Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:)

share|cite|improve this question
up vote 3 down vote accepted

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.