Metrizable space with countable dense subset

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In summary, the solution to showing that every metrizable space with a countable dense subset has a countable basis involves taking a countable collection of open balls with radius 1/n around each point in the dense subset, and showing that this collection covers the entire space and forms a basis. To do this, one must use the fact that the dense subset is countable and the triangle inequality to show that the basis generates the correct topology.
  • #1
radou
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Homework Statement



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

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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  • #2
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
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
radou said:
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.
 
  • #5
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
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 doesn't have to A. And in that case B doesn't have to lie in our basis. So you'll need to find another candidate for B.

I hope I didnt misunderstood your post...
 
  • #7
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
You'll need to use density of A again, and probably the triangle inequality...
 
  • #9
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?
 
  • #10
Yes, that is true for any metric space!
 

Related to Metrizable space with countable dense subset

1. What is a metrizable space?

A metrizable space is a topological space that can be described using a metric, which is a function that assigns a distance between any two points in the space. This allows for the concept of open and closed sets, which are fundamental in topology.

2. What is a countable dense subset?

A countable dense subset is a subset of a topological space that contains a countable number of points and is also dense, meaning that every point in the space is either in the subset or a limit point of the subset. In other words, the subset is "spread out" throughout the space.

3. How do you prove that a space is metrizable with a countable dense subset?

To prove that a space is metrizable with a countable dense subset, you need to find a metric that satisfies the four axioms of a metric space: non-negativity, symmetry, the triangle inequality, and the definition of a metric for a single point. Then, you must show that the metric generates the same topology as the original space, and that the subset is both countable and dense in the space.

4. What are some examples of metrizable spaces with countable dense subsets?

Some examples of metrizable spaces with countable dense subsets include the real line, the unit interval, and Euclidean spaces of any dimension. Other examples include discrete spaces, metric spaces, and compact spaces.

5. Why is the concept of a metrizable space with countable dense subset important in mathematics?

Metrizable spaces with countable dense subsets are important in mathematics because they provide a way to study and understand topological spaces using the familiar concept of a metric. They also allow for the use of techniques and results from metric spaces in more general topological spaces, making them a powerful tool for understanding and solving problems in mathematics.

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