Definition: Let $U\subseteq \mathbb{R}$ open. We say $A\subseteq \mathbb{R}$ is categorically dense in $U$ iff $A \cap V$ is of second category for every nonempty open $V\subseteq U$
My question arises whether the following is true:
If $A$ is of second category, then there exists some nonempty open interval $(a,b)$ such that $A$ is categorically dense in (a,b)
I believe yes, so I'm trying by contradiction, but got that $A \cap (c,d)$ must be of first category for some open interval $(c,d)\subseteq (a,b)$ (and this holds for every open interval $(a,b)$)
Not really sure how to continue. Any hint? Or a counterexample?
Let $A$ be 2nd category in $\Bbb R.$ Let $B$ be the set of non-empty bounded open intervals with rational end-points. Let $C$ be the set of those $b\in B$ for which $A\cap b$ is 1st category in $\Bbb R.$ Let $D=\cup \{b\cap A: b\in C\}.$
Then $D$ is 1st category in $\Bbb R.$ So $E=A\setminus D$ is 2nd category in $\Bbb R.$
So $E$ must be dense in some $U\in B.$
Now for any non-empty open $V\subset U,$ if $V\cap A$ were 1st category in $U$ then $V\cap A$ would be 1st category in $\Bbb R,$ so $A\cap b$ would be 1st category in $\Bbb R$ for any open $b\subset V$, so $$V\cap A=\cup \{b\cap A: V\supset b\in B\}\subset \cup C= D.$$ But then $V\cap E=V\cap (A\setminus D)=\emptyset,$ in contradiction to the denseness of $E$ in $U$.