Vtyjkyuk

I have written an answer for the problem 20, chapter 3 of Walter Rudin's Principle of Mathematical Analysis. I think the proof is correct, but since I am new with this kind of proofs, I am skeptical about the subtle points. So it would be really helpful to me if someone can comment on my proof, and maybe give some hints if some error is found. Any help is appreciated. Here is my proof along with the original question and hint:

Exercise 22: Suppose $X$ is a nonempty complete metric space, and $G_n_nge 1$ be a sequence of open dense subsets of $X$. Prove Baire's theorem, namely, that $ bigcap_n=1^inftyG_n$ is nonempty. (In fact, it is dense in $X$). Hint: Find a shrinking sequence of neighborhoods $E_n$, such that $overlineE_nsubset G_n$ and then apply exercise $21$.

Exercise $21$ asks to prove that it $E_n$ is a decreasing sequence of closed , nonempty and bounded sets in a complete metric space with $lim_nto inftymboxdiam E_n=0$, then $bigcap_1^infty E_n$ contains exactly one point.

I have proven exercise 21 and I know that that proof is okay. So, I will show here how I constructed the sequence of $E_n$ for exercise $22$. Also I have tried to prove that $bigcap_1^infty G_n$ is in fact dense in $X$.

Let $xin X$. Let $B_epsilon(x)$ be an open ball around $x$.

Claim 1: $forall n$, $B_epsilon(x)cap G_nne emptyset$

Proof: If $exists n$ such that $xin G_n$ then it is obvious. Otherwise, as $G_n$ is dense in $X$, $x$ is a limit point of $G_n$ which implies the result.$blacksquare$

Now, since each $G_n$ is open and so is $B_epsilon(x)$, $B_epsilon(x)cap G_n$ is open $forall n$. Now, let $p_1in B_epsilon(x)cap G_n$. Then, $exists delta>0$ such that $B_delta(p_1)subset B_epsilon(x)cap G_1$. Choose $0<delta_1<delta$ and set $E_1=B_delta_1(p_1)$.

Claim 2: $overlineE_1subset B_delta(p_1)$

Proof: Clearly, $E_1subset B_delta(p_1)$. Now, let $q$ be a limit point of $E_1$. Then, $forall epsilon'>0, exists p'in E_1$ such that $d(p',q)<epsilon'implies forall epsilon'>0, d(p_1,q)le d(p_1,p')+d(p',q)<delta_1+epsilon'implies d(p_1,q)le delta_1<delta$ from which the claim follows.$blacksquare$

Hence it follows that $overlineE_1subset B_epsilon(x)cap G_1$.

Let us assume that we have constructed $E_k$ such that $E_1supset E_2supsetcdotssupset E_k$ and $overlineE_isubset B_epsilon(x)cap G_i, i=1,2,cdots, k$. Now, $E_kcap G_k+1ne emptyset$ and is open by construction. Let $p_k+1in E_kcap G_k+1$, then, $exists delta'>0$ such that $B_delta'(p_k+1)subset E_kcap G_k+1$. Then, choose $0<delta_k+1<delta'$ and set $E_k+1=B_delta_k+1(p_k+1)subset E_kcap G_k+1subset B_epsilon(x)cap G_k+1$, so that $overlineE_k+1subset B_epsilon(x)cap G_k+1$. Then, from the Exercise $21$ of the book, it follows that $bigcap_1^inftyG_n$ is dense in $X$.$blacksquare$

Your proof is correct.

I would perhaps make the conclusion a bit more explicit. You skip a few steps. They are simple, but an inexperienced reader might be confused. For example, Exercise 21 shows that there exists some point $yin bigcap overline E_k$, so there exists some point $yin bigcap G_n$ such that $yin B_epsilon(x)$. Because $epsilon$ is arbitrary, every open ball around $x$ contains a point of the intersection $bigcap G_n$. Since $x$ is arbitrary, this shows $bigcap G_n$ is dense.