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Initially, I was wondering if $T_4$ (normal and $T_1$) P-spaces (each countable intersection of open sets is open) are $T_5$ (completely normal and $T_1$), or if anyone knows a counterexample of such. Thinking further about the hypotheses, I read this theorem, which has redirected my interest in knowing if regular P-spaces are normal (as regular and P-space are hereditary properties), or there exists a counterexample of this aswell.

Each way that I have tried to reason, I have reached a weaker conclusion (pseudonormality), or I have reached normality after having required an additional hypothesis depending on the reasoning way. However, I haven't found any counterexamples yet neither.

Does anyone know an example of a regular P-space that isn't normal?

Almanzoris
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1 Answers1

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Here is an example:

For a limit ordinal $\lambda$ let $B(\lambda) = \{\alpha +1: \alpha < \lambda\} \cup \{\lambda\}$ with the subspace topology of $\lambda +1$ with the order topology. Of course, $B(\lambda)$ is $T_4$, in particular regular, $T_1$.
Each point of $B(\lambda)$ is isolated, except $\lambda$. If cf$(\lambda) > \omega$, then $\lambda$ is a P-point in $\lambda +1$, hence also in $B(\lambda)$. Thus, $B(\lambda)$ is P-space.

Now let $X = (B(\omega_2) \times B(\omega_1)) \setminus \{(\omega_2, \omega_1)\}$. Since finite products and subspaces of P-spaces are P-spaces, $X$ is a P-space. Of course, it is regular, $T_1$.

Let $A = B(\omega_2) \times \{\omega_1\}$, $B = \{\omega_2\} \times B(\omega_1)$. Then $A\cap X, B\cap X$ are closed in $X$, but cannot be separated by open sets. The proof is analogously to the one for the deleted Tychonoff plank. Hence, $X$ is not normal.

Steven Clontz
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Ulli
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  • Thank you so much, Ulli. – Almanzoris Oct 01 '24 at 21:39
  • Do you know a $T_4$ P-space that isn't $T_5$ too? – Almanzoris Oct 01 '24 at 21:40
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    I have to think tomorrow about it: have you checked, whether $B(\omega_2) \times B(\omega_1)$ is normal? If yes, it would be an example for a $T_4$ P-space, not $T_5$. – Ulli Oct 01 '24 at 21:47
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    @Almanzoris: yes, indeed, $B(\omega_2) \times B(\omega_1)$ is normal, even ultraparacompact: define $X$ is almost discrete, if $X$ is $T_1$ and has at most one non-isolated point. It is easy to see that almost discrete spaces are ultraparacompact. Moreover, a product of an ultraparacompact $T_2$ and an almost discrete space is ultraparacompact (but, I think, the proof is too lengthy to record it here). – Ulli Oct 02 '24 at 05:43
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    Nice example. Please double check me on this. It seems that $B(\lambda)$ would also be homeomorphic to the full ordinal space $\lambda+1$ with a finer topology obtained by make all points $<\lambda$ isolated. In some cases that would also be homeomorphic to one of "generalized Fort/Fortissimo spaces" from https://math.stackexchange.com/questions/4833761. For example, $B(\omega_1)\cong F_{\omega_1,\omega}$ and $B(\omega_2)\cong F_{\omega_2,\omega_1}$. And your example $X$ is an interesting variation of the Deleted Dieudonne plank. – PatrickR Oct 02 '24 at 07:03
  • Do you know of an accepted name in the literature for the $B(\lambda)$ or the generalized Fort/Fortissimo spaces $F_{\lambda,\kappa}$? – PatrickR Oct 02 '24 at 07:07
  • (Hmm, I got mixed up in my notation. $B(\omega_n)\cong F_{\omega_n,\omega_n}$.) – PatrickR Oct 02 '24 at 07:16
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    @PatrickR: yes, for the topology you described,
    $\lambda +1 \rightarrow B(\lambda), \alpha \mapsto \alpha +1 (\alpha < \lambda), \lambda \mapsto \lambda$ is a homeomorphism. And no, I'm not aware of a common name for these spaces.     – Ulli Oct 02 '24 at 07:52
  • Thanks again for your answer, Ulli. – Almanzoris Oct 02 '24 at 13:26