Converse of the Urysohn metrization theorem 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.  
 A: 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.
A: 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
