Exercise $3.22$ of baby Rudin is to prove the Baire's theorem:
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 $\cap_{n=1}^{\infty}G_n$ is not empty. (In fact, it is dense in $X$.) Hint: Find a shrinking sequence of neighborhoods $E_n$ such that $\overline{E_n}\in G_n$, and apply Exercise $21$.
And the result of Exercise $21$ is:
If $\{E_n\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_n\supset E_{n+1}$, and if $$\lim_{n\to\infty}\operatorname{diam} E_n=0,$$ then $\cap_{n=1}^{\infty}E_n$ consists of exactly one point.
My try:
Let $A\subset X\implies \exists g_1\in G_1$ such that $g_1\in A$, since $G_1$ is dense in $X$. ($A\not=\varnothing$, open, bounded.)
Take open nbhd $E_1$ of $g_1$ such that $g_1\in E_1\subset G_1\cap A$ with $$\operatorname{diam}E_1=\frac{1}{2}\operatorname{diam}(G_1\cap A)\leq \frac{1}{2}\operatorname{diam}A.$$
Since $E_1$ is open and $G_2$ is dense in $X$, $\exists g_2\in G_2$ such that $g_2\in E_1.$
Take open nbhd $E_2$ of $g_2$ such that $g_2\in E_2\subset G_2\cap E_1$ with
$$\operatorname{diam}E_2=\frac{1}{2}\operatorname{diam}(G_2\cap E_1)\leq \frac{1}{2}\operatorname{diam}E_1.$$
Continuing this, we get a sequence of closed nonempty bounded sets $\{\overline{E_n}\}$, such that $\overline{E_n}\supset \overline{E_{n+1}}$, and
$$\operatorname{diam}E_{n+1}\leq \left(\frac{1}{2}\right)^n\operatorname{diam}A<\infty,$$
so $$\lim_{n\to\infty}\operatorname{diam}\overline{E_n}=0.$$
Now we apply the result of Exercise $21$ and get $\varnothing\not=\cap\overline{E_n}\subset \cap G_n$.
Since this is true for all (bounded) nonempty open subset $A$ of $X$, $\cap G_n$ dense in $X$.
Is this correct? Could not-covering the unbounded cases cause a problem? Or elsewhere contains an error..?
Best Answer
Your idea is correct but you used the fact that $\overset {-} {E_N} \subset G_n$ at the end without justification. When you choose your neighborhoods $E_1,E_2,...$ choose them to be open balls whose closures are contained in $G_1 \cap A, G_1\cap E_1,...$. This is possible because the closure of $B(x,r)$ is contained in $B(x,r')$ whenever $r <r'$.