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Baire’s theorem | Functional Analysis

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 v1, v2,… 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 v1 is dense, W ∩ V1 is a non – empty open set. Hence ∃ x1 ∈ W ∩ V1 and

0 < r1<1 such that S(x1, r1) ⊂ W ∩ V1 ——————– (ii)

If n ≥ 2 and xn – 1 and rn – 1 are chosen. The denseness and open of vn shows that

Vn ∩ S(xn – 1, rn – 1) is non empty. We can therefore find xn and rn  such that

S (xn, rn) ⊂ Vn ∩ S(xn – 1, rn – 1) , 0 <rn<1/n ——————- (iii)


By induction this processes a sequence {xn} in X.

If i,j>n the construction lies in S(xn,rn).

So that d(xi, xj) <2rn < 2/n and hence {xn} is a Cauchy sequence in X, since X is a complete metric ∃ a point x ∈ X such that

Since xi lies in closed set S (xn, rn) if i > n, it follows x lies in S (xn, rn) and (iii) shows x lies in each Vn. By (ii) we have x ∈ W.

This completes the proof of the theorem.    (Proved)

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