Example of an easy non-pseudotopological convergence Recently, I have had an introduction to convergence spaces and I was wondering if there is an easy example of a non pseudotopological convergence space.

Thank you for your help.
I think the following is the most basic example of a convergence space which is not (equivalent to) a pseudotopological space:

Let $X$ be an infinite set, and fix some $\infty \in X$.  Define a convergence on $X$ so that


$\dot{x} \rightarrow x$ for each $x \in X$;
if $\mathcal{U}$ is a free (nonprincipal) ultrafilter on $X$, then $\mathcal{U} \rightarrow \infty$.


(And there are no other convergences.)

It is fairly straightforward to show that this defines a convergence on $X$.  However, if $\mathcal{F}$ is a (free) filter (but not an ultrafilter) on $X$ extending the co-finite filter, then $\mathcal{F} \not\rightarrow \infty$ even though $\mathcal{U} \rightarrow \infty$ for each ultrafilter $\mathcal{U} \supseteq \mathcal{F}$.
I think the following is the most basic example of a convergence space which is not (equivalent to) a pseudotopological space:

Let $X$ be an infinite set, and fix some $\infty \in X$.  Define a convergence on $X$ so that


$\dot{x} \rightarrow x$ for each $x \in X$;
if $\mathcal{U}$ is a free (nonprincipal) ultrafilter on $X$, then $\mathcal{U} \rightarrow \infty$.


(And there are no other convergences.)

It is fairly straightforward to show that this defines a convergence on $X$.  However, if $\mathcal{F}$ is a (free) filter (but not an ultrafilter) on $X$ extending the co-finite filter, then $\mathcal{F} \not\rightarrow \infty$ even though $\mathcal{U} \rightarrow \infty$ for each ultrafilter $\mathcal{U} \supseteq \mathcal{F}$.