Meagerness on an open subset with the relative topology (Kechris)

Question

I'm still working on meagerness, and what I post is something that I had achieved yesterday but today I don't remember how I did that. At pp. $48$, Kechris (Classical Descriptive Set Theory) defines, for a topological space $X$, a meager set $A\subseteq X$ in an open subset $U\subseteq X$ to be a subset s.t. $A\cap U$ is meager in $X$.

And he says that it is equivalent to require that $A\cap U$ be meager in $U$ w.r.t. the relative topology.

It's clear that meagerness in $X$ implies meagerness in $U$ (essentially by monotonicity of inclusion), but I have some problems to prove the converse. Indeed, meager in $U$ means that there is $F_n$ s.t. $$\emptyset=\mathrm{Int_U}[\mathrm{Cl_U}(F_n\cap U)]=(\overline{F_n\cap U})^°\cap U,$$ from which I deduce that either $(\overline{F_n\cap U})^°=\emptyset$ or they are disjoint.

Well, if I can prove that disjointness is impossibile, I'm done, but I'm not sure that this holds in general.

Notation: $\overline{(.)}$ is the closure operator on $X$, while $(.)^°$ is the interior operator on $X$.

Answer 1

If $E \subset U$ and if the closure of $E$ in $U$ has empty interior in $U$ then the closure of $E$ in $X$ has empty interior in $X$. This is because If $V$ is a non-empty open set in $X$ contained in the $X-$ closure of $E$ then $V\cap U$ cannot be empty: $V$ has to intersect $E$ (since any point of $V$ is in the $X-$ closure of $E$, so the neighborhood $V$ of that point has to intersect $E$) and $E \subset U$ so $E$ intersects $U$. Thus $V \cap U$ is a non-empty open set in $U$ contained in the closure of $E$ in $U$, contradicting the hypothesis. Taking $E=A\cap U$ shows that if $A\cap U$ is meager in $X$ the it is meager in $U$.