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I want to know is there a topology on the countable set which makes the space is not first countable but has countable pseudocharacter?

Thanks for any help:)

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The pseudo-character is defined as $$\psi(p, X) = \min\{ |\mathcal A| : \mathcal A \subseteq \tau_X\text{ is a pseudo-base at }p\text{ (i.e.} \bigcap \mathcal A = \{p\})\},$$ $$\psi(X)=\sup\{\psi(p,X) : p\in X\}+\omega$$ see e.g. here.


The space $S_2^-$ from this answer is one possible example of a space which has countable pseudo-character and it is not first countable.

It is not sequential, hence it is not first countable.

Each point has a countable pseudo-base. This is obvious for isolated points. If we use the same notation as in the linked answer then $\{\omega\}=\bigcap_{n=0}^\infty (\{\omega\}\cup \{n,n+1,\dots\}\times\omega)$. So we see that every point is a countable intersection of open sets.


This example is very similar to Arens-Fort space, so you can try to have a look at this space, if you're more familiar with it.


Another standard example of a space which is not first-countable is the quotient space $\mathbb R/\mathbb N$. Again, since $\mathbb N= \bigcap_{k\in\mathbb N} (\bigcup_{n\in\mathbb N} \left(n-\frac1k,n+\frac1k\right)$, this space has countable pseudo-character.

EDIT: Sorry, I've forgotten that you want countable spaces only; $\mathbb R/\mathbb N$ is of course not countable. But $\mathbb Q/\mathbb N$ should work for the same reasons.

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  • $\begingroup$ Mm, it's a good example! $\endgroup$ – Paul Jun 20 '12 at 12:36
  • $\begingroup$ I'm not familar with Arens-Fort space. Is this space also has countable pseudocharacter? $\endgroup$ – Paul Jun 20 '12 at 12:46
  • $\begingroup$ The proof for Arens-Fort space is almost the same as for $S_2^-$. $\endgroup$ – Martin Sleziak Jun 20 '12 at 12:49

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