# Confused by baby Rudin chapter 3.22

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 $$0,and choose $$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$$

• can somebody take a look Mar 5, 2019 at 0:18
• Thanks a lot for your advice Mar 5, 2019 at 0:18
• math.stackexchange.com/questions/221423/… Mar 5, 2019 at 17:23

## 1 Answer

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.