Metric Spaces

Let $X$ be a set. A metric on $X$ is a function $d:X\times X\rightarrow\mathbb{R}$ that assigns a distance to each pair of points and satisfies the metric axioms for all $x,y,z\in X:$

$$d(x,y)\geq 0,\,\,\mathrm{with}\,\,d(x,y)=0\,\,\mathrm{if}\,\mathrm{and}\,\mathrm{only}\,\mathrm{if}\,\,x=y$$ $$d(x,y)=d(y,x)$$ $$d(x,z)\leq d(x,y)+d(y,z).$$

The third axiom is the triangle inequality. A metric space $(X,d)$ is a set equipped with a metric, and is written $X$ when the metric is understood. Any subset $A\subseteq X$ is a metric space with the restriction of $d$ to $A\times A,$ the metric subspace $A.$ The real numbers with $d(x,y)=\lvert x-y\rvert$ and $\mathbb{R}^n$ with the Euclidean distance are metric spaces, the latter being the Euclidean space $\mathbb{R}^n.$ Any set is a metric space under the discrete metric, which is $0$ when $x=y$ and $1$ otherwise.

A normed vector space is a vector space $\mathcal{V}$ over $\mathbb{R}$ or $\mathbb{C}$ with a function $\lVert\cdot\rVert:\mathcal{V}\rightarrow\mathbb{R},$ its norm, that satisfies the norm axioms for all $\underline{u},\underline{v}\in\mathcal{V}$ and scalars $\lambda:$

$$\lVert\underline{v}\rVert\geq 0,\,\,\mathrm{with}\,\,\lVert\underline{v}\rVert=0\,\,\mathrm{if}\,\mathrm{and}\,\mathrm{only}\,\mathrm{if}\,\,\underline{v}=\underline{0}$$ $$\lVert\lambda\,\underline{v}\rVert=\lvert\lambda\rvert\lVert\underline{v}\rVert$$ $$\lVert\underline{u}+\underline{v}\rVert\leq\lVert\underline{u}\rVert+\lVert\underline{v}\rVert.$$

The norm of an inner product space satisfies these axioms by the Cauchy-Schwarz inequality. Every normed vector space is a metric space with the induced metric $d(\underline{u},\underline{v}):=\lVert\underline{u}-\underline{v}\rVert.$ For $p\geq 1$ the $p$-norm and the maximum norm on $\mathbb{R}^n$ are

$$\lVert\underline{x}\rVert_p:=\left(\sum_{i=1}^{n}\lvert x_i\rvert^p\right)^{\frac{1}{p}},\qquad\lVert\underline{x}\rVert_\infty:=\max_{1\leq i\leq n}\lvert x_i\rvert,$$

of which $\lVert\cdot\rVert_2$ is the Euclidean norm, $\lVert\cdot\rVert_1$ is the Manhattan norm and $\lVert\underline{x}\rVert_p\rightarrow\lVert\underline{x}\rVert_\infty$ as $p\rightarrow\infty.$ The triangle inequality for the $p$-norm is the Minkowski inequality.

Let $(X,d)$ be a metric space, $x\in X$ and $r>0.$ The open ball, closed ball and sphere of radius $r$ centred at $x$ are the sets of points at distance less than, at most and exactly $r$ from $x.$

$$B(x,r):=\left\{y\in X:d(x,y)<r\right\},\qquad\overline{B}(x,r):=\left\{y\in X:d(x,y)\leq r\right\},\qquad S(x,r):=\left\{y\in X:d(x,y)=r\right\}$$

A set $U\subseteq X$ is open if every point of $U$ is the centre of some open ball contained in $U,$ and a set $F\subseteq X$ is closed if its complement $X\setminus F$ is open.

$$U\,\,\mathrm{is}\,\mathrm{open}\Leftrightarrow\forall\,x\in U\,\exists\,r>0:B(x,r)\subseteq U$$

In $\mathbb{R}^n$ with the Euclidean distance this is the earlier definition of an open set. Open balls are open and closed balls, spheres and finite sets are closed; open intervals are open in $\mathbb{R}$ and closed intervals are closed. A set can be both open and closed, as $\varnothing$ and $X$ are, or neither, as $\left[0,1\right)$ is in $\mathbb{R}.$ Under the discrete metric every set is open. Open and closed sets have the following properties:

• $\varnothing$ and $X$ are open and closed;
• the union of any collection of open sets is open and the intersection of finitely many open sets is open;
• the intersection of any collection of closed sets is closed and the union of finitely many closed sets is closed.

The finiteness conditions cannot be dropped: $\bigcap_{n=1}^{\infty}\left(-\frac{1}{n},\frac{1}{n}\right)=\left\{0\right\}$ is an intersection of open sets that is not open.

Let $A\subseteq X$ and $x\in X.$ $x$ is an interior point of $A$ if some open ball centred at $x$ is contained in $A,$ a limit point of $A$ if every open ball centred at $x$ contains a point of $A$ other than $x,$ and an isolated point of $A$ if $x\in A$ and some open ball centred at $x$ contains no other point of $A.$ The interior $\mathrm{int}\,A$ of $A$ is the set of its interior points, the closure $\overline{A}$ of $A$ is the union of $A$ and its limit points and the boundary $\partial A$ of $A$ is the closure without the interior.

$$\mathrm{int}\,A\subseteq A\subseteq\overline{A},\qquad\partial A:=\overline{A}\setminus\mathrm{int}\,A$$

The interior is the largest open set contained in $A$ and the closure is the smallest closed set containing $A,$ so $A$ is open if and only if $A=\mathrm{int}\,A$ and closed if and only if $A=\overline{A},$ which holds if and only if $A$ contains all of its limit points. The boundary is closed and $x\in\partial A$ if and only if every open ball centred at $x$ meets both $A$ and $X\setminus A.$ For example, in $\mathbb{R}$ the interval $\left[0,1\right)$ has interior $\left(0,1\right),$ closure $\left[0,1\right]$ and boundary $\left\{0,1\right\},$ whereas the rational numbers $\mathbb{Q}$ have empty interior and closure $\mathbb{R}.$ A set $A$ is dense in $X$ if $\overline{A}=X,$ as $\mathbb{Q}$ is in $\mathbb{R}.$

A set $A\subseteq X$ is bounded if it is contained in some ball, and its diameter is the supremum of the distances between its points.

$$\mathrm{diam}\,A:=\sup\left\{d(x,y):x,y\in A\right\}$$

A sequence $\left(x_n\right)$ of points of $X$ converges to $x\in X,$ written $x_n\rightarrow x,$ if and only if the distances $d\left(x_n,x\right)$ converge to $0$ as a sequence of real numbers.

$$\forall\varepsilon>0\,\exists N\in\mathbb{N}:d\left(x_n,x\right)\leq\varepsilon\,\,\forall\,n\geq N$$

Equivalently, every open ball centred at $x$ contains all but finitely many terms of the sequence. A convergent sequence has exactly one limit and is bounded. In $\mathbb{R}^n$ a sequence converges if and only if each of its coordinates converges, and in a metric subspace convergence means convergence in the whole space with the limit lying in the subspace. Let $n_1<n_2<\cdots$ be an increasing sequence of indices. The sequence $\left(x_{n_k}\right)$ is a subsequence of $\left(x_n\right),$ and every subsequence of a convergent sequence converges to the same limit.

Convergence determines the closed sets: $x\in\overline{A}$ if and only if $x$ is the limit of a sequence of points of $A,$ so $A$ is closed if and only if it contains the limit of every convergent sequence of its points.

Let $\left(X,d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces, $f:X\rightarrow Y$ be a function and $x_0\in X.$ $f$ is continuous at $x_0$ if and only if points close to $x_0$ are mapped to points close to $f\left(x_0\right),$ and continuous on $X$ if it is continuous at every point.

$$\forall\varepsilon>0\,\exists\delta>0:d_Y\left(f(x),f\left(x_0\right)\right)\leq\varepsilon\,\,\forall\,x\in X,\,d_X\left(x,x_0\right)<\delta$$

For real functions of one or several real variables this is the earlier definition of continuity. The following are equivalent:

• $f$ is continuous on $X;$
• $f\left(x_n\right)\rightarrow f(x)$ whenever $x_n\rightarrow x$ in $X;$
• the inverse image $f^{-1}(V)$ of every open set $V\subseteq Y$ is open in $X;$
• the inverse image of every closed set in $Y$ is closed in $X.$

A composition of continuous functions is continuous. The metric $d:X\times X\rightarrow\mathbb{R}$ is itself continuous in each argument, since $\lvert d(x,z)-d(y,z)\rvert\leq d(x,y),$ and in a normed vector space the norm, addition and scalar multiplication are continuous.

$f$ is uniformly continuous if $\delta$ can be chosen to depend on $\varepsilon$ alone, Lipschitz continuous with constant $L\geq 0$ if $d_Y\left(f(x),f(y)\right)\leq L\,d_X(x,y)$ for all $x,y\in X$ and an isometry if it preserves distances, $d_Y\left(f(x),f(y)\right)=d_X(x,y).$ An isometry is injective and Lipschitz continuous with constant $1,$ and Lipschitz continuity implies uniform continuity, which implies continuity.

Two metrics $d$ and $d'$ on $X$ are equivalent if they have the same open sets, which holds if and only if they have the same convergent sequences with the same limits, and strongly equivalent if there are constants $0<c\leq C$ with

$$c\,d(x,y)\leq d'(x,y)\leq C\,d(x,y)\,\,\forall\,x,y\in X.$$

Strongly equivalent metrics are equivalent and have the same bounded sets, Cauchy sequences and uniformly continuous functions, whereas equivalent metrics need not: $d'(x,y)=\min\left\{1,\lvert x-y\rvert\right\}$ is equivalent to the usual metric on $\mathbb{R}$ but makes $\mathbb{R}$ bounded. Two norms are equivalent if their induced metrics are strongly equivalent. Any two norms on $\mathbb{R}^n$ are equivalent, since

$$\lVert\underline{x}\rVert_\infty\leq\lVert\underline{x}\rVert_2\leq\lVert\underline{x}\rVert_1\leq n\lVert\underline{x}\rVert_\infty,$$

so open sets, convergence and continuity in $\mathbb{R}^n$ do not depend on which norm is used.