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I'm looking for an example for a second countable topological space $T$ such that there exist a quotient structure $T/\sim$ which is not second countable.

Does there exist an example where $T$ is a manifold?

user267839
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  • Have you tried proving that a quotient of a second countable space is second countable? – Thibaut Dumont Feb 10 '19 at 15:09
  • @ThibautDumont: I don't see any chance to do it: If I start with a countable base $B_T$ of $T$ then every image of an open $U \in B_T$ is "scattered" by the identifications in $T / \sim$ so there is in general no possibility (at least I don't see some) to reach a new basis generated by images of $B_T$ since it's not fine enough to form a basis. – user267839 Feb 10 '19 at 15:33
  • @ThibautDumont: btw the images do not have to be open if the quotient map isn't saturated. So I guess that a good candidate for a conterexample for the statemant above is a not saturated quotient space :) – user267839 Feb 10 '19 at 15:38

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Let $T=\mathbb{R}$ in its usual topology (a second countable manifold).

Let $\sim$ be the equivalence relation that identifies $\mathbb{Z}$ to a point and leaves other points untouched: $x \sim y$ iff $x,y \in \mathbb{Z}$ and $x \sim y$ iff $x=y$ otherwise. So the classes are exactly $\{x\}$ for all $x \notin \mathbb{Z}$ adn $\mathbb{Z}$. The map sending $x$ to its class I call $q$ and the set of classes $T/ \sim$ gets the quotient topology w.r.t. $q$.

Then the class $\mathbb{Z} \in T/\sim$ does not have a countable local base (several proofs of this fact can be found on this site, a good one is here, e.g.) so a fortiori this quotient is not second countable (second countable implies first countable).

Henno Brandsma
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