Let $S$ be an uncountable subset of $[0,1]$ such that:
$S$ is dense in $[0,1]$;
as a topological space, $S$ is Baire.
Is it true that $S$ is of second category as a subset of $[0,1]$?
Let $S$ be an uncountable subset of $[0,1]$ such that:
$S$ is dense in $[0,1]$;
as a topological space, $S$ is Baire.
Is it true that $S$ is of second category as a subset of $[0,1]$?
The answer is yes.
Lemma. Suppose $X$ is a topological space and $S$ is dense in $X$. If $U$ is open and dense in $X$, then $U \cap S$ is open and dense in the relative topology on $S$.
Proof. $U \cap S$ is open in $S$ (i.e. is an open set in the relative topology of $S$) by definition of the relative topology. Suppose $V \subset S$ is open in $S$ and nonempty. Then $V$ is of the form $V = W \cap S$ for some (nonempty) open $W \subset X$. Now since $U$ is open and dense, $W \cap U$ is a nonempty open set in $X$, so it intersects the dense set $S$. Thus $S \cap (W \cap U) = V \cap (U \cap S)$ is nonempty. Since $V$ was arbitrary, $U \cap S$ is dense in $S$.
Now to your question. Let $U_n$ be any sequence of dense open sets in $X = [0,1]$. Then each $U_n \cap S$ is dense and open in $S$. Since $S$ is Baire, we have $S \cap \bigcap_n U_n = \bigcap_n (U_n \cap S) \ne \emptyset$. So $S$ intersects every countable intersection of dense open sets, and is therefore second category in $X$.