Metric Spaces

Let $X$ be a set. A metric on $X$ is a function $d:X\times X\rightarrow\mathbb{R}$ that 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.

Let $X$ be a set and $d$ be a metric on $X.$ The pair $(X,d)$ is a metric space, and is written $X$ when the metric is understood. The real numbers with $d(x,y)=\lvert x-y\rvert$ form a metric space, and $\mathbb{R}^n$ with the Euclidean distance is the metric space called Euclidean space.

Let $(X,d)$ be a metric space and $A\subseteq X.$ The metric subspace $A$ is the metric space $(A,d_A)$ where $d_A$ is the restriction of $d$ to $A\times A.$

Let $X$ be a set. The discrete metric on $X$ is the metric that is $0$ when $x=y$ and $1$ otherwise. Any set is a metric space under the discrete metric.

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

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

Let $\mathcal{V}$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$ and $\lVert\cdot\rVert$ be a norm on $\mathcal{V}.$ The pair $(\mathcal{V},\lVert\cdot\rVert)$ is a normed vector space. The norm of an inner product space satisfies the norm axioms by the Cauchy-Schwarz inequality, so every inner product space is a normed vector space.

Let $\mathcal{V}$ be a normed vector space. The induced metric on $\mathcal{V}$ is the metric given by the norm of the difference of two vectors.

$$d(\underline{u},\underline{v}):=\lVert\underline{u}-\underline{v}\rVert$$

Every normed vector space is a metric space with its induced metric.

Let $\underline{x}\in\mathbb{R}^n$ and $p\geq 1.$ The $p$-norm of $\underline{x}$ is the $p$-th root of the sum of the $p$-th powers of the absolute values of its coordinates, and the maximum norm of $\underline{x}$ is the largest absolute value of its coordinates.

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

The $2$-norm is the Euclidean norm, the $1$-norm 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 of radius $r$ centred at $x$ is the set of points at distance less than $r$ from $x.$ The closed ball of radius $r$ centred at $x$ is the set of points at distance at most $r$ from $x.$ The sphere of radius $r$ centred at $x$ is the set of points at distance exactly $r$ from $x.$

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

Let $(X,d)$ be a metric space and $U\subseteq X.$ The set $U$ is open if and only if every point of $U$ is the centre of some open ball contained in $U.$

$$\forall\,x\in U\,\exists\,r>0:B(x,r)\subseteq U$$

Let $(X,d)$ be a metric space and $F\subseteq X.$ The set $F$ is closed if and only if its complement $X\setminus F$ is open.

In $\mathbb{R}^n$ with the Euclidean distance this is the earlier definition of an open set. Open and closed sets have the following properties:

• 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 in $\mathbb{R};$
• 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;
• 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 $(X,d)$ be a metric space, $A\subseteq X$ and $x\in X.$ The point $x$ is an interior point of $A$ if and only if some open ball centred at $x$ is contained in $A.$ The point $x$ is a limit point of $A$ if and only if every open ball centred at $x$ contains a point of $A$ other than $x.$ The point $x$ is an isolated point of $A$ if and only if $x\in A$ and some open ball centred at $x$ contains no other point of $A.$

Let $(X,d)$ be a metric space and $A\subseteq X.$ The interior of $A,$ written $\mathrm{int}\,A,$ is the set of interior points of $A.$ The closure of $A,$ written $\overline{A},$ is the union of $A$ and its limit points. The boundary of $A,$ written $\partial A,$ is the closure of $A$ without the interior of $A.$

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

The interior, closure and boundary have the following properties:

• the interior is the largest open set contained in $A,$ so $A$ is open if and only if $A=\mathrm{int}\,A;$
• the closure is the smallest closed set containing $A,$ so $A$ is 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 $\mathbb{R}.$

Let $(X,d)$ be a metric space and $A\subseteq X.$ The set $A$ is dense in $X$ if and only if $\overline{A}=X.$ The rational numbers are dense in $\mathbb{R}.$

Let $(X,d)$ be a metric space and $A\subseteq X.$ The set $A$ is bounded if and only if it is contained in some ball. The diameter of $A$ is the supremum of the distances between its points.

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

Let $(X,d)$ be a metric space, $\left(x_n\right)$ be a sequence of points of $X$ and $x\in X.$ The sequence $\left(x_n\right)$ converges to $x,$ 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. Convergent sequences in a metric space have the following properties:

• 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;
• in a metric subspace a sequence converges if and only if it converges in the whole space with the limit lying in the subspace;
• every subsequence of a convergent sequence converges to the same limit;
• $x\in\overline{A}$ 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 $x_0\in X.$ The function $f$ is continuous at $x_0$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that points within $\delta$ of $x_0$ are mapped to points within $\varepsilon$ of $f\left(x_0\right).$

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

Let $\left(X,d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f:X\rightarrow Y$ be a function. The function $f$ is continuous on $X$ if and only if it is continuous at every point of $X.$ 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 $X.$

Continuous functions on metric spaces have the following properties:

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

Let $\left(X,d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f:X\rightarrow Y$ be a function. The function $f$ is uniformly continuous if and only if for every $\varepsilon>0$ there exists $\delta>0,$ depending on $\varepsilon$ alone, such that $d_Y\left(f(x),f(y)\right)\leq\varepsilon$ for all $x,y\in X$ with $d_X(x,y)<\delta.$

Let $\left(X,d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f:X\rightarrow Y$ be a function. The function $f$ is Lipschitz continuous with constant $L\geq 0$ if and only if $d_Y\left(f(x),f(y)\right)\leq L\,d_X(x,y)$ for all $x,y\in X.$

Let $\left(X,d_X\right)$ and $\left(Y,d_Y\right)$ be metric spaces and $f:X\rightarrow Y$ be a function. The function $f$ is an isometry if and only if it preserves distances.

$$d_Y\left(f(x),f(y)\right)=d_X(x,y)\,\,\forall\,x,y\in X$$

An isometry is injective and Lipschitz continuous with constant $1.$ Lipschitz continuity implies uniform continuity, which implies continuity.

Let $X$ be a set and $d$ and $d'$ be metrics on $X.$ The metrics $d$ and $d'$ are equivalent if and only if they have the same open sets. Two metrics are equivalent if and only if they have the same convergent sequences with the same limits.

Let $X$ be a set and $d$ and $d'$ be metrics on $X.$ The metrics $d$ and $d'$ are strongly equivalent if and only if there are constants $0<c\leq C$ such that each metric is bounded above and below by constant multiples of the other.

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

Strongly equivalent metrics are equivalent, and have the same bounded sets, Cauchy sequences and uniformly continuous functions. 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.

Let $\mathcal{V}$ be a vector space and $\lVert\cdot\rVert$ and $\lVert\cdot\rVert'$ be norms on $\mathcal{V}.$ The norms are equivalent if and only 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.