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Given a first countable space it is always possible to build a neighborhood basis of a point for which the inclusion is a total order. Let's say $\lbrace V_n \rbrace _{n \in \mathbb{N}}$ is any countable neighborhood basis of that point, then define $\lbrace U_n \rbrace _{n \in \mathbb{N}}$ as follows:

$U_1=V_1$

$U_k=V_k \cap V_{k-1} \quad \mbox{if $k>1$}$

It's easy to check that $\lbrace U_n \rbrace _{n \in \mathbb{N}}$ equipped with the inclusion relation is a totally ordered set.

Now, if we have a space which is not $N_1$ we may not able to use this inductive construction, so my question is if there is some other technique we can use for a (possible) non-countable neighborhood basis of a point or if it is proven that is not possible to construct such a basis.

Thanks

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  • $\begingroup$ How about transfinite induction? $\endgroup$ – sqtrat Jun 15 '16 at 13:20
  • $\begingroup$ @sqtrat The problem of transfinite induction is that infinite intersection of open sets need not to be open. $\endgroup$ – Crostul Jun 15 '16 at 13:20
  • $\begingroup$ @Crostul Oh yes, of course. Completely slipped my mind. In that case, I think that this won't be possible. $\endgroup$ – sqtrat Jun 15 '16 at 13:23
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    $\begingroup$ A counterexample is $\Bbb{R}$ with the cofinite topology. Any basis of neighbourhoods of $0$ cannot be a totally ordered set. $\endgroup$ – Crostul Jun 15 '16 at 13:24
  • $\begingroup$ @Crostul That seems to be the correct idea. $\endgroup$ – sqtrat Jun 15 '16 at 13:26

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