Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following is problem 3.22 from Rudin's Princples of Mathematical Analysis:

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 $\bigcap_{n=1}^\infty G_n$ is not empty. Hint: find a shrinking sequence of neighbourhoods $E_n$ such that $\overline{E}_n\subset G_n$.

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 $i$, $\overline{E}_i\subset G_i$ (I had convinced myself that this would be true but I am now having doubts). Let $e\in \overline{E}_i$. 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?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer

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 $x \in \overline{E_i} \setminus G_i$.

In order to prove this, note the following facts.

  • If $G$ is a dense open set, and $U$ is any nonempty open set, then $G \cap U$ is a nonempty open set.
  • The intersection of finitely many dense open sets is itself dense open.

Using the above, we can construct a sequence $\langle x_n \rangle_n$ in $X$ such that for each $n$ there is a $\delta_n$ with $0 < \delta_n \leq 2^{-n}$ such that

  1. $\overline{B}(x_n;\delta_n) = \{ x \in X : d(x,x_n) \leq \delta_n \} \subseteq G_n$;
  2. if $m > n$, then $B(x_m;\delta_m) \subseteq B(x_n;\delta_n)$.

To recursively construct such a sequence:

Suppose that $x_1 , \ldots , x_n$ have been appropriately chosen (with associated positive reals $\delta_1 , \ldots , \delta_n$). As $B ( x_n , \delta_n ) \cap G_{n+1}$ is a nonempty open set we may pick some $x_{n+1} \in B ( x_n , \delta_n ) \cap G_{n+1}$. Then there is a $\varepsilon > 0$ such that $B ( x_{n+1} , \varepsilon ) \subseteq B ( x_n , \delta_n ) \cap G_{n+1}$, so set $\delta_{n+1} = \min \{ 2^{-(n+1)} , \frac{\varepsilon}{2} \}$.

To see how such a sequence proves the result:

Since for $m > n$ we have that $d(x_n,x_m) < \delta_n \leq 2^{-n}$ it follows that such a sequence (if constructed) must be Cauchy, and so has a limit, $x$. Furthermore for each $n$ as the tail $\langle x_k \rangle_{k=n}^\infty$ of the sequence is contained in $B ( x_n , \delta_n )$, then $x \in \overline{B ( x_n , \delta_n )} \subseteq \overline{B} ( x_n , \delta_n ) \subseteq G_n$. Therefore $x \in \bigcap_n G_n$.

As for the nature of dense open subsets complete metric spaces, note that for the real line $\mathbb{R}$ the following are examples 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 in $\mathbb{R}$ a dense open set an open set (so a union of open intervals) whose complement includes no "non-degenerate intervals" (intervals of non-zero length).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.