THM. If (X.d) is metric separable, then it possesses a countable base.

Proof. Let {x_n : n ∈ N} be dense in X, and K_n, r denote an open ball with a center x_n and positive rational radius r. Then B = {K_n, r : n ∈ N,  r ∈ Q₊} is a countable family of sets.

To show that B is a base of (X,d) we need only to prove that for every x ∈ X, ε > 0 there is some K_n, r ∈ B which fulfills x ∈ K_n, r ⊂ K(x, ε).

Let x ∈ X, ε > 0. By the density of {x_n : n ∈ N} in X, there is x_n , such that d(x,x_n) < ε/3.

Let r ∈ Q₊ be such that d(x,x_n) < r < ε/3. Then x ∈ K_n, r. Now we prove that K_n, r ⊂ K(x,ε).

Assume that z ∈ K_n, r. Hence d(z,x_n) < r. Moreover, by the triangle inequality:

$d\left( z,x \right) \leq d\left( z,x_{n} \right) + d\left( x_{n},x \right) < r + r < \left( \frac{2}{3} \right)\varepsilon < \varepsilon$.

Remark. In the above proof we use the fact from our first lecture:

If (X, θ) is a topological space then

B ⊂ θ is a base of (X,θ)  ⇔  (∀x∈X)(∀U_x∈θ)(∃C∈B)  x ∈ C ⊂ U .

(U_x denotes a neighborhood of x)