We recall that a topological space $(X,\tau)$ is Pseudo-finite, if each compact subset of $X$ is finite.


One of the classical example of Pseudo-finite topological spaces can be considered as an uncountable set $X$ with the co-countable topology.(i.e.each subset with countable complement is open)


The above topology has no isolated point but it fails to be at least Hausdorff. On the base of my Knowledge there are two **Tychonoff** Pseudo-finite topological spaces as follows:


A. All discrete spaces are trivial examples of these spaces.


B. Consider the set $\Sigma=\mathbb{N}$$\cup$ {$p$}, and topologize it as follows:


* Consider a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$.
* All points of $\mathbb{N}$ are isolated.
* The Neighborhoods of $p$ are of the form: $U$$\cup$ {$p$}, where $U \in \mathcal{U}$.


We must recall that Case "B" is a special Example of maximal Hausdorff topologies on a set.


But I think there is no example of a Pseudo-finite Tychonoff space without isolated point !. and I guess the following statement:


**Statement**:Every Pseudo-finite Tychonoff space has an isolated point.


Is there a counterexample of the above statement?




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On Pseudo-finite topological spaces
Every $P$-space (a $P$-space is a completely regular space where every G\_{
\delta}-set is open) is pseudofinite since one can easily show that every subspace of a $P$-space is a $P$-space and every compact $P$-space is finite. However there are many $P$-spaces with no isolated points. For instance, take the $P$-space coreflection of an uncountable product of the space {0,1} and you will get a pseudofinite space with no isolated points.


