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Let $\beta \mathbb{N}$ denote the Stone-Cech compatification of the natural numbers and $\beta \mathbb{N} \setminus\mathbb{N}$ denote the reminder of this compactification. I wonder if there is a characterization in ZFC of the continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$. I mean: Which compact Hausdorff spaces are continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$?

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  • $\begingroup$ In restriction to totally disconnected spaces, this is equivalent, by Stone duality, to classifying Boolean subalgebras of $2^\mathbf{N}/(\mathrm{fin})$ up to isomorphism. Here $(\mathrm{fin})$ is the ideal of finite subsets. $\endgroup$
    – YCor
    Commented Mar 21, 2018 at 17:11
  • $\begingroup$ @YCor Since every compact space is the continuous image of an extremally disconnected compact space, it is enough to only consider totally disconnected images; the two problems are equivalent. $\endgroup$
    – Not Mike
    Commented Mar 21, 2018 at 20:11
  • $\begingroup$ @NotMike I don't think that the fact you mention make the problems equivalent. $\endgroup$
    – YCor
    Commented Mar 21, 2018 at 20:48
  • $\begingroup$ @YCor let $B$ be the completion of $A=\mathcal{P}(\omega)/fin$ then any continuous map taking $S(A)$ onto $X$ can be pulled-back to a map of $S(B)$ onto $X$. Let $G$ be the extermally disconnected Gleason space associated with $X$; since extremally disconnected compact spaces are projective, either $G$ maps onto $S(B)$ or vice versa; without loss of generality, $S(B)$ maps onto $G$, in this case we can find some $A_0 \subset A$ which generates a sub-algebra isomorphic to a dense sub-algebra of $G$. $\endgroup$
    – Not Mike
    Commented Mar 21, 2018 at 21:08
  • $\begingroup$ @YCor whoops meant $CO(G)$ instead of $G$. $\endgroup$
    – Not Mike
    Commented Mar 21, 2018 at 21:33

1 Answer 1

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This is a great question. There has been quite a bit of work done to figure out what the continuous images of $\beta \mathbb N \setminus \mathbb N$ are. I'll do my best to summarize some of that work here.

In some sense the answer to your question is yes: a space $X$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ if and only if it is the remainder $X = \gamma \mathbb N \setminus \mathbb N$ of some compactification $\gamma \mathbb N$ of $\mathbb N$.

This answer is a bit unsatisfying, though, because what one really wants is an internal characterization of the images of $\beta \mathbb N \setminus \mathbb N$, some recognizable property of $X$ that tells us straightaway whether $X$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ or not.

There is no complete characterization of this kind that I am aware of. However, there are plenty of theorems that give sufficient conditions:

$\bullet$ Every separable compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$

$\bullet$ Every compact Hausdorff space of weight$^1$ $\leq\! \aleph_1$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (Parovicenko, 1963)$^2$

$\bullet$ Every compact Hausdorff space of weight $<\! \mathfrak{p}$ is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (van Douwen and Przymusinski, 1980)$^3$

$\bullet$ Every perfectly normal$^4$ compact space is a continuous image of $\beta \mathbb N \setminus \mathbb N$ (Przymusinki, 1982)

Notice that Parovicenko's theorem implies that, under ZFC+CH, a compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$ if and only if it has weight $\leq\! \aleph_1$. This gives you a very nice characterization of the continuous images of $\beta \mathbb N \setminus \mathbb N$ under ZFC+CH, but you may already know this -- you asked for a ZFC characterization.

Just to give you an idea of how hard it is, in general, to decide whether something is a continuous image of $\beta \mathbb N \setminus \mathbb N$, let me point out that it was an open question for years whether (in ZFC) every first countable compact Hausdorff space is a continuous image of $\beta \mathbb N \setminus \mathbb N$. This was finally solved by Murray Bell in a 1990 paper, where he showed that (in the Cohen model) there is a first countable space that is not a continuous image of $\beta \mathbb N \setminus \mathbb N$.

Along similar lines, there is also an interesting paper of Dow-Hart from 1999 entitled "$\omega^*$ has almost no continuous images" where they show . . . well, they show that many things you might suspect are continuous images of $\beta \mathbb N \setminus \mathbb N$ just aren't (assuming OCA). Work of Ilijas Farah indicates that OCA+MA is something of an optimal hypothesis for ensuring $\beta \mathbb N \setminus \mathbb N$ has as few continuous images as possible (see his book Analytic Quotients for the full story, or his much more accessible article "The fourth head of $\beta \mathbb N$ for a taste).

Finally, let me recommend Jan van Mill's survey article on $\beta \mathbb N$ from the Handbook of set-theoretic topology (link), which looks at this question from several angles and gives a lot of good information on the topic.


  1. Recall that the weight of $X$ is the smallest cardinality of a basis for $X$.

  2. The $\aleph_1$ in this result is optimal. Kunen showed in his thesis that in the Cohen model, the weight-$\aleph_2$ space $\omega_2+1$ is not a continuous image of $\beta \mathbb N \setminus \mathbb N$.

  3. Again, the cardinality bound is optimal. It is consistent with $\mathfrak{p} = \mathfrak{c}$ that there is a compact Hausdorff space of weight $\mathfrak{c}$ that is not a continuous image of $\beta \mathbb N \setminus \mathbb N$. A construction is outlined in van Mill's survey article on $\beta \mathbb N$ from the Handbook of set-theoretic topology.

  4. Recall that a perfectly normal space is one in which every closed subset is a $G_\delta$.

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    $\begingroup$ With respect to oddness surrounding continuous images of $\omega^\ast$ caused by $OCA$, another great resource is Ilijas Farah's "Analytic Quotients." $\endgroup$
    – Not Mike
    Commented Mar 21, 2018 at 2:52
  • $\begingroup$ @NotMike: Absolutely -- I added that to my answer. $\endgroup$
    – Will Brian
    Commented Mar 21, 2018 at 13:39
  • $\begingroup$ Thanks very much, Will Brian! Following the references you suggested I found exactly what I needed. My conjecture was that every first countable linearly ordered compact space is a continuous image of beta N\N. And it seems to be true. It is stated in Corollary 3.2 of Bell´s paper: "A first countable compact space, not an N* image". Unfortunately, I could not find Maurice´s book: Compact Ordered spaces, to understand the proof of this result. $\endgroup$ Commented Mar 21, 2018 at 18:28
  • $\begingroup$ I found Maurice´s book! $\endgroup$ Commented Mar 21, 2018 at 18:47
  • $\begingroup$ @ClaudiaCorrea: Great -- glad I could help! $\endgroup$
    – Will Brian
    Commented Mar 21, 2018 at 18:57

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