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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Question about the Nagata-Smirnov metrization theorem

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