt \frac{\varepsilon}{2}.$$ Moreover, since $(a_{n_k}) \to a$, there exists $M \in \mathbb{N}$ such that for all $k \geq M$, $$\rho(a_{n_k}, a) \lt \frac{\varepsilon}{2}.$$

If $K > \max\{N, M\}$, then for all $n \geq K$, and since $n_K \geq K$, we have $$ \begin{align} \rho(a_n, a) &\leq \rho(a_n, a_{n_K}) + \rho(a_{n_K}, a) \\ &\lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align} $$

Therefore $(a_n) \to a$.

Proof.

$(1) \implies (2).$ Suppose $\mathfrak{F}$ is a collection of closed sets in $X$ with the finite intersection property. Suppose further that $\displaystyle\bigcap_{F \in \mathfrak{F}} F$ is nonempty. Then if we define $\mathfrak{U}$ as the collection of the complements $U := X \setminus F$ for each $F \in \mathfrak{F}$, we have $$ \bigcup_{U \in \mathfrak{U}} U = \bigcup_{F \in \mathfrak{F}} (X \setminus F) = X \setminus \bigcap_{F \in \mathfrak{F}} F = X \setminus \varnothing = X, $$ so that $\mathfrak{U}$ is an open cover of $X$. By hypothesis, there exists a finite subcover $\{U_1, U_2, \dots, U_n\}$ of $\mathfrak{U}$ and thus $$ X = \bigcup_{i=1}^n U_i = \bigcup_{i=1}^n (X \setminus F_i) = X \setminus \bigcap_{i=1}^n F_i, $$ so that $\displaystyle\bigcap_{i=1}^n F_i = \varnothing$. This contradicts the fact that $\mathfrak{F}$ has the finite intersection property. Therefore $\mathfrak{F}$ has a nonempty intersection.

$(2) \implies (3).$

$(3) \implies (4).$

$(4) \implies (5).$

$(5) \implies (1).$