A Tychonoff topological space X is called pseudocompact if every continuous real-valued function with domain X is bounded.

A space is called feebly compact if every locally finite family of open sets in the space is finite. It is clear that feeble compactness and pseudocompactness coincide in the class of Tychonoff spaces.

Please, I need a feebly compact space that is not pseudocompact.

  • $\begingroup$ What separation axioms do your definitions of feebly compact, and pseudocompact imply? And does Tychonoff imply Hausdorff in your definition? $\endgroup$ – user642796 May 18 '14 at 0:29
  • $\begingroup$ Sorry, I had made a mistake in the translation. $\endgroup$ – user151661 May 19 '14 at 7:03

Every feebly compact space is pseudocompact (at least in the sense that all continuous $f : X\to \mathbb{R}$ are bounded). If $f : X \to \mathbb{R}$ is continuous, consider the family of all sets of the form $U_n = \{ x \in X : |f(x)|>n \}$ for $n \in \mathbb{N}$. This is locally finite (for $x\in X$ take an open neighbourhood $V$ of $x$ such that $|f(y)-f(x)|<1$ for all $y\in V$, then $V\cap U_n=\varnothing$ for all $n>\lceil |f(x)|\rceil+1$), and so only finitely many of the $U_n$ are nonempty, which then implies that $f$ is bounded.

On the other hand any trivial space of size at least two is feebly compact but not Tychonoff (well, not Hausdorff).

(Added 2014-05-19)

The following is a description of a feebly compact Hausdorff space which is not (completely) regular.

Let $X = \{ \langle n,j \rangle \in \mathbb{N} \times \mathbb{N} : n > 0 \} \cup \{ \infty \}$. For each point of $X$ we describe its basic open neighbourhoods:

  • every $\langle n , j \rangle$ with $j > 0$ is isolated;
  • the basic open neighbourhoods of $\langle n,0 \rangle$ are of the form $\{ \langle n,0 \rangle \} \cup \{ \langle n,j \rangle : j \geq k \}$ for some $k > 0$;
  • the basic open neigbourhoods of $\infty$ are of the form $\{ \infty \} \cup \{ \langle n,j \rangle : n \geq \ell , j > 0 \}$ for some $\ell > 0$.

It is easy to see that this space is Hausdorff. Note that $U = \{ \infty \} \cup \{ \langle n,j \rangle : n,j > 0 \}$ is an open neighbourhood of $\infty$, and given any open $V$ with $\infty \in V \subseteq U$ there must be an $n > 0$ such that $\langle n,0 \rangle \in \overline{V}$. Therefore $X$ is not regular.

To see that $X$ is feebly compact, suppose that $\mathcal{U}$ is an infinite family of nonempty open subsets of $X$. Without loss of generality we may assume that each set in $\mathcal{U}$ is a basic open set.

  • If $\mathcal{U}$ is not point finite, then it cannot be locally finite. Therefore we may assume that only finitely many sets in $\mathcal{U}$ contain $\infty$, and that for each $n > 0$ only finitely many set in $\mathcal{U}$ contain $\langle n,0 \rangle$.
  • If there are infinitely many $n > 0$ such that a some set in $\mathcal{U}$ contains $\langle n,0 \rangle$, then every neighbourhood of $\infty$ meets infinitely many sets in $\mathcal{U}$, and so we may assume that for only finitely many $n > 0$ is $\langle n,0 \rangle$ an element of some set in $\mathcal{U}$.
  • It follows that there are infinitely many singletons $\{ \langle n,j \rangle \}$ ($n,j > 0$) in $\mathcal{U}$. If there is an $n > 0$ such that for infinitely many $j > 0$ does $\{ \langle n,j \rangle \}$ belong to $\mathcal{U}$, then every neighbourhood of $\langle n,0 \rangle$ meets infinitely many sets in $\mathcal{U}$. If there are infinitely many $n > 0$ such that some singleton $\{ \langle n,j \rangle \}$ belongs to $\mathcal{U}$, then every neighbourhood of $\infty$ meets infinitely many sets in $\mathcal{U}$.

Therefore no infinitely family of nonempty open subsets of $X$ is locally finite.


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