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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.

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

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  • $\begingroup$ Every relatively open set has the form $U\cap S$, where $U$ is open. $\endgroup$
    – Asaf Karagila
    Commented Dec 1, 2020 at 17:19

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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 $X$.

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  • $\begingroup$ Thanks, I delete the comments that are not relevant anymore. $\endgroup$ Commented Dec 1, 2020 at 17:47

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