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Metrizable space with countable dense subset

by radou
Tags: countable, dense, metrizable, space, subset
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radou
#1
Nov16-10, 02:14 PM
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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.
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micromass
#2
Nov16-10, 03:44 PM
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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...
radou
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Nov16-10, 03:49 PM
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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.

micromass
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Nov16-10, 03:52 PM
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Metrizable space with countable dense subset

Quote Quote by radou View Post
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.
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.
radou
#5
Nov16-10, 03:55 PM
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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?
micromass
#6
Nov16-10, 04:01 PM
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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 [tex]x\in B\subseteq G[/tex]. 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...
radou
#7
Nov16-10, 04:16 PM
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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.
micromass
#8
Nov16-10, 04:20 PM
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You'll need to use density of A again, and probably the triangle inequality...
radou
#9
Nov16-10, 05:14 PM
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Just a small sub-question. The metric topology for a metric space X consists of all unions of elements of the family {B(x, ε) : x is in X and ε > 0}, right? This is true for any metric space, right?
micromass
#10
Nov16-10, 05:15 PM
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Yes, that is true for any metric space!


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