In topological space, does first countable+ separable imply second countable? If not, any counterexample?
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Martin Sleziak
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Some counterexamples are listed in one of the answers to this previous Question. – hardmath Jan 21 '16 at 04:02
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However, the implication does hold for topological groups. – Thomas Winckelman Apr 27 '21 at 04:07
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It’s false in general. A simple counterexample is the Sorgenfrey line, also known as $\Bbb R$ with the lower limit topology: $\Bbb Q$ is still a dense set, and each point $x$ has a countable local base of sets of the form $\left[x,x+\frac1n\right)$ for $n\in\Bbb Z^+$, but the space is not second countable.
The Sorgenfrey plane, the Cartesian product of the Sorgenfrey line with itself, is even worse: $\Bbb Q\times\Bbb Q$ is a dense subset, so it’s separable, and as a product of two first countable spaces it is certainly first countable, but the reverse diagonal, $\{\langle x,-x\rangle:x\in\Bbb R\}$, is an uncountable closed discrete set. It’s easy to see that no space with an uncountable closed discrete subset can be second countable.
Brian M. Scott
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