Homeomorphic Image of Nowhere Dense Set is Nowhere Dense

Theorem
Let $X$ and $Y$ be topological spaces.

Let $f : X \to Y$ be a homeomorphism.

Let $A$ be a nowhere dense subset of $X$.

Then $f \sqbrk A$ is a nowhere dense subset of $Y$.

Proof
Since $A$ is nowhere dense, we have:
 * $\paren {A^-}^\circ = \O$

where $A^-$ is the closure of $A$ and $\paren {A^-}^\circ$ is the interior of $A^-$.

From Image of Empty Set is Empty Set, we have:
 * $\O = f \sqbrk {\paren {A^-}^\circ}$

So, we have:

Hence $f \sqbrk A$ is a nowhere dense subset of $Y$.