Statement: If X is a complete metric space, the intersection of every countable collection of dense open subsets of X is dense in X.

Proof: Suppose v_{1}, v_{2},… are dense in and open in X. We have to prove that

is dense in X.

It is suffices to show that every non empty open set intersect.

Let W be a non empty open set and ρ be the metric of α and let

S(x, r) = {y∈X: ρ(x,y)<r} ———————- (i)

and __S__(x,r) be the closure of S(x,r).

Since v_{1} is dense, W ∩ V_{1} is a non – empty open set. Hence ∃ x_{1} ∈ W ∩ V_{1} and

0 < r_{1}<1 such that __S__(x_{1}, r_{1}) ⊂ W ∩ V_{1} ——————– (ii)

If n ≥ 2 and x_{n – 1 }and r_{n – 1 }are chosen. The denseness and open of v_{n} shows that

V_{n} ∩ S(x_{n – 1}, r_{n – 1}) is non empty. We can therefore find x_{n} and r_{n } such that

__S__ (x_{n}, r_{n}) ⊂ V_{n} ∩ S(x_{n – 1}, r_{n – 1}) , 0 <r_{n}<1/n ——————- (iii)

By induction this processes a sequence {x_{n}} in X.

If i,j>n the construction lies in S(x_{n},r_{n}).

So that d(x_{i}, x_{j}) <2r_{n} < 2/n and hence {x_{n}} is a Cauchy sequence in X, since X is a complete metric ∃ a point x ∈ X such that

Since x_{i} lies in closed set __S__ (x_{n}, r_{n}) if i > n, it follows x lies in __S__ (x_{n}, r_{n}) and (iii) shows x lies in each V_{n}. By (ii) we have x ∈ W.

This completes the proof of the theorem. (Proved)