Regular + second countable implies normal. But the Urysohn Metrization Theorem shows that regular + second countable also implies metrizable. And metrizable implies normal. But I wonder what is the additional condition for normal space to be metrizable. Thanks!
Additional condition for normal space to be metrizable
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general-topology
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1The existence of a $\sigma$-locally finite basis: see the Nagata-Smirnov Metrization Theorem https://en.wikipedia.org/wiki/Nagata%E2%80%93Smirnov_metrization_theorem – 2017-02-25
1 Answers
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The following are equivalent for a space $X$:
- $X$ is metrisable.
- $X$ is $T_3$ and has a $\sigma$-locally finite base.
- $X$ is $T_4$ and has a $\sigma$-locally finite base.
1 implies 3 is trivial (as soon as you know that all metric spaces have such a base, and are normal), 3 implies 2 is trivial as $T_4$ implies $T_3$ and 2 implies 1 is the harder part of the Nagata-Smirnov theorem.
So the additional condition is the same for $T_4$ and $T_3$.