I'm asking this question with this other on my mind:
Second category subset vs subspace
On it, a counterexample of a 1st category set which is not of the 1st category on itself is given: the Cantor set, considered within $\mathbb{R}$ with the standard topology.
I'm opening this new question since I couldn't find an instance for a counterexample of the converse: a set which is of the 2nd category but which is not of the 2nd category on itself. I wonder if there's $X$ a topological space which has some $A\subset X$ which is of the 2nd category in $X$ but not in itself. Is there some known example of such a thing?
Thank you in advance and my excuses if is this is a duplicate, I wasn't able to find this answered here before.
There is no such set.
Suppose that $A$ is first category in itself; then there is a countable family $\mathscr{D}$ of nowhere dense subsets of $A$ such that $\bigcup\mathscr{D}=A$. That is, $\operatorname{cl}_{\tau_A}D$ has empty interior in $A$ for each $D\in\mathscr{D}$.
Suppose that $V\subseteq\operatorname{cl}_{\tau_X}D$ for some $V\in\tau_X$ and $D\in\mathscr{D}$. Then $V\cap A\subseteq\operatorname{int}_{\tau_A}\operatorname{cl}_{\tau_A}D=\varnothing$, and I claim that in fact $V=\varnothing$. For if not, then $V\cap\operatorname{cl}_{\tau_X}D\ne\varnothing$, so $V\cap D\ne\varnothing$, since $V$ is open in $X$, and thus finally $V\cap A\ne\varnothing$, which is impossible. $\mathscr{D}$ is therefore a family of sets that are nowhere dense in $X$, and so $A$ is first category in $X$.