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I apologize that I cannot post the exact question and I know this is too vague, but I am asked to show that, around some point $x$ in some quotient space, it does not have a countable basis around that point. I was guessing that, then around $x$, it does have to have an uncountable basis. But I'm getting nowhere from here, and I cannot really think of any trick.

Maybe I need to use the contradiction, but I'm not sure what "having countably many basis" will imply. Could someone give me any input?

able20
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    It will depend on knowing what defines your topology. Of course metric topologies cannot satisfy your condition. – hardmath Feb 08 '21 at 03:29
  • @hardmath It is a quotient space formed from Euclidean space, with some identification. Could you just throw out some keywords just as an inspiration? – able20 Feb 08 '21 at 03:32
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    A likely approach is to let $\mathscr{B}={B_n:n\in\Bbb N}$ be an arbitrary countable family of open nbhds of the point and show how to construct an open nbhd of the point that does not contain any of the sets $B_n$. Here is an answer in which I used that idea. – Brian M. Scott Feb 08 '21 at 03:56

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The commonest way I've seen is a kind of diagonalisation argument: given countably many $(B_n)_n$ open neighbourhoods of $x$ we can find an open neighbourhood $O$ of $x$ such that $B_n \nsubseteq O$ for all $n$.

You could also contradict a consequence of first countability: e.g. $x \in \overline{A}$ for some set $A$ while there is no sequence from $A$ converging to $x$.

It all depends on who the topology is defined. For quotients the first way is the most common, maybe because open sets are easy to recognise.

Henno Brandsma
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