A homeomorphism of a topological space with itself maps a set into one of the same category

Question

Prove: If $h$ is a homeomorphism of $S$ onto $S$ and if $E\subset S$, then $E$ and $h(E)$ have the same category in $S$.

Rudin, Functional Analysis, 2/e, p.43.

(My own answer follows below.)

Answer 1

We only need to show if $E$ is of the first category in $S$, then $h(E)$ is also of the first category in $S$. Let $(E_i)_{i=1}^\infty$ be nowhere dense sets s.t. $E=\cup_i E_i$. Then $h(E)=\cup_i h(E_i).$ We only need to show the $h(E_i)$'s are nowhere dense. By way of contradiction, suppose some $h(E_i)$ is not nowhere dense, i.e. $\overline{h(E_i)}$ has nonempty interior, which let's call $O$. Then $h^{-1}(O)$ is nonempty, open, and is a subset of $h^{-1}(\overline{h(E_i)}) \subset \overline{h^{-1}(h(E_i))}=\overline{E_i}$. (By Theorem 18.1 in Munkres 2/e, $f:X\to Y$ is continuous iff for every subset $A$ of $X$, one has $f(\bar{A})\subset \overline{f(A)}.$) Hence $E_i$ is not nowhere dense---a contradiction.