Exercise 3.22

Question

Exercise 22:
    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 $\cup_1^\infty G_n$ is not
    empty. (In fact, it is dense in $X$.)

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon}

Since $G_1$ is open, we can find a neighborhood $E_1$ such that $\bar E_1 \subset G_1$,
    and $\operatorname{diam}E_1 < 1$ by letting $E_1 = N_\epsilon(x)$ for a sufficiently small $\epsilon$ around $x \in G_1$.

Assume $E_n$ is open, that $\bar E_n \subset G_n$, that $\bar E_{n} \subset \bar E_{n-1}$
    and that $\operatorname{diam} E_n < \frac 1 n$. Since $G_{n+1}$ is dense in $X$ and open,
    $E_n \cap G_{n+1}$ is non-empty and open. We can therefore choose a neighborhood 
    $E_{n+1}$ in the same manner as above such that $\operatorname{diam} E_{n+1} < \frac 1 {n+1}$,
    $\bar E_{n+1} \subset G_{n+1}$ and $\bar E_{n+1} \subset \bar E_n$. 
    This gives us a sequence of sets with the desired properties.

By ex. 21, $\bigcap \bar E_n$ contains a point $x$. Since $x \in G_n$ for all $n$,
    $\bigcap E_n$ is non-empty as well, which is what we wanted to show.

Amit Awasthi · Answer

\begin{proof}
      We will construct a sequence $\{V_n\}$ of neighborhoods as follows. Let $V_1$ be any neighborhood of $x_1 \in G_0 \bigcap G_1$, which is possible since $G_0$ is a dense open subset of $X$ and $G_1$ is nonempty and open; their intersection is nonempty and also open, so we can construct a neighborhood inside of it. Suppose $V_n$ has been constructed such that $\overline{V}_n \subset \overline{V}_{n-1} \bigcap G_n$. Then, there is a neighborhood $V_{n+1}$ such that $\overline{V}_{n+1} \subset V_n \bigcap G_{n+1}$, an our induction can proceed. Then, if we take the sequence $\{\overline{V}_n\}$, we know that $\bigcap^\infty_1 \overline{V}_n$ contains a point $y$ by Exercise ef{3.21} since $\overline{V}_n$ is closed and bounded for all $n$ and $X$ is complete. Since $\overline{V}_n \subset G_n$, we know that $y \in \bigcap^\infty_1 G_n$ by our construction and so $\bigcap^\infty_1 G_n 
eq \emptyset$.
    \end{proof}

Exercise 3.22

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Exercise 3.22

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