Let $S$ be an uncountable subset of $[0,1]$ such that:


1. $S$ is dense in $[0,1]$;
2. as a topological space, $S$ is [Baire](https://en.wikipedia.org/wiki/Baire_space).



> 
> Is it true that $S$ is of second category as a subset of $[0,1]$?
> 
> 
> 



Is this subset of $[0,1]$ of second category?
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 $X$.


