Given $D$ is a dense set of a topological space $X$, prove that a set $B$ intersect $D$ if and only if interior of $B$ is not empty.

Question

Given $D$ is a dense set of a topological space $X$, prove that a set $B$ intersect $D$ if and only if interior of $B$ is not empty.

The left direction is very easy to prove, however I find it quite hard to show the right direction. Can anyone help?

Thanks!

Answer 1

This is not if and only if. If the interior of $B$ is non-empty, it's a non-empty open set and as such intersects $D$ as $D$ is dense. But if $B = \{1\}$ in the reals (usual topology) then it intersects the dense set $D = \mathbb{Q}$, but it does not have non-empty interior, as it contains no interval.

Answer 2

That is a direct result of the theorem:
D is dense iff for all not empty open U, U $\cap$ D is not empty.

That theorem follows from the closure of D is the space and
x in $\overline D$ implies for all open nhoods of x, the
nhood intersects with D.