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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.

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 $d(x,x_k)<r_n$, 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 $x_n$.

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With a minor change the $r_{n+1}$ can be chosen to be also strictly decreasing, which would easily imply that the sequence in Cauchy. –  Arthur Fischer Aug 13 '12 at 19:37
    
@Arthur Can you please explain in detail? I don't know about you, but it's extremely hard for me..actually –  Katlus Aug 13 '12 at 19:39
    
By the way, this works also for non-separable spaces. And you could try to show the complement of this statement which says that a countable intersection of open dense sets is dense. It's quite simple, too. You begin with an arbitrary open ball and start choosing closed balls inside it inductively so that the radius goes to zero and the $n$:th step ball is a subset of $n$ first members of the dense open sets. The intersection of these closed balls is a singleton by completeness: it belongs to the intersection of the open dense sets and our original ball, proving the claim. –  Thomas E. Aug 13 '12 at 19:55
    
@Thomas: The context here is without the axiom of choice. You need a well-orderable dense set to apply this sort of argument. –  Asaf Karagila Aug 13 '12 at 20:01
1  
@AsafKaragila. True. The ZF part was so well hidden in the link that I totally missed it. –  Thomas E. Aug 13 '12 at 20:04

1 Answer 1

up vote 1 down vote accepted

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.


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!

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How do in know there exists such k satisfies both a) and b) –  Katlus Aug 13 '12 at 20:33
    
@Katlus: Both (a) and (b) have the property that if any $k$ satisfies the condition, then any integer greater than $k$ also satisfies it. As both (a) and (b) can be individually satisfied, they can thus be jointly satisfied. –  Arthur Fischer Aug 13 '12 at 20:37
    
Got it. I think $F_n$ should be changed to $F_{n+1}$ in (3) then –  Katlus Aug 13 '12 at 20:42
    
I don't understand the next part too.. Help. What does $k$ designate?? (in the link) –  Katlus Aug 13 '12 at 21:06
    
@Katlus: What $k$ are you talking about? –  Arthur Fischer Aug 13 '12 at 21:14

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