Let us define the Parovichenko cardinal $\mathfrak{P}$ as the largest cardinal $\kappa$ such that each compact Hausdorff space $K$ of weight $w(K)<\kappa$ is the continuous image of the remainder $\beta\mathbb N\setminus\mathbb N$ of the Stone-Cech compactification of the discrete space of positive integers $\mathbb N$.

By a classical theorem of Parovichenko, $\mathfrak P\ge\aleph_2$.

On the other hand, Theorem 2.7 in this paper of van Douwen and Przymusinski implies that $\mathfrak P\ge\mathfrak p$ where $\mathfrak p$ is the well-known pseudointersection number.

These two results yield the inequality $\mathfrak P\ge\max\{\aleph_2,\mathfrak p\}$.

So, under CH we have $\mathfrak P=\aleph_2>\mathfrak c=\mathfrak p=\aleph_1$.

By a result of Kunen of 1968 in the Cohen model $\mathfrak P=\aleph_2=\mathfrak c>\mathfrak p=\aleph_1$.

Finally, PFA implies $\mathfrak P=\aleph_2=\mathfrak c=\mathfrak p>\aleph_1$, see Corollary 4.6 in Baumgartner's survey "Applications of the Proper Forcing Axiom" in the "Handbook of Set-Theoretic Topology".

Let us also mention a result that follows from Theorems 2.1 and 2.2 of van Douwen and Przymisinski:

$\mathfrak P\le\mathfrak c$ if one of the following conditions holds:

$\bullet$ $\aleph_2\le\mathfrak c<2^{\aleph_1}=\aleph_{\omega_2}$ or

$\bullet$ $\mathfrak c=2^{\aleph_1}$ and $\mathfrak q_0=\aleph_1$.

Here $\mathfrak q_0$ is the largest cardinal $\kappa$ such that each subset $X\subset \mathbb R$ of cardinality $|X|<\kappa$ is a $Q$-set (which means that each subset of $X$ is an $F_\sigma$-set in $X$).

Problem. Is it consistent that $\mathfrak P>\max\{\aleph_2,\mathfrak p\}$?

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    $\begingroup$ Taras, you have bumped twelve old questions to the front page in the last 14 minutes. Please stop, and don't do it again. $\endgroup$ Commented Mar 12, 2019 at 11:13
  • $\begingroup$ @GerryMyerson I am trying to create a new tag (small-uncountable-cardinals). Looking at Meta, I have seen a proposal of Marcin Sleziak to create such a tag. So I upvoted for this proposal and recalled that many my questions fall under that tag. So, I created the new tag "small-uncountable-cardinal" and tagged some of my question with this new tag, but then I observed that I have forgotten to write the letter "s" at the end of "cardinal", so I tried to create a new tag "small-uncountable-cardinals", but the system told me that a tag "small-uncountable-cardinal" already exists. $\endgroup$ Commented Mar 12, 2019 at 11:16
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    $\begingroup$ I don't care what you are trying to do, Taras, I know what you are doing: you are bumping a dozen new questions off the front page by editing tags on old questions. You want to create a new tag? Fine: retag three or four questions a day until you've finished the job, not a dozen in a matter of minutes. Three or four a day, please! $\endgroup$ Commented Mar 12, 2019 at 11:19
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    $\begingroup$ @GerryMyerson Ok. Then I will follow you advise: I will stop retagging new questions, but should at least to finish those that are already bumped up. Sorry for this mess. $\endgroup$ Commented Mar 12, 2019 at 11:22
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    $\begingroup$ @YCor I have just written a post on Meta concerning this new tag "small-uncountable-cardinals", see meta.mathoverflow.net/q/4154/61536 $\endgroup$ Commented Mar 12, 2019 at 11:39

1 Answer 1


No -- it is consistent that $\mathsf{CH}$ fails, and that every compact Hausdorff space of weight $\leq\!\mathfrak{c}$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$. (This is due to Baumgartner, who mentions it off-hand in his article in the Handbook of Set Theoretic Topology; the mutual consistency with $\mathsf{MA}_{\sigma\text{-linked}}$ was established later by Baumgartner, Frankiewicz, and Zbierski in this paper.) In such a model, $\mathfrak{P} = \mathfrak{c}^+ > \max\{\aleph_2,\mathfrak{p}\}$.

  • $\begingroup$ In the model of Baumgartner, Frankiewicz and Zbierski $\mathfrak P=\mathfrak c^+>\mathfrak p=\mathfrak c>\aleph_2$. Right? $\endgroup$ Commented Nov 12, 2018 at 8:38
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    $\begingroup$ @TarasBanakh: Yes. Every $\sigma$-centered partial order is $\sigma$-linked, so $\mathsf{MA}_{\sigma\text{-linked}}$ implies $\mathsf{MA}_{\sigma\text{-centered}}$, which is equivalent to $\mathfrak{p} = \mathfrak{c}$ by a result of Murray Bell. $\endgroup$
    – Will Brian
    Commented Nov 12, 2018 at 10:12

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