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It is easy to show that not every pseudompact space is countably compact, i. e. particular point topology on $\mathbb{R}$. $$ \mathcal{P}=\langle \mathbb{R},\tau=\{A\subset\mathbb{R}\mid 0\in A\}\rangle $$

I am wondering if there exists a Hausdorff space with the same properties. I have tried proving that there does not, but got no result, so I think there exists a counterexample.

Definition 1. A topological space $X$ is said to be pseudocompact iff every continuous real-valued function is bounded. By continuous real-valued I mean a countinuous function between topological spaces $X$ and $\mathbb{R}$ with standart topology. It can be shown that every continuous function form $\mathcal{P}$ to $\mathbb{R}$ is constant, hence bounded.

Definition 2. A topological space $X$ is said to be countably compact iff you can choose a finite subcover from any countable open cover of $X$. It is obvious, that $\mathcal{P}$ does not satisfy this definition.

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    $\begingroup$ Mrówka’s space $\Psi$ is an example; it’s discussed in detail here in Dan Ma’s Topology Blog. It’s not just Hausdorff: it’s Tikhonov (though not normal). $\endgroup$ – Brian M. Scott Sep 27 '16 at 18:38

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