# Metrizable space with countable dense subset

1. Nov 16, 2010

1. The problem statement, all variables and given/known data

Show that every metrizable space with a countable dense subset has a countable basis.

3. The attempt at a solution

Let A be a countable dense subset of the metric space (X, d). For any x in A, take the countable collection of open balls {B(x, 1/n) : n is a positive integer}. Since A is countable, the family (call it F) of all such open balls is countable. We claim that it is a basis for X.

First of all, F covers X. Since Cl(A) = X, for any x0 in X, there exists a sequence xn of points of A such that xn --> x. So, for any ε > 0, there exists a positive integer N such that for n >= N, xn is in the open ball B(x0, ε). But now, for any such xn, the open ball B(xn, ε) contains x0, so F indeed covers all of X.

Clearly if x is in the intersection of any two elements of F, there exists another element of F (i.e. some positive integer N) such that B(x, 1/N) is contained in the intersection.

2. Nov 16, 2010

### micromass

Staff Emeritus
All is good. But your not done. You still need to show that your basis generates the correct topology. All you've proven now is that it's a basis. But the topology could be coarser then the one you want...

3. Nov 16, 2010

OK, but this seems very obvious, doesn't it? (unless I'm missing something, of course)

In the metric topology, a set is open iff for every x in the set, there exists some open ball B(x, ε) contained in it. Take any integer N such that 1/N < ε. Then the open ball B(x, 1/N) is definitely contained in our open set.

Hence, our collection is a basis "almost" by hypothesis.

4. Nov 16, 2010

### micromass

Staff Emeritus
It's a little trickier then that... You don't know that x is in A. Thus B(x,1/N) is not necessairily in the basis.

5. Nov 16, 2010

Hm, why would x be in A? Haven't I shown that my family covers X? Since I have shown that for any x in X there is some open ball containing it, and by definition, there must be an open ball around x contained in that very open ball, right?

6. Nov 16, 2010

### micromass

Staff Emeritus
Hmm, I'm just saying that if you take an open set G, then for every x in that open set, you need to find a basiselement B such that $$x\in B\subseteq G$$. But if you take B=B(x,1/N), then this is not the correct choice since x doesnt have to A. And in that case B doesnt have to lie in our basis. So you'll need to find another candidate for B.

I hope I didnt misunderstood your post...

7. Nov 16, 2010

Hm, OK, I think I get it now.

Let U be an open subset of X, and let x be in U. Out family F is a cover for X, so take an element of the form B(x0, 1/n) containing x. Now we need to find an open ball around x itself contained in B(x0, 1/n), but if x is not in A...hmm...OK, I'll think about it some more.

8. Nov 16, 2010

### micromass

Staff Emeritus
You'll need to use density of A again, and probably the triangle inequality...

9. Nov 16, 2010