Every separable Metric space has countable base.

[rumjum]

Homework Statement

Every separable metric space has countable base, where base is collection of sets {Vi} such that for any x that belongs to an open set G (as subset of X), there is a Vi such that x belongs to Vi.

Homework Equations

Hint from the book of Rudin: Center the point in a countable dense subset of the metric space and have a union of all "rational" radius.

The Attempt at a Solution

I need to understand this problem a bit more, so would appreciate any hints. I am mainly unclear as to why one needs to choose rational radius. Is it because rational numbere are countable? So we can have finitely many Vi.

My understanding so far is that if x belongs to G , then we need to prove that G is a union of Vi , where i belongs to N.

Now, based on the hint, we can find a point "p" in the dense subset A Intersection X, and have neighborhoods of all rational radii. Let any such neighborhood be Vi. Any neighborhood is an open set and the union of such open sets is open. Let this set be G , then any element that belongs to G , belongs to Union of Vi. But Vi is a collection of rational radii neighborhood and rational numbers are countable. Hence, the collection Vi is countable. In other words, {Vi} is the base.

Kind of vague?...& stuck

 

[mighty2000]

...

First of all, great job on starting to think about the problem and trying to understand it! It's always important to have a good understanding of the problem before attempting a solution.

To answer your question about choosing rational radii, yes, it is because rational numbers are countable. This allows us to have a countable collection of neighborhoods that cover the dense subset of the metric space. This is important because we want to show that any open set G can be written as a union of elements from this countable collection.

Now, let's try to break down the problem a bit more. We want to show that for any open set G, there exists a countable collection of sets {Vi} such that G is a union of elements from this collection. We can start by choosing a point p in the dense subset A ∩ X. Then, for each rational number r, we can construct a neighborhood of p with radius r. Let's call this neighborhood Vr. Since we are using only rational numbers, we can have a countable collection of neighborhoods {Vr}.

Now, let's consider any open set G. Since G is open, for any point x in G, there exists some neighborhood of x contained in G. But since our dense subset A ∩ X is dense, there exists some point p in A ∩ X that is also in this neighborhood. And since our collection of neighborhoods {Vr} covers the dense subset, there exists some neighborhood Vr that contains p. This means that x is also in Vr, and therefore in G. This shows that G is a union of the neighborhoods {Vr}, which are elements of our countable collection.

I hope this helps clarify the problem a bit more. Keep up the good work!