Let $(X, \tau_1)$ be a Hausdorff second countable topological space and $(X, \tau_2)$ be a space with a coarser topology, but still Hausdorff. Is $\tau_2$ still second countable?
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1$\begingroup$ By "coarser", you mean $\tau_2 \subset \tau_1$, right? $\endgroup$– MPWCommented 10 hours ago
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$\begingroup$ @MPW Yes, exactly $\endgroup$– CarcassiCommented 8 hours ago
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2 Answers
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Quite often, it won't be. Let $\tau_1$ be the norm topology of a separable Hilbert space $X$, and $\tau_2$ the weak topology. Then, if $X$ is not finite-dimensional, $\tau_2$ is not even first-countable, a fortiori it's not second-countable.
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Let $\tau_1$ be the discrete topology on $\mathbb N$ and let $\tau_2$ be any non-first-countable Hausdorff topology on $\mathbb N$, for example, the collection of all sets $U\subseteq\mathbb N$ such that either $1\notin U$ or else $U$ has asymptotic density $1$.
