Is a comeager set also comeager with respect to any closed set containing it?

Question

Let $X$ be a topological space. Let $Y,Z$ be two subsets of $X$. If $Z$ is comeager in $\overline Y$ and $Z\subseteq Y$ then $Z$ is comeager in $Y$ too?

Answer 1

Since $Z$ is comeagre in $\operatorname{cl}Y$, $(\operatorname{cl}Y)\setminus Z=\bigcup_{n\in\Bbb N}A_n$, where each $A_n$ is nowhere dense in $\operatorname{cl}Y$. For $n\in\Bbb N$ let $G_n=(\operatorname{cl}Y)\setminus\operatorname{cl}A_n$; each $G_n$ is a dense open set in the relative topology that $\operatorname{cl}Y$ inherits from $X$. For $n\in\Bbb N$ let $U_n=G_n\cap Y$; each $U_n$ is open in the relative topology that $Y$ inherits from $\operatorname{cl}Y$, which is the same as the relative topology that it inherits from $X$. Moreover, each $U_n$ is dense in $Y$. (Why?) Thus, $\bigcap_{n\in\Bbb N}U_n$ is comeagre in $Y$. But

$$\begin{align*} \bigcap_{n\in\Bbb N}U_n&=Y\cap\bigcap_{n\in\Bbb N}G_n\\ &=Y\cap\bigcap_{n\in\Bbb N}\big((\operatorname{cl}Y)\setminus\operatorname{cl}A_n\big)\\ &=Y\cap\left((\operatorname{cl}Y)\setminus\bigcup_{n\in\Bbb N}\operatorname{cl}A_n\right)\\ &=Y\setminus\bigcup_{n\in\Bbb N}\operatorname{cl}A_n\\ &\subseteq Y\setminus\bigcup_{n\in\Bbb N}A_n\\ &=Y\cap Z\\ &=Z \end{align*}$$

and a superset of a comeagre set is comeagre, so $Z$ is comeagre in $Y$ as well.

Answer 2

Yes!

By definition, Z is the intersection of countably many open dense sets, $Z=\bigcap_n Z_n$. Be careful that $Z_n$ are open in the (subspace) topology on $\bar{Y}$ inherited by $X$.

Now consider $Z'_n=Z_n\cap Y$. Since $Z\subset Y$ and $Z=\bigcap_n Z_n$, it follows thatr $Z=\bigcap_n Z'_n$. Also, the $Z_n'$ are open in the (subspace) topology on $Y$ and they are dense in $Y$, because the $Z_n$ are dense in $\bar{Y}$.