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So I know how to prove that the complement of a set of first category is of second category. But, I have difficulty showing that it is dense too. Let $X$ be a complete metric space and $A$ a set of first category in $X$.

Then $A = \bigcup_{n=1}^\infty U_n$ such that $int\, \overline U_n =\emptyset, \forall n.$ This means that for each $n$, $X \setminus \overline U_n$ is both open and dense in $X$. So I should connect it to Baire's theorem from here. That is, if I could write the complement ($X\setminus A$) as the intersection of such a family of open dense sets, then it would be dense as well. Now, $$X\setminus A = X\setminus \bigcup_{n=1}^\infty U_n = \bigcap_{n=1}^\infty X\setminus U_n. $$ But I actually needed the term $\bigcap_{n=1}^\infty X\setminus \overline U_n$ here to use Baire's theorem. What am I missing here? I'd appreciate any help or hint.

Edit: May I argue that $(\bigcap_{n=1}^\infty X\setminus \overline U_n) \subseteq (X\setminus \bigcup_{n=1}^\infty U_n)$, and, because $\bigcap_{n=1}^\infty X\setminus \overline U_n$ is dense in $X$, its superset, $X\setminus \bigcup_{n=1}^\infty U_n$, is dense too?

Neutrino
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    Yes, you may. Why do you doubt that argument? – Kavi Rama Murthy Jun 09 '21 at 05:07
  • @KaviRamaMurthy Thanks. I was wondering maybe there is a better (more elegant) way of proving the statement (?) – Neutrino Jun 09 '21 at 05:09
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    I was wondering maybe there is a better (more elegant) way of proving the statement --- Whether better/elegant is in the eye of the beholder, but since you're in a metric space, you can work more concretely with open balls in the complement of the set (see this answer) instead of the seemingly more abstract notion of a topological closure operator, which allows for a proof of what you want that is almost the same as the nested interval proof of the uncountability of the reals (get denseness by starting in an arbitrary nonempty open ball). – Dave L. Renfro Jun 09 '21 at 07:32

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Actually much more is true, you call residual or co-meager a set that is the complement of a first category (also called meager) set.

What's true is that a set is co-meager $\iff$ it contains a dense $G_\delta$.

I'll sketch the proof, you can fill the details:
If $A$ is co-meager then $A^c=\cup_{n\in\mathbb{N}}A_n$, with $A_n$ nowhere dense.
Thus as $int(\overline{A_n})=\varnothing$ for each $n\in\mathbb{N}$, then $int(\cup_{n\in\mathbb{N}}\overline{A_n})=\varnothing$ (this is actually equivalent to being a Baire space).
$int(\cup_{n\in\mathbb{N}}\overline{A_n})=\varnothing$ $\iff$ the complement of ${\cup_{n\in\mathbb{N}}\overline{A_n}}$ is dense.
Conclude that this is a $G_\delta$ contained in $A$.
On the other hand suppose that $A$ contains a dense $G_\delta$ then it's complement is contained in a union of closed set say $A^c\subseteq\cup_n K_n$ and $int(U_n K_n)=\varnothing$, this implies easily that $A^c$ is meager.
Hope it helps.

Speripro
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