Probs. 21 and 22 in Baby Rudin: Baire's Theorem

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.

And, here's Prob. 22, Chap 3:

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.

Finally, here's Theorem 3.10 :

(a) If $\overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$\mathrm{diam} \ \overline{E} = \mathrm{diam} \ E.$$

(b) If $K_n$ is a sequence of compact sets in $X$ such that $K_n \supset K_{n+1}$ ($n = 1, 2, 3, \ldots$) and if $$\lim_{n \to \infty} \mathrm{diam} \ K_n = 0,$$ then $\cap_1^\infty E_n$ consists of exactly one point.

Now here's my proof of the result stated in Prob. 21:

Let's take a point $x_n \in E_n$. Then $\left\{ x_n \right\}$ is a sequence in $X$. Let $\varepsilon > 0$ be given. Then since $$\lim_{n \to \infty} \mathrm{diam} \ E_n = 0,$$ we can find a natural number $N$ such that $$\mathrm{diam} \ E_n < \varepsilon \ \mbox{ for all natural numbers } n > N.$$ So, if $m$ and $n$ are natural numbers such that $n\geq m > N$, then since $E_m \supset E_n$, we note that both $x_m$ and $x_n$ belong to $E_m$ and therefore $$d \left( x_m, x_n \right) < \varepsilon.$$ This shows that $\left\{ x_n \right\}$ is a Cauchy sequence in $X$, which is a compleste metric space, implying the existence of a point $x \in X$ to which this sequence converges.

We now show that this point $x \in \cap_1^\infty E_n$. Let $k$ be any natural number. Then $x_n \in E_k$ for any natural number $n \geq k$ and the sequence $$\left\{ x_k, x_{k+1}, x_{k+2}, \ldots \right\}$$ is a sequence of points of $E_k$ which converges to $x$, and since $E_k$ is a closed set in $X$, this point $x \in E_k$ also. Thus $x$ is in every $E_k$ and hence in $\cap_1^\infty E_n$.

Now let $y$ be any other point of $X$. Then since $d(x, y) > 0$, we can find a natural number $N^\prime$ such that $$\mathrm{diam} \ E_n < \frac{ d(x,y)}{2} \ \mbox{ for any natural number } n > N^\prime.$$ Thus, $$x \in E_{N^\prime + 1}$$ and $$\mathrm{diam} \ E_{N^\prime + 1} < \frac{ d(x,y)}{2}.$$ So, for all $p \in E_{N^\prime + 1}$, we have $$d(x,p) \leq \mathrm{diam} \ E_{N^\prime + 1} < \frac{ d(x,y)}{2},$$ which shows that $y \not\in E_{N^\prime + 1}$ and hence $y \not\in \cap_1^\infty E_n$.

Is this solution correct?

Now here's my solution of Prob. 22:

Let $F_n = X- G_n$ for each $n$. Then, as each $G_n$ is a dense open subset of $X$, so each $F_n$ is closed and contains no nonempty open subset of $X$. Let $U$ be any nonempty open subset of $X$. Then there is a point $x_1 \in U-F_1$ and since $U-F_1 = U\cap G_1$ is open, we can find a real number $r_1 > 0$ such that $$N_{r_1}\left( x_1 \right) \subset U-F_1.$$ Here $$N_{r_1}\left(x_1 \right) = \left\{ \ p \in X \ \colon \ d\left( p, x_1 \right) < r_1 \ \right\}.$$ Let $$E_1 = \left\{ \ p \in X \ \colon \ d\left( p, x_1 \right) < \frac{r_1}{2} \ \right\}.$$ Then $$\overline{E_1} \subset N_{r_1}\left( x_1 \right) \subset U-F_1 = U \cap G_1 \subset G_1.$$

Now as $E_1$ is open in $X$, so $E_1$ is not contained in $F_2$; so we can find a point $x_2 \in E_1-F_2 = E_1 \cap G_2$, which is open, and so there is a real number $r_2 > 0$ such that $$N_{r_2}\left( x_2 \right) \subset E_1-F_2.$$ Let $$E_2 = N_{\frac{r_2}{2}}\left( x_2 \right).$$ Then we note that $$\overline{E_2} \subset N_{r_2}\left( x_2 \right) \subset E_1 - F_2 = E_1 \cap G_2 \subset E_1.$$ Suppose $E_n$ has been found. Then $E_n$ is not contained in $F_{n+1}$ and so we can find a point $x_{n+1} \in E_n - F_{n+1} = E_n - G_{n+1}$, which is open, and so we can find a real number $r_{n+1} > 0$ such that $$N_{r_{n+1}} \left( x_{n+1} \right) \subset E_n \cap G_{n+1}.$$
Let $$E_{n+1} = N_{\frac{ r_{n+1}}{2} } \left( x_{n+1} \right).$$ Then we see that $$\overline{E_{n+1}} \subset E_n \cap G_{n+1},$$ and thus we have a sequence $\left\{ \overline{ E_n } \right\}$ of nonempty, closed, and bounded subsets of $X$ such that $$\overline{E_n} \supset \overline{E_{n+1}},$$ $$\overline{E_n} \subset G_n,$$ and $$\mathrm{diam}\ \overline{ E_n} = \frac{r_n}{2}$$ for each $n$.

Am I right so far? If so, then what next?

• The first argument is correct. In the second argument you want to choose the numbers $r_n$ so that they converge to $0$; then you can apply the previous result. – Brian M. Scott Jan 3 '17 at 20:11
• A common notation for $\{p\in X: d(p,x)<r\}$ is $B_d(x,r).$ – DanielWainfleet Jan 4 '17 at 3:41
• For the Baire Category Theorem, observe that each positive $r_n$ can be arbitrarily small. So take $r_{n+1}<r_n/2$ so the diameters of the $\overline {E_n}$ converge to $0.$ – DanielWainfleet Jan 4 '17 at 3:46