Baire Category Theorem

Question

This is Asaf's argument; (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior

Suppose that $(X,d)$ is a separable complete metric space, and $\{a_k\mid k\in\omega\}=D\subseteq X$ is a countable dense subset.

By contradiction assume that we can write $X=\bigcup F_n$ where $F_n$ are closed and with empty interior, we can further assume that $F_n\subseteq F_{n+1}$.

Define by recursion the following sequence:

1. $x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$;
2. $r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.
3. $x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;
4. $r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.

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I understand his proof till here, ($x_n,r_n$ are well-defined recursively) Here, I don't know how to show that $\{x_n\}$ is a Cauchy Sequence and don't understand the rest of the proof..

$\{r_n\}$ is not monotonic.

This is the first time im facing with somewhat recursion and sequence are mixed.. and i really want to understand this. Please explain the rest of the proof in detail..

I think 'understanding different proofs' is 'learning techiniques'. I don't insist on this proof, but do want to learn this technique.

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Latter part of the argument;

Note that $x_n$ is a Cauchy sequence, therefore it converges to a point $x$. If $x\in F_n$ for some $n$, first note that $d(x_k,F_n)\leq d(x_k,x)$, by the definition of a distance from a closed set.

If so, for some $k$ we have that $d(x,x_k)<r_n$, in particular $d(F_n,x_k)<r_n$. First we conclude that $n<k$, otherwise $d(F_n,x_n)>r_n$. Now we note that:

$$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$$

It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.

Answer 1

You can change the set-up slightly so that in the recursive choices we instead take

• [4.] $r_{n+1} = 2^{-k}$ where $k \in \mathbb{N}$ is least such that (a) $2^{-k} < r_n$; (b) $2^{-k} < r_n - d( x_{n+1}, x_n )$; and (c) $d ( F_{n+1} , x_{n+1} ) > k^{-1}$.

(Note that since the $r_n$ are all of the form $2^{-k}$, then (a) implies that that $r_{n+1} \leq \frac{r_n}{2}$; (b) will imply that $B (x_{n+1},r_{n+1}) \subseteq B ( x_n,r_n )$, and therefore for $m < n$ we have that $d ( x_m , x_n ) \leq r_m$.)

Induction should show that $r_n \leq 2^{-n}$, and as $x_{n+1} \in B ( x_n, r_{n} )$ we have that $d ( x_{n+1} , x_n ) < r_n \leq 2^{-n}$. From here you should be able to show that the sequence $(x_n)_n$ is Cauchy.

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The basic idea is that since $(x_k)_k$ is Cauchy, it converges to some $x$. We now want to show that $x \notin \bigcup F_n$. So assume that $x \in F_n$ for some $n$.

As $( x_k )_k$ converges to $x$, there must be a $K$ such that $d ( x_k , x ) < r_n$ for all $k \geq K$. Take $k = \max ( K , n + 1 )$. Note that we also have that $d( x_k , F_n ) \leq d ( x_k , x ) < r_n$.

We'll derive a contradiction by estimating $$d ( F_n , x_n ) \leq d ( x , x_n ) \leq d ( x , x_k ) + d ( x_k , x_n ).$$

• From the above we have $d ( x_k , x ) < r_n$.
• Since $k > n$, by an observation above we have $d ( x_k , x_n ) \leq r_n$.

Therefore $d(F_n,x_n) \leq 2 r_n$. However it is relatively easy to show that $2 r_n \leq n^{-1} < d ( F_n , x_n )$, as stipulated in the recursive choices!