The following example with a little modified from the handbook of set theoretic topology, Page 574:

Let $\kappa$ be any cardinal for which there exists a family $\{H_\alpha: \alpha < \kappa\}$ of infinite subset of $\omega$ such that

1) $\beta < \alpha$ implies $|H_\beta \setminus H_\alpha|<\omega$ and

2) for every $H \in [\omega]^\omega$, there exists $\alpha <\kappa$ such that $ |H\setminus H_\alpha|=\omega$.

The underlying set for the desired space $X$ is the set $\kappa \cup \omega$, where we consider $\kappa$ and $\omega$ to be disjoint. The topology is definied as follows: For each $\alpha < \kappa$ we define a basic neighbourhood of $\alpha$ for each $\beta < \alpha$ and each $F\in [\omega]^{<\omega}$ by $$ N(\alpha: \beta, F)=\{ \text{ any set } A \subset \kappa \text{ with } \alpha \in A\} \cup ((H_\beta \setminus H_\alpha)\setminus F).$$ Points in $\omega$ are declared to be isolated. With this topology on $X=\kappa \cup \omega$ it is clear that $\omega$ is a countable dense set of isolated points, and $\kappa$ is a closed discrete subset of $X$ whose subspace topology is the usual discrete topology on $\kappa$.

Question: Is $X$ pseudocompact?

Thanks for your help.

  • $\begingroup$ What role does $F$ play in the definition of $N ( \alpha : \beta , F )$? $\endgroup$ – user642796 May 15 '13 at 9:07
  • $\begingroup$ @Arthur Fischer: Sorry. It is a wrong type. $\endgroup$ – Paul May 15 '13 at 9:54
  • $\begingroup$ Thanks. I'm also not exactly sure what you mean by "$\{ \text{ any set } A \subset \kappa \text{ with } \alpha \in A\}$". Could you perhaps mean to say that the basic open neighbourhoods of $\alpha$ are of the form $A \cup ( H_\beta \setminus ( H_\alpha \cup F ) )$ where $A \subseteq \kappa$ contains $\alpha$, $\beta < \alpha$ and $F \subseteq \omega$ is finite? (Note that in this case $\{ \alpha , \beta \} \cup ( H_\beta \setminus H_\alpha )$ would be an open set containing $\beta$, but includes no basic open neighbourhood of $\beta$.) $\endgroup$ – user642796 May 15 '13 at 10:03
  • $\begingroup$ @Arthur Fischer: Yes. Just as explains, I modified the example in which $\kappa$ is hoped to be a closed discrete subset of $X$ whose subsapce topology is the discrete topology on $\kappa$. I don't know whether there is a contradiction while constucting the example. $\endgroup$ – Paul May 15 '13 at 10:43
  • $\begingroup$ @Paul, as written conditions (1) and (2) are contradictory. (Assuming (1), take $H = H_0$ in (2).) Did you mean to have the roles of $\alpha$ and $\beta$ switched in (1)? $\endgroup$ – Paul McKenney May 15 '13 at 20:41

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