The present problem is realted to the following posting in MSE.
Recall that for a given a topological space $(X,\tau)$, a set $E\subset X$ is nowhere dense in $X$ if $\operatorname{Int}(\overline{E})=\emptyset$. A subset of $X$ that can be written as a countable union of nowhere dense sets in $X$ is said to be a set of first category. The space $(X,\tau)$ is said to be of first category if $X$, as a subset of itself, is set of first category.
A set $A$ may be of first category in some space $(X,\tau)$ but not a subspace of first category. For example, $\mathbb{Z}\subset\mathbb{R}$ is a set of first category in $\mathbb{R}$ (the latter endowed with the usual Euclidean topology) but, $\mathbb{Z}$ as a subspace of $\mathbb{R}$ is not of first category, for the topology on $\mathbb{Z}$ inherited by that on $\mathbb{R}$ is the discrete topology in which every subset of $\mathbb{Z}$ is both open and closed.
The question that I have is: Suppose $A$ is considered as a subspace of $X$, that is $A$ is endowed with the topology $\tau_A$ described above. If $(A,\tau_A)$ is a space of first category, is $A$ a set of first category in $(X,\tau)$?
- At first this may appear as a problem of semantics, but I think it is more nuanced than that.
- I think that in general, the answer to my question is no, but I have not found any counter-example yet.
Edit: following the reasoning of @Ruy, it turns out my guess was wrong, and that in fact $A$ is a set of first category in $X$.
Thanks for the interest and the answer!
