Suppose $X$ is a nonempty complete metric space, and $\left\{ G_n \right\}$ is a sequence of dense open subsets of $X$. Prove Baire's Theorem, namely that $\cap_1^\infty G_n$ is not empty. (In fact, it is dense in $X$.) Hint: Find a shrinking sequence of neighborhoods $E_n$ such that $\overline{E_n} \subset G_n$, and apply Exercise 21.

Here's Prob. 21, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Prove the following analogue of Theorem 3.10 (b): If $\left\{ E_n \right\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_n \supset E_{n+1}$, and if $$ \lim_{n \to \infty} \mathrm{diam} \ E_n = 0,$$ then $\cap_1^\infty E_n$ consists of exactly one point.

I want to check my proof whether right or not? Proof

Let $F_n$ be the complement of $G_n$,for any nonempty open set U ,since $F_n$ doesn’t contains any open interval .(if L$\subset F_n,$ then there exist $O_{r_1}(y)\subset L\subset F_n$,then $G_n$ is not a dense set).so I can choose $x_1\in U-F_1$,there exist $E_1=O_{r_1}(x_1)\subset U-F_1$,let $0<r_n<\frac{1}{n}$,and choose $x_2 \in E_1-F_2$, $E_2’=O_{r_2}(x_2)\subset E_1$,and let $E_2=\overline O_{\frac{r_2}{2}}(x_2),E_2\subset \overline E_1$…we can do this over and over ,and use exercise 21,$E=\bigcap_{n=1}^{+\infty}E_n$nonempty, ,so there exist y$\in U-(F_1\cup F_2…)$,so y $\in \cap_1^\infty G_n$


1 Answer 1


Your solution seems correct. Although I have some minor remarks. The existence of the set $E_1$ is granted because $G_1 \cap U$ is an open set. This is not so clear from the way you wrote it. If $x_1 \in U \cap F_{1}^{C}$ it just means that $x_1 \in U \cap G_1$. As it is an open set, the existence of $E_1$ follows from definition. Same follows for the last bit of you solution.


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