Let $(X,T)$ be a topological space. When we say $A$ is first category in $(X,T)$ we mean that it is the union of (or is covered by) countably many sets which are nowhere dense in $(X,T)$. $A$ is second category iff it is not first category.
Does $A$ being first category in $(X,T)$ imply that $A$ is first category in the subspace $(A,T_A)$, or vice versa?
Does $A$ being second category in $(X,T)$ imply that $A$ is second category in the subspace $(A,T_A)$, or vice versa?
I could not get anywhere in the proof because if $B\subseteq(X,T)$, for closure $\overline{B_A}\subseteq\overline{B}$ while for interior $B^\circ\cap A\subseteq B_A^\circ$ where $B_A=B\cap A$ as a set in $(A,T_A)$. So I think there should be a counterexample.
Let $C$ be the middle-thirds Cantor set in $\Bbb R$; $C$ is closed and nowhere dense in $\Bbb R$, so it’s a first category (or meagre) subset of $\Bbb R$. Considered as a space in its own right, however, it’s a Baire space, meaning that if $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a countable collection of dense open subsets of the space $C$, then $\bigcap\mathscr{U}$ is dense in $C$; in particular, $C$ is therefore not of first category in itself.