A question about the Urysohn metrization theorem

[radou]

Here's just a conceptual question about this proof.

One can show a space is X metrizable by embedding it into a metric space, i.e. proving it is homeomorphic to a subset of a metric space (Y, d). The metric d induces the metric topology on Y, and a homeomorphism from X to Y (i.e. from Y to X) preserves open, closed sets, and other topological properties.

This may be a stupid question, but I don't directly see how this metric induces the given topology T on X.

 

[Tinyboss]

You're done on the first line: if X embeds homeomorphically into a metric space Y, then the topology on X is metrizable. You just define a metric d on X using the metric d' on Y by setting d(x,x') = d'(f(x),f(x')) where f is the homeomorphism. The work is in showing that f exists. In a sense, it's more like the original topology on X is inducing the map f and therefore the metric on the image of X, rather than the other way around.

 

[radou]

You're done on the first line: if X embeds homeomorphically into a metric space Y, then the topology on X is metrizable. You just define a metric d on X using the metric d' on Y by setting d(x,x') = d'(f(x),f(x')) where f is the homeomorphism. The work is in showing that f exists. In a sense, it's more like the original topology on X is inducing the map f and therefore the metric on the image of X, rather than the other way around.

Hm, I see.. Btw, the defined metric is an isometry, right? So in a way it preserves the "shapes" of the open sets in Y when mapped into X?

 

[radou]

Hm, I see.. Btw, the defined metric is an isometry, right? So in a way it preserves the "shapes" of the open sets in Y when mapped into X?

Pardon me, not the metric, but the function f.

 

[radou]

OK, so, just to check if my reasoning is correct, I found an answer to my question in the first theorem on http://drexel28.wordpress.com/2010/03/10/thoughts-about-separation-urysohns-metrization-theorem/" page.

Now, regarding the "<==" direction and its ending "the conclusion follows":

We need to show that the metric d' induces the topology T on X. So, let U be an open set in T. Then f^-1(U) = V is some open set in (E, d), so it can be written as a union of open balls in E, i.e. V = U Bd(x, ε). But then f(V) = U = U f(Bd(x, ε)) = U Bd'(f(x), ε), and hence we have shown that any open set in X can be represented as a union of open balls in the metric d'. So, the metric d' induces the topology T.

Is this correct?

(Note: the notation isn't ideal, since U represents both the open set U and the union of sets, but I think it's clear enough what I'm referring to.)

 

[klackity]

I think you are making it more complicated than you need to.

DEFINITION: A space is metrizable if it is homeomorphic to a metric space.

PROPOSITION: Any subset of a metric space is a metric space.

Proof: all properties of a metric space hold for arbitrary subsets

Right from this we have that:

X is metrizable <==> X is homeomorphic to a subset of a metric space.

Your question is: "How does the metric on the metric space induce a topology T on X?"

DEFINITION: A homoemorphism f: X -> Y is a bijective, bicontinuous map between topological spaces.

PROPOSITION: If f:X -> Y is a homeomorphism, and S is a topology on Y, then there is a unique topology T on X induced by S.

Proof: Again, it's obvious. A homeomorphism (and it's inverse) maps open sets to open sets (by definition!) Thus a subset U of X is in T if and only if f(U) is in S.

This leaves no room for error: the open sets in X are precisely defined by the homeomorphism f from the open sets in Y.

In fact, a homeomorphism is characterized by these two properties:
1) It is a bijection between the sets
2) It is a "bijection" between the topologies

In your case, the space Y is the subset of the metric space which X is homeomorphic to. We start with these two spaces:

(X,T)

and

(Y,d)

Where d is the metric topology (inherited from the metric space of which Y is a subset).

Once we have a homeomorphism f between them, we know that the "open" sets in X are in bijective correspondence with the "open" sets in Y. (i.e., if we know the open sets in one space, we know the open sets in the other).

In other words: once we have a homeomorphism between (X,T) and (Y,d), they are exactly the same topological space!

 

[radou]

klackity, thanks for the reply.

You laid all this out pretty nice, but I'm still interested if what I wrote is true.

 

[lavinia]

Here's just a conceptual question about this proof.

One can show a space is X metrizable by embedding it into a metric space, i.e. proving it is homeomorphic to a subset of a metric space (Y, d). The metric d induces the metric topology on Y, and a homeomorphism from X to Y (i.e. from Y to X) preserves open, closed sets, and other topological properties.

This may be a stupid question, but I don't directly see how this metric induces the given topology T on X.

the metric defines the open sets by open balls of some radius. the open sets in any subset are unions of the intersections of these open balls with the subset.

 

[micromass]

OK, so, just to check if my reasoning is correct, I found an answer to my question in the first theorem on http://drexel28.wordpress.com/2010/03/10/thoughts-about-separation-urysohns-metrization-theorem/" page.

Now, regarding the "<==" direction and its ending "the conclusion follows":

We need to show that the metric d' induces the topology T on X. So, let U be an open set in T. Then f^-1(U) = V is some open set in (E, d), so it can be written as a union of open balls in E, i.e. V = U Bd(x, ε). But then f(V) = U = U f(Bd(x, ε)) = U Bd'(f(x), ε), and hence we have shown that any open set in X can be represented as a union of open balls in the metric d'. So, the metric d' induces the topology T.

Is this correct?

Yes, this is correct!
In fancy language, you have just proven that "metrizability is inherited by subspaces"; This is quite important in metrizability theory... But you probably already noticed that