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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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Yet another spelling of Tychonoff. –  Martin Brandenburg Jun 9 '12 at 10:58
    
I am sorry about my wrong spelling. –  Ali Reza Jun 9 '12 at 11:02
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4 Answers

up vote 4 down vote accepted

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.

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Hello dear Joseph. Thank you very much for your nice description. Yes, the fact that you mentioned Is well-Known. I must point the fact that it may be useful. When I saw you answer, another Question came in my mind that if there is a non P-space without isolated point which is also Pseudo-finite. The answer in this case is also no. It Suffices to consider the topological space $X$ which you described as above and multiply it with the space $\Sigma$(i.e. $Y=X \times \Sigma$) (Best Wishes) –  Ali Reza Jun 11 '12 at 7:09
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The counterexamples so far depend on AC, but one can have such spaces just from ZF.

In particular, instead of an ultrafilter on ${\Bbb N}$ as in your example $B$, one can use the filter of subsets with asymptotic density 1. The rest follows along the lines of my previous answer (and David Milovich's details).

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The rationals are pseudo-finite.

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The set of $\{1/n\}\cup\{0\}$ is compact but not finite. –  Jim Conant Jun 10 '12 at 20:27
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I think Warren was confused The concept of Pseudo-finiteness with the concept "pseudo-discreteness".The topological space $X$ is pseudo-discrete if if every compact subset of $X$ has finite interior. Every pseudo-finite space is a pseudo-discrete space, but not conversely. The space $\mathbb{Q}$ is a pseudo-discrete space which is not pseudo-finite. –  Ali Reza Jun 11 '12 at 8:51
    
Yup thanks for fixing that. How about the set of weak P-points of in \Beta N - N. Or any other infinite weak crowded weak P-space. –  Warren McG Jun 11 '12 at 13:08
    
It might be best if you edited your answer to fix the problem. –  S. Carnahan Jun 12 '12 at 7:32
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I believe you can grow your example B into a counterexample.

Stage 0 is $\emptyset$

Stage 1 is $\{p\}$.

Stage 2 is your $B$: you've added a copy of ${\Bbb N}$ for each point newly added in the previous stage (all one of them) with an ultrafilter to define neighborhoods of the old point(s).

And the recipe for Stage $n+1$ adds a copy of ${\Bbb N}$ for each point newly added in Stage $n$ with an ultrafilter to define neighborhoods of the old points.

Natural numbers suffice to index the stages (no transfinite induction necessary).

Clearly we kill all the isolated points in the limit. I believe you get pseudo-finiteness much as you get it for $B$, but there are details.

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Details: Let $\mathcal{U}$ be a free ultrafilter on $\omega$; let $X$ be the tree $\omega^{<\omega}$ with the topology generated by the set $\mathcal{B}$ consisting of all subsets $Y$ such that $Y$ has a unique root and $\forall y\in Y\ \forall^{\mathcal{U}}n\ \ y^\frown n\in Y$. Check that $X$ is $T_1$ and $\mathcal{B}$ is a clopen base, implying $X$ is $T_{3.5}$. By Konig's Lemma, if $A\subset X$ is infinite, then $A$ contains an infinite chain or an infinite set of the form $\\{s^\frown n:n\in I\\}$. In either case, check that $A$ is not compact. –  David Milovich Jun 10 '12 at 19:31
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