# Question about the Nagata-Smirnov metrization theorem

1. Dec 30, 2010

So, the Nagata-Smirnov metrization theorem states that a space X is metrizable iff it is regular and has a basis that is countably locally finite.

Obviously, this is a stronger theorem than the Urysohn metrization theorem, since it gives not only conditions for metrizability, but says that a metrizable space must be regular and have a countably locally finite basis. Also, it is a stronger statement because of the fact that X itself need not have a countable basis (like in the Urysohn theorem) - a collection may be conuntably locally finite, but still be uncountable.

Now, the concept of the proof is to imbed X into the metric space (R^J, ρ), where ρ is the uniform metric on R^J. Now, after a few steps, one arrives at a function F : X --> [0, 1]^J defined with F(x) = (fn,B(x)), where (n, B) is in J, and where a pair (n, B) means that the basis element B is in Bn (B' = U Bn is a countably locally finite basis for X) . By an earlier theorem, relative to the product topology, the map F is an imbedding of X into [0, 1]^J. Now, the uniform metric is finer than the product topology, so images of open sets in X will be open in F(X). One still needs to show that F is continuous with respect to the uniform topology.

Now, my question is: why didn't we imbed F into the same space, but with the metric which induces the product topology? Wouldn't the map F then automatically be continuous?

2. Dec 30, 2010

### micromass

Staff Emeritus
The reason is simply: the space $$[0,1]^J$$ with the product topology is not a metric space. In fact, we have that $$[0,1]^J$$ is a metric space if and only if J is countable.
In general, the product space $$\prod_{i\in I}{X_i}$$ is metrizable if and only if I is countable.

So, while you could easily embed the space in $$[0,1]^J$$ with the product topology, it does not follow that our space is metrizable. That's why Munkres puts another topology on $$[0,1]^J$$, namely the uniform topology. Then metrizability follows easily, but it's a bit harder to show that our embedding is continuous...

The embedding of the space into $$[0,1]^J$$, with the product topology, can be useful however. For example, when trying to find a compactification of the space...

3. Dec 30, 2010

Ahh, it's fascinating how things slip my mind! Of course, it's stated nicely in Theorem 20.5. that it applies to Rω!

Yes, showing continuity of this map is a bit tricky, I'm going through the proof right now, I'll post again if there'll be any questions.

4. Dec 30, 2010

Btw, the Urysohn metrization theorem, the Nagata-Smirnov metrization theorem, and the Nagata metrization theorems are basically all we know about metrizability, right? (I did not yet reach the Nagata theorem, I see it follows after the chapter on paracompactness in the book)

5. Dec 30, 2010

### micromass

Staff Emeritus
Hmm, metrizability is not at all my specialty, but I think there are a lot of metrization theorems out there. For example, exercise 5 states the Bing metrization theorem. And I've also seen a metrization theorem for uniform spaces (a uniform space is something similar to a topological space, but with more structure). So there are much more metrization theorems than those 3.

However, the Urysohn, Nagata-Smirnov and Smirnov theorems are by far the most useful! For example, the Smirnov metrization theorem is often used in differential geometry. So if you want to show the metrizability of some space, then these 3 are the most useful...

6. Dec 30, 2010

OK, so just to jump back to the upper posts and sum up:

One constructed a imbedding F into [0, 1]^J in the product topology, i.e. a homeomorphism. Now, we consider the same space with the uniform topology. The topological structure with respect to the mapping stays unchanged, but continuity becomes a question, since the topology of the codomain is changed, right? i.e. all the properties of the imbedding are preserved except continuity, because we changed the topology on [0, 1]^J? That's why one needs to check continuity...

7. Dec 30, 2010

### micromass

Staff Emeritus
Yes, that is correct. We make the topology of the codomain finer, but that means that continuity becomes an issue. So you need to check it...

8. Dec 30, 2010

OK, actually, here's the clearest possible way (I think) to express what I was saying:

A homeomorphism is a bijective correspondence f : X --> Y such that U is open iff f(U) is open. When we change the topology in the codomain, pointwise, it doesn't affect the bijective correspondence, but the topology may be affected, in the "direction" that if f(U) is open, then U must be open. And exactly that is what's checked in the proof.

9. Dec 30, 2010

### micromass

Staff Emeritus
Well, take a homeomorphism f:X-->Y. Then f is a homeomorphism iff the following holds: U is open if and only if f(U) is open.

However, when you change the topology on Y, then both directions of the "if and only if" can be disturbed. So both continuity of f and $$f^{-1}$$ can be disturbed.
However, if you take a coarser topology on Y, then the continuity of f remains true. It is the continuity of $$f^{-1}$$ that forms the problem.
But, in this case, you have finer topology. Then the continuity of f becomes a problem, but the continuity of $$f^{-1}$$ might not be true anymore.

But all this is probably what you meant...

10. Dec 30, 2010

OK, just to make sure...

If we take a coarser topology on Y, it basically means that we "remove" some open sets from the previous topology on Y. But it seems to me that if we take U open in X, then f(U) needn't be anymore open in Y. Then again, if we tak any set form the coarser topology on Y, it was in the previous topology, so f^-1 of this set is open in X.

A finer topology on Y means that we "added" some open sets to Y. The way I see it, the continuity of f remains OK, since if U is open, f(U) is open in Y, we still have that set in Y! But since we added sets to Y, it may not be true that f^-1 of such sets is open in X.

Sorry if I'm misunderstanding something, but I'd really like to get this straight.

11. Dec 30, 2010

### micromass

Staff Emeritus
Wow, physicsforums is being wierd today. It doesnt even let me post properly

Yes, this is correct.

No, the continuity of f doesn't remain OK. What remains OK is "if U is open, then f(U) is open". Continuity states "if U is open, then $$f^{-1}(U)$$ is open", and this doesn't remain OK...

12. Dec 30, 2010