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In Rudin's Real and Complex Analysis, Theorem 5.6 (i.e., Baire's Theorem, page 97) states that

If X is a complete metric space, then the intersection of every countable collection of dense open subsets of X is dense in X.(1)

After the proof, he says that Theorem 5.6 is equivalent to the statement:

No (non empty) complete metric space is of the first category.(2)

And he points out that to see this, only need to take complements in the statement of Theorem 5.6. But I do not see how that works, by taking complements in the statement of Theorem 5.6, I can only have that Theorem 5.6 is equivalent to the statement:

If X is a complete metric space, $\{A_i\}$ is a countable collection of nowhere dense closed subsets of X, then int$(\cup A_i)=\emptyset$.(3)

(Or see corollary to baire category theorem)

Of course, (3) or (1) implies (2), but I don't know how to prove that (2) implies (1) or (3).

I also know that Theorem 5.6 is equivalent to the statement:

If X is a complete metric space, then every non empty open subset of X is of second category.(4)

(See About the equivalence of definitions of a Baire Space)

But I do not see this would be helpful.

How to prove that (2) implies (1) or (3) or (4)?

Any help would be appreciated.

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I think your issue is the following:

A space $X$ that only satisfies (2) (i.e. it is not of first category) ,cannot be shown to be Baire (the countable intersection of dense open sets is dense) alone using those properties.

E.g. $X= [0,1] \cup ([2,3]\cap\mathbb{Q})$ is not of first category (or $[0,1]$ would be, contradiction), but the countable family of open dense sets $O_q = [0,1] \cup ([2,3]\cap \mathbb{Q})\setminus\{q\}$, for $q \in \mathbb{Q} \cap [2,3]$ has intersection $[0,1]$ which is not dense. Note that this space also does not satisfy (3), as witnessed by the sets $\{q\}, q \in [2,3]\cap\mathbb{Q}$, whose union is open in $X$.

In fact you need all open sets to be of second category (4) (Not just $X$) to have true equivalence on a space to space basis.

So for full equivalence we need to consider the statement quantified over all complete metric spaces, and then Nate's argument is a fine way to use that we know stuff for all complete metric spaces.

So for all complete metric spaces (2), implies for all complete metric spaces (1). But a metric space with (2), does not need to obey (1) (Baire).

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  • $\begingroup$ Yes, that's exactly what I want! Thanks! $\endgroup$ – DL Qian May 3 '17 at 11:47
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By taking complements, (2) is equivalent to (2'): "If $X$ is a complete metric space, then the intersection of a countable collection of dense open subsets of $X$ is not empty." We want to upgrade "not empty" to "dense".

Suppose (2') holds. Let $X$ be a complete metric space and let $U_n$ be a countable collection of open dense subsets of $X$. Set $G = \bigcap_n U_n$.

Let $B(x,r)$ be an arbitrary open ball in $X$. Let $E = \overline{B(x,r/2)}$ (where this means the closure of the ball of radius $r/2$, which is not necessarily equal to the closed ball of radius $r/2$). Verify that $E \subset B(x,r)$. Now $E$ is a complete metric space, and $U_n \cap E$ is open in $E$ (by definition of the subspace topology). Prove that $U_n \cap E$ is dense in $E$ (exercise). Now applying (2') on $E$, we conclude that $\bigcup_n (U_n \cap E) = G \cap E$ is nonempty. We have thus shown that $B(x,r)$ contains a point of $G$. Since the ball was arbitrary, $G$ must be dense in $X$.

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