Definition A metric space (X, d) consists of a set X together with a distance function d: X × X → [0, +∞) on X satisfying the following axioms:

(i) d(x, y) ≥ 0 for all x, y ∈ X,

(ii) d(x, y) = d(y, x) for all x, y ∈ X,

(iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X,

(iv) d(x, y) = 0 if and only if x = y.

The quantity d(x, y) should be thought of as measuring the distance between the points x and y. The inequality d(x, z) ≤ d(x,y)+d(y, z) is referred to as the Triangle Inequality. The elements of a metric space are usually referred to as points of that metric space.

An n-dimensional Euclidean space Rn is a metric space with with respectto the Euclidean distance function d, defined by

for all x, y ∈ Rⁿ. Any subset X of Rⁿ may be regarded as a metric space whose distance function is the restriction to X of the Euclidean distance function on R n defined above.

Definition: Let (X, d) be a metric space. Given a point x of X and r ≥ 0,  the open ball BX(x, r) of radius r about x in X is defined by

BX(x, r) ≡ {x΄ ∈ X : d(x΄, x) < r}.

Definition: Let (X, d) be a metric space. A subset V of X is said to be an open set if and only if the following condition is satisfied:

• given any point v of V there exists some δ > 0 such that BX(v, δ) ⊂ V .

By convention, we regard the empty set ∅ as being an open subset of X.

(The criterion given above is satisfied vacuously in this case.)

Lemma 1.1 Let X be a metric space with distance function d, and let x₀ be a point of X. Then, for any r > 0, the open ball BX(x₀, r) of radius r about

x₀ is an open set in X.

Proof Let x ∈ BX(x₀, r). We must show that there exists some δ > 0 such that BX(x, δ) ⊂ BX(x₀, r). Now d(x, x0) < r, and hence δ > 0, where

δ = r − d(x, x₀). Moreover if x΄ ∈ BX(x, δ) then

                                             d(x΄, x₀) ≤ d(x΄, x) + d(x, x₀) < δ + d(x, x₀) = r,

by the Triangle Inequality, hence x΄ ∈ BX(x₀, r). Thus BX(x, δ) ⊂ BX(x₀, r),  showing that BX(x₀, r) is an open set, as required.

Proposition 1.2 Let X be a metric space. The collection of open sets in X has the following properties:—

(i) the empty set ∅ and the whole set X are both open sets;

(ii) the union of any collection of open sets is itself an open set;

(iii) the intersection of any finite collection of open sets is itself an open set.

Proof The empty set ∅ is an open set by convention. Moreover the definition of an open set is satisfied trivially by the whole set X. Thus (i) is satisfied.

Let A be any collection of open sets in X, and let U denote the union of all the open sets belonging to A. We must show that U is itself an open set.

Let x ∈ U. Then x ∈ V for some open set V belonging to the collection A.  Therefore there exists some δ > 0 such that BX(x, δ) ⊂ V . But V ⊂ U, and thus BX(x, δ) ⊂ U. This shows that U is open. Thus (ii) is satisfied.

Finally let V1, V2, V3, . . . , Vk be a finite collection of open sets in X, and let V = V₁ ∩ V₂ ∩ · · · ∩ V_k. Let x ∈ V . Now x ∈ V_j for all j, and therefore there exist strictly positive real numbers δ₁, δ₂, . . . , δ_k such that BX(x, δ_j ) ⊂ V_j for j = 1, 2, . . . , k. Let δ be the minimum of δ₁, δ₂, . . . , δ_k. Then δ > 0.

(This is where we need the fact that we are dealing with a finite collection of open sets.) Moreover BX(x, δ) ⊂ BX(x, δ_j ) ⊂ V_j for j = 1, 2, . . . , k, and thus BX(x, δ) ⊂ V . This shows that the intersection V of the open sets V₁, V₂, . . . , V_k is itself open. Thus (iii) is satisfied.

Any metric space may be regarded as a topological space. Indeed let X be a metric space with distance function d. We recall that a subset V of X is an open set if and only if, given any point v of V , there exists some δ > 0 such that {x ∈ X : d(x, v) < δ} ⊂ V . Proposition 1.2 shows that the topological space axioms are satisfied by the collection of open sets in any metric space. We refer to this collection of open sets as the topology generated by the distance function d on X.