# The Parovichenko cardinal, is it equal to $\max\{\aleph_2,\mathfrak p\}$?

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 $$\mathfrak P>\max\{\aleph_2,\mathfrak p\}$$?

• Taras, you have bumped twelve old questions to the front page in the last 14 minutes. Please stop, and don't do it again. Mar 12, 2019 at 11:13
• @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. Mar 12, 2019 at 11:16
• 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! Mar 12, 2019 at 11:19
• @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. Mar 12, 2019 at 11:22
• @YCor I have just written a post on Meta concerning this new tag "small-uncountable-cardinals", see meta.mathoverflow.net/q/4154/61536 Mar 12, 2019 at 11:39

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, $$\mathfrak{P} = \mathfrak{c}^+ > \max\{\aleph_2,\mathfrak{p}\}$$.
• In the model of Baumgartner, Frankiewicz and Zbierski $\mathfrak P=\mathfrak c^+>\mathfrak p=\mathfrak c>\aleph_2$. Right? Nov 12, 2018 at 8:38
• @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. Nov 12, 2018 at 10:12