Metric Space Definition

Scientific claim

Metric Space Definition

Claim

A metric space is an ordered pair $(X,d)$ , where $X$ is a set and $d:X\times X\to\mathbb{R}$ is a function such that, for all $x,y,z\in X$ :

d(x,y)\geq0

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

d(x,y)=d(y,x)

d(x,z)\leq d(x,y)+d(y,z).

Theoretical: Not applicableExperimental: Not applicable

Assumptions

No assumptions recorded.

Proof

This is a definition, so it is not proved from more fundamental propositions within this node.

Its purpose is to formalise the idea of distance in a way that supports rigorous reasoning about points becoming arbitrarily close.

Let $X$ be a set of objects called points. Let $d:X\times X\to\mathbb{R}$ be a function, where $d(x,y)$ denotes the distance assigned to the ordered pair of points $x,y\in X$ .

The first condition requires distances to be non-negative:

d(x,y)\geq0.

This prevents a separation between two points from being assigned a negative size.

The second condition specifies exactly when the distance is zero:

d(x,y)=0 \iff x=y.

The implication $x=y\Rightarrow d(x,y)=0$ ensures that a point has zero distance from itself. The reverse implication $d(x,y)=0\Rightarrow x=y$ ensures that two distinct points cannot be treated as having no separation.

The third condition requires symmetry:

d(x,y)=d(y,x).

This ensures that distance depends on the two points being compared, not on the direction in which the comparison is written.

The fourth condition is the triangle inequality:

d(x,z)\leq d(x,y)+d(y,z).

This ensures that travelling directly from $x$ to $z$ cannot be longer than travelling from $x$ to $y$ and then from $y$ to $z$ . It prevents inconsistent distance assignments in which an indirect route would be shorter than the direct separation.

Together, these conditions ensure that $d$ provides a coherent notion of closeness. In particular, they allow statements such as $d(x_n,x_m)<\varepsilon$ to have a stable meaning for every chosen tolerance $\varepsilon>0$ .

A metric space is any collection of things for which you can consistently say how far apart any two things are.

Examples

(Valid) Real Numbers with Ordinary Distance
Let $X=\mathbb{R}$ and let $x,y\in X$ .
$d(x,y)=|x-y|.$
This is a valid metric because the absolute difference is always non-negative, is zero only when $x=y$ , is symmetric, and satisfies the triangle inequality.
(Valid) Plane with Euclidean Distance
Let $X=\mathbb{R}^2$ . For points $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ in $X$ :
$d(\mathbf{x},\mathbf{y}) = \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}.$
This is a valid metric because it is the ordinary straight-line distance, which satisfies all four metric-space conditions.
(Valid) Plane with Manhattan Distance
Let $X=\mathbb{R}^2$ . For points $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ in $X$ :
$d(\mathbf{x},\mathbf{y}) = |x_1-y_1|+|x_2-y_2|.$
This is a valid metric because it measures the total horizontal and vertical separation between two points and satisfies all four metric-space conditions.
(Valid) Discrete Metric
Let $X$ be any set. For $x,y\in X$ :
$d(x,y) = \begin{cases} 0, & x=y, \\ 1, & x\neq y. \end{cases}$
This is a valid metric because a point is distance $0$ from itself, every distinct pair is distance $1$ apart, and the triangle inequality always holds.
(Invalid) Signed Difference
Let $X=\mathbb{R}$ and let $x,y\in X$ .
$d(x,y)=x-y.$
This is not a valid metric because a distance cannot be negative. For example:
$d(1,2)=-1.$
(Invalid) Squared Distance
Let $X=\mathbb{R}$ and let $x,y\in X$ .
$d(x,y)=|x-y|^2.$
This is not a valid metric because it violates the triangle inequality. For $x=0$ , $y=1$ , and $z=2$ :
$d(0,2)=4.$ $d(0,1)+d(1,2)=1+1=2.$
Therefore:
$d(0,2)>d(0,1)+d(1,2).$
(Invalid) Zero Distance for Every Pair
Let $X$ be a set containing at least two distinct elements. For all $x,y\in X$ :
$d(x,y)=0.$
This is not a valid metric because two distinct points must have a positive distance from one another. If $x\neq y$ , then:
$d(x,y)=0.$
This violates:
$d(x,y)=0 \iff x=y.$
(Invalid) Asymmetric Distance
Let $X=\mathbb{R}$ . For $x,y\in X$ :
$d(x,y) = \begin{cases} 0, & x=y, \\ y-x, & xy. \end{cases}$
This is not a valid metric because distance must be the same in either direction. For example:
$d(1,2)=1.$ $d(2,1)=2.$
Therefore:
$d(1,2)\neq d(2,1).$

Evidence

01

metric spaceUrs Schreiber · 2026-01-27 · supporting

Related Concepts

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