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:
- $x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$;
- $r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.
- $x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;
- $r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.
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.
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)
$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$
It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.