1
$\begingroup$

Urysohn metrization theorem says that every regular and second countable topological space is metrizable. My question, is the converse of this theorem ture ? If not, what are the counter examples?

Any reply kindly appreciated. Thanks.

$\endgroup$
0
$\begingroup$

The converse of this theorem does not hold.

As an example consider the set of all real numbers $ \mathbb{R} $ with the discrete topology $ \tau_{d} $. Clearly $ (\mathbb{R},\tau_{d}) $ is metrizable and the discrete metric $ \rho_{d} $ is the metrization of $ \tau_{d} $. Also $ \mathbb{R} $ is regular with respect to $ \tau_{d} $ since every closed subset of $ \mathbb{R} $ is open with respect to $ \tau_{d} $.

Notice that if a collection $ \mathcal{B} $ of subsets of $ \mathbb{R} $ is a base for $ (\mathbb{R},\tau_{d}) $ then $ \mathcal{B} $ contains $ \{\{x\}:x\in \mathbb{R}\} $. But $ \{\{x\}:x\in \mathbb{R}\} $ is uncountable. Hence $ \mathcal{B} $ is also uncountable. Therefore $ (\mathbb{R},\tau_{d}) $ has no countable base even though $ (\mathbb{R},\tau_{d}) $ is metrizable. Therefore the converse of the Urysohn metrization theorem does not hold.

$\endgroup$
  • 1
    $\begingroup$ All metric spaces are regular (in fact, perfectly normal), so the last sentence of the second paragraph is somewhat superfluous. $\endgroup$ – user642796 Jan 29 '15 at 9:32
  • $\begingroup$ @ArthurFischer you are correct :) $\endgroup$ – ASB Jan 29 '15 at 9:35
0
$\begingroup$

A metrizable space is second countable if and only if it is separable.There are many examples of non-separable metrizable spaces,e.g.,in the previous answer, the discrete topology on any uncountable set

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.