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:

\(\ds f \sqbrk {\paren {A^-}^\circ}\)

\(=\)

\(\ds \paren {f \sqbrk {A^-} }^\circ\)

Image of Interior of Set under Homeomorphism is Interior of Image

\(\ds \)

\(=\)

\(\ds \paren {\paren {f \sqbrk A}^-}^\circ\)

Homeomorphism iff Image of Closure equals Closure of Image

\(\ds \)

\(=\)

\(\ds \O\)

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

$\blacksquare$