Metric Spaces

This scene axiomatizes distance. Compare how the shape of a ball changes when the metric changes.

A metric obeys only three rules. Distances are never negative. The distance is the same both ways. A detour is never shorter than the direct way. Same rules, different measurement: balls of equal radius take different shapes.

For example, change how a 📏 ruler measures between two points. Measure straight, and you get a disk. Measure along grid lines only, and you get a diamond. Same radius, different reach.

Examples & Exercises

Prove the discrete metric $d(x,y)=0$ if $x=y$ , $1$ otherwise, satisfies the metric axioms.

d(x,y) \geq 0,\quad d(x,y)=0 \iff x=y

d(x,y)=d(y,x),\quad d(x,z) \leq d(x,y)+d(y,z)

Discrete metric,

r = 0.5

—

B(p_0, 0.5) = \{p_0\}

Show proof

$d(x,y)$ takes only the values $0$ or $1$ , so $d(x,y) \geq 0$ . By definition $d(x,y) = 0$ exactly when $x = y$ , and $d(x,y) = 1$ when $x \neq y$ , so $d(x,y) = 0 \iff x = y$ .
Symmetry: if $x = y$ then $d(x,y) = 0 = d(y,x)$ , and if $x \neq y$ then $d(x,y) = 1 = d(y,x)$ , so $d(x,y) = d(y,x)$ .
For the triangle inequality, argue by cases. If $x = z$ then the left side is $d(x,z) = 0 \leq d(x,y) + d(y,z)$ .
If $x \neq z$ then the left side is $d(x,z) = 1$ . For any $y$ : if $y$ equals $x$ the right side is $0 + 1 = 1$ ; if $y$ equals $z$ it is $1 + 0 = 1$ ; if $y$ equals neither it is $1 + 1 = 2$ . In every case $1 \leq d(x,y) + d(y,z)$ .
Hence $d(x,z) \leq d(x,y) + d(y,z)$ holds for all $x, y, z$ — the discrete metric satisfies all the metric axioms.

Prove every subset of a discrete metric space is open.

B(x, 1/2) = \{x\}

r = 0.5 < 1

— every singleton is an open ball.

Show proof

Let $A$ be any subset of a discrete metric space. If $A = \emptyset$ , then $A$ is open vacuously — there is no point of $A$ at which openness could fail.
Otherwise take $x \in A$ .
With $r = 1/2$ we have $B(x, 1/2) = \{x\}$ : if $d(x,y) < 1/2$ then, since $d(x,y)$ is only $0$ or $1$ , we must have $d(x,y) = 0$ , i.e. $y = x$ .
Hence $B(x, 1/2) = \{x\} \subseteq A$ . Since $x$ was arbitrary, every point of $A$ is an interior point — $A$ is open.

Prove a convergent sequence in a discrete metric space is eventually constant.

d(x_n, x) < 1/2 \implies x_n = x

r = 1.5 > 1

— the ball becomes all six points.

Show proof

Let $x_n \to x$ . By the definition of convergence, with $\varepsilon = 1/2$ there is an $N$ such that $d(x_n, x) < 1/2$ for all $n \geq N$ .
In the discrete metric, $d(x_n, x) < 1/2$ forces $d(x_n, x) = 0$ since $0$ is the only attainable value, and that means $x_n = x$ .
Thus $x_n = x$ for all $n \geq N$ — the sequence is eventually constant.

Part of these learning paths

Examples & Exercises