METRIC AND HILBERT SPACES CONSULTATION 27/10/2015
JOSEPH CHAN
Proposition 1. Let (X, g) be a metric space. Let U and V be open sets dense in X
equiped with the metric topology. Then U ∩ V is open and dense in X.
Proof. Before starting, note that since V is dense, V = X and any x ∈ U ⊂ X is a
close point of V . This means that any neighbourhood N of x satisfies N ∩ V 6= ∅.
In particular, since U is open, there exists an open ball B (x) which is contained
in U . This open ball is a neighbourhood of x so B (x) ∩ V 6= ∅. Thus U ∩ V ⊃
B (x) ∩ V 6= ∅ so U and V have non-empty intersection.
First, show that the intersection U ∩ V is open. Let x ∈ U ∩ V be a point in the
intersection. Then there needs to be an ∈ R>0 such that the radius open ball
B (x) = {y ∈ X | g(x, y) < }
is in U ∩ V .
Since U is open, there exists 1 ∈ R>0 such that B1 (x) is in U . Since V is open,
there exists 2 ∈ R>0 such that B2 (x) is in V .
Let = min{1 , 2 } so that ≤ i for i = 1, 2. Then if y is in B (x), then for
i = 1, 2,
g(x, y) < ≤ i
and so y is in Bi (x).
Thus,
B (x) ⊂ B1 (x) ⊂ U and B (x) ⊂ B2 (x) ⊂ V
Thus B (x) ⊂ U ∩ V .
Next, show that U ∩ V is dense. That is, show that U ∩ V = X. This means
that any point x ∈ X is a close point of U ∩ V . Then we want to show that for any
neighbourhood N of x, N ∩ (U ∩ V ) 6= ∅.
Let x ∈ X and since U is dense, U = X so x is a close point of U . Let N be any
neighbourhood of x, then W = N ∩ U 6= ∅.
Since W is non-empty, let y ∈ W . Since U is open, there is an open set U0 in U
containing y. Since N is a neighbourhood of y, N contains an open set N0 containing
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METRIC AND HILBERT SPACES CONSULTATION 27/10/2015
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y. The intersection W0 = U0 ∩ N0 is the intersection of two open sets so it too is an
open set.
Since V is dense, V = X so y is a close point of V . Since W0 is a neighbourhood
of y ∈ X, W0 ∩ V 6= ∅. But
W0 ∩ V ⊂ (N ∩ U ) ∩ V
thus N ∩ U ∩ V is non-empty as required.