Proving Urysohn's Metrization Theorem on X

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In summary, to prove Urysohn's metrization theorem, we assume that X is normal and second countable and show that there is a homeomorphism between X and a subspace of the Hilbert cube, [0,1]^w. To show that X is metrizable, we first assume that X is infinite and then show that it has a countably infinite basis. Then, using the fact that X is normal and second countable, we can prove that the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk is countably infinite. This demonstrates that X is metrizable.
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math8
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the question is to prove Urysohn's metrization theorem. But there some steps I need to show first.
Assuming X is normal, second countable. We show there is a homeomorphism of X onto a subspace of [0,1]^w (the Hilbert cube which is metrizable), so X is metrizable.

We first show we can assume that X is infinite.
Suppose X is finite. Since X is normal, it is T1, so every singleton is closed. Hence we can see that every subset of X is closed so every subset is open. So X has the discrete topology. So from here, how can I show that X is metrizable (what would be a homeomorphism between X and [0,1]^w ?).

Now, I can show that X has a countably infinite basis {B1,B2,...} where each Bn is neither X nor the empty set. How do I show that the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk is countably infinite?
I can show it's countable because that set is a subset of {B1,B2,...} X {B1,B2,...} which is countable. But how do I show it is infinite?
 
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To show that the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk is countably infinite, you will need to use the fact that X is normal and second countable. Since X is normal, we know that for each Bn there is a open set Un containing its closure. Since X is second countable, we can assume that there is a countable basis {Un1,Un2,...} of open sets containing the closures of Bn. So, given any Bn, its closure is contained in some Unm and so (Bn,Unm) is an element of the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk. Thus, this set contains at least one element for each Bn and so it is countably infinite.
 

Related to Proving Urysohn's Metrization Theorem on X

1. What is Urysohn's Metrization Theorem on X?

Urysohn's Metrization Theorem is a mathematical theorem that states that any topological space X that is regular and second-countable can be metrized, meaning it can be equipped with a metric space structure that induces the same topology on X.

2. Why is Urysohn's Metrization Theorem important?

Urysohn's Metrization Theorem is important because it provides a way to characterize and understand topological spaces by equipping them with a metric structure. This allows for the use of tools and techniques from metric spaces, which are often more familiar and easier to work with.

3. How is Urysohn's Metrization Theorem proved?

Urysohn's Metrization Theorem is typically proved using a combination of techniques from general topology, including properties of regular and second-countable spaces, as well as techniques from metric spaces, such as constructing a metric that induces the desired topology on X.

4. What are the applications of Urysohn's Metrization Theorem?

Urysohn's Metrization Theorem has applications in various areas of mathematics, including functional analysis, differential geometry, and topology. It also has practical applications in fields such as computer science, where metric spaces are commonly used to analyze and model data.

5. Are there any limitations to Urysohn's Metrization Theorem?

Yes, there are some limitations to Urysohn's Metrization Theorem. It only applies to regular and second-countable spaces, so it cannot be used to metrize all topological spaces. Additionally, the metric structure induced by the theorem may not be unique, so there may be multiple ways to metrize a given topological space.

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