# Baby Rudin problem 3.22: prove Baire's theorem. Am I going in a reasonable direction?

Suppose $X$ is a nonempty complete metric space, and $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove Baire's theorem, namely, that $\bigcap_1^\infty G_n$ is not empty. Hint: find a shrinking sequence of neighbourhoods $E_n$ such that $\bar{E}_n\subset G_n$ [$\bar{E}_n$ denotes the closure of $E_n$].

Here's what I've tried so far: let $\{r_n\}$ be a Cauchy sequence of positive real numbers converging to 0. Fix $x\in X$ and define $E_i=\{g\in G_i:d(g,x)<r_i\}$, which is nonempty since $G_i$ is dense. I would like to show that for all $i$, $\bar{E}_i\subset G_i$ (I had convinced myself that this would be true but I am now having doubts). Let $e\in \bar{E}_i$. Then either $e\in E_i$ or $e$ is a limit point of $E_i$. If $e\in E_i$ then $e\in G_i$. Otherwise, every neighbourhood of $e$ contains a point in $E_i$. I thought that I should be able to then choose some point $e'\in E_i$ in a neighbourhood of $e$ and, since $G_i$ is open, it'll have a neighbourhood $N\subset G_i$ which contains $e$, but this is proving to be difficult and I'm worried that it's not true. If I can show that this is true then the rest will follow from results I've already proven.

Does my approach make any sense?

Incidentally, as a secondary question, what type of a thing would $G_n$ be? A sequence of dense open subsets seems weird to me—at first I was thinking of some sequence of infinite subsets of rational numbers in the real numbers but I realized that those aren't open. Is there anything which would be familiar to my little undergrad brain which would be analogous to this problem?

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Do you want the $\{ r_n \}$ to be a sequence of positive reals converging to $0$? –  Arthur Fischer Oct 26 '12 at 8:58
@ArthurFischer yes. I suppose I should add that. –  crf Oct 26 '12 at 9:02

It would probably be easier to start as follows: Notice that the $G_i$ being dense and open guarantees that $G_1 \cap G_2 \neq \emptyset$. Now choose an $x$ in so that there is a ball $E_1$ in completely contained in the intersection. Shrinking the ball if necessary, you can assume that $\overline{E_1}$ is completely contained in the intersection. Now the intersection of $E_1$ with $G_3$ is non-empty and so you can choose some $\overline{E_2}$ completely contained in $E_1 \cap G_3$ and hence in $E_1$. If you go on like this, you will have a decreasing (with respect to containment) sequence of closed and bounded sets which has non-empty intersection.

Now how do you prove this last assertion? You can either use the theorem in chapter 2 on intersection of compact sets (notice the nested bit guarantees that the intersection of finitely many of them is non-empty) or you can go straight up from the definition of sequential compactness.

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Your approach won't work, since your $x$ might not belong to $G_i$, but then it would be a limit point of $E_i$, and so $x \in \overline{E_i} \setminus G_i$.

Hint 1: If $U$ is open and $G$ is dense-open, then $U \cap G$ is a nonempty open set (and so there must be an $x$ and an $\epsilon > 0$ such that $B ( x , \epsilon ) \subseteq G \cap U$).

Hint 2: Given $z \in X$ and $\delta > 0$, how do $\overline{ B ( z , \delta ) }$ and $\overline{B} ( z , \delta ) = \{ x \in X : d ( x , z ) \leq \delta \}$ compare?

As for the nature of dense open subsets of $\mathbb{R}$, note that the following types of sets would be of this kind:

• The complement of any finite set.
• The complement of the integers.
• The complement of any convergent sequence (including its limit point).
• The complement of the Cantor ternary set.
• If you enumerate the rational numbers as $\{ q_i : i \in \mathbb{N} \}$ and let $\{ \epsilon_i : i \in \mathbb{N} \}$ be any sequence of positive reals, then the set $\bigcup_i ( q_i - \epsilon_i , q_i + \epsilon_i )$.

Basically it is an open set (so a union of open intervals) whose complement includes no "non-degenerate intervals" (intervals of non-zero length).

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