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Let X be topological space. The subset B is meager, not open and non-empty. How can you prove that the complement is dense?

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This is not true in general. For example if $X=\mathbb Q$ (with the topology inherited from the real line) then every subset of $X$ is countable, and thus meagre. To get a counterexample to your claim, just take an arbitrary subset of $X$, which is not open, and has non-empty interior. For example, take $B=(-\infty,0]\cap\mathbb Q$.


However, the claim is true if $X$ is a Baire space. Baire spaces are spaces in which Baire Category Theorem holds. For example, every complete metric space and every locally compact Hausdorff space is a Baire space.

Indeed, suppose that $X$ is a Baire space and $B$ is meagre. Then $B=\bigcup_{i=1}^\infty B_i$, where each $A_i$ is nowhere-dense. From this we get $B\subseteq \bigcup_{i=1}^\infty \overline{B_i}$, where $\operatorname{Int\overline{B_i}}=\emptyset$.

Now we get for the complement $$X\setminus B\supset \bigcap_{i=1}^\infty (X\setminus\overline{B_i}).$$ Each of the sets $X\setminus\overline{B_i}$ is dense in $X$ (since $\operatorname{Int\overline{B_i}}=\emptyset$). In a Baire space, intersection of countably many dense open sets is dense. Therefore $\bigcap\limits_{i=1}^\infty (X\setminus\overline{B_i})$ and, consequently, $X\setminus B$ is dense.

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  • $\begingroup$ In the first version of my post I was missing the condition that $B$ should have non-empty interior. I have edited the post after this was pointed out in a (now deleted) comment. $\endgroup$ – Martin Sleziak May 3 '14 at 11:09

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