Subset of Nowhere Dense Subset is Nowhere Dense

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $A \subseteq S$ be nowhere dense in $T$.

Let $B \subseteq A$.

Then $B$ is nowhere dense in $T$.

Proof

Aiming for a contradiction, suppose it is not the case that $B$ is nowhere dense in $T$.

Then by definition of nowhere dense:

$B^-$ contains some open set of $T$ which is non-empty.

From Set Closure Preserves Set Inclusion, we have:

$B^- \subseteq A^-$

So:

$A^-$ contains some open set of $T$ which is non-empty.

So $A$ is not nowhere dense in $T$.

This contradicts our assertion that $A$ is nowhere dense in $T$:

The result follows by Proof by Contradiction.

$\blacksquare$