I believe that anyone who can answer this question knows the terminology, but $\beta\omega$ is the Čech-Stone compactification of the integers and a point $p$ is a weak P-point in a space $X$ iff it is not in the closure of any countable subset of $X-\{p\}$. I admit to lacking intuition with $\beta\omega$ and maybe this question has a trivial answer, apologies if it is the case.

My motivation is the following. I want to find a countably compact space $X$ and a surjective map $f:X\to[0,1]$ such that $f(C)\not=[0,1]$ if $C\subset X$ is compact. There might be a simple example that escaped me, but if the question in the title has a positive answer, taking $X$ to be $\beta\omega$ minus its weak P-points yields such a space:

-It is countably compact because $\beta\omega$ is compact, hence any countable subset has a cluster point which cannot be a weak P-point,

-It maps onto $[0,1]$ because we may map $\omega$ onto the rational numbers in $[0,1]$ and extend the function to all of $\beta\omega$. The image of $X$ is countably compact, hence compact in $[0,1]$ and is thus all of $[0,1]$.

-If $f(E)=[0,1]$ with $E\subset X$, $E$ has cardinality $\mathfrak{c}$ and its closure non-compact.

But maybe I am trying to kill a mosquito with an atomic bomb and a much simpler example exists.

  • $\begingroup$ The answer to your question, as it is phrased, is no, because every infinite closed subset of $\beta \omega$ has cardinality $>\! \mathfrak{c}$. So for example, if $\{x_1,x_2,\dots\}$ is any countably infinite set of non-weak-P-points, then its closure $X$ contains no weak $P$-points. Any $Y \subseteq X$ of cardinality $\mathfrak{c}$ therefore provides a negative answer to your question. Is this what you are looking for, or am I confused about what you are trying to ask? $\endgroup$ – Will Brian May 6 '18 at 13:49
  • $\begingroup$ Oh, I just totally forgot the fact that the closure of countable sets are so big in $\beta\omega$. So my remark about lacking intuition was quite right. Thanks for your comment. But is it possible to have countably compact subspace of $\beta\omega$ in which any subset of cardinality $\mathfrak{c}$ is non-compact ? $\endgroup$ – Mathieu Baillif May 6 '18 at 13:57
  • $\begingroup$ Infinite subsets of $\beta \omega$ with size $\mathfrak{c}$ are always non-compact, because every infinite closed subset of $\beta \omega$ has cardinality $>\!\mathfrak{c}$ (more precisely, it has cardinality $2^\mathfrak{c}$, and it even contains a topological copy of $\beta \omega$). $\endgroup$ – Will Brian May 6 '18 at 14:04
  • $\begingroup$ Thanks for your answer. I have to get more used to $\beta\omega$ to avoid asking trivial questions in the future. $\endgroup$ – Mathieu Baillif May 6 '18 at 14:07
  • $\begingroup$ No worries -- $\beta \omega$ takes some getting used to. $\endgroup$ – Will Brian May 6 '18 at 14:19

The answer is no, because every infinite closed subset of $\beta \omega$ has cardinality $2^\mathfrak{c}$. So for example, if $\{x_1,x_2,\dots\}$ is any countably infinite set of non-weak-$P$-points, then its closure $X$ contains no weak $P$-points either. Any $Y \subseteq X$ of cardinality $\mathfrak{c}$ provides a negative answer to the question.


To get a countably compact space $X$ such that there is a continuous surjection $f \colon X \to [0,1]$ but if $C$ is a compact subset of $X$, $f(C) \ne [0,1]$, let $X$ be a Bernstein-type set in $\omega^*$, that is, a subset of $\omega^*$ such that neither $X$ nor $\omega^* \setminus X$ contains an infinite compact set. (This can be done by mimicking the construction of a Bernstein set in $\mathbb R$: Well-order the infinite compact subsets of $\omega^*$ and inductively choose pairs of points one of which is in $X$ and the other of which is not in $X$.) Then $X$ is countably compact because any infinite subset has a limit point in $X$--all limit points cannot be in $\omega^* \setminus X$. It follows from the fact that fibers of continuous functions from $\omega^*$ to $[0,1]$ have non-empty interiors that if $F$ maps $\omega^*$ onto $[0,1]$, then the restriction $f$ of $F$ to $X$ maps $X$ onto $[0,1]$. Finally, since no compact subset of $X$ is infinite, no compact subset of $X$ maps onto $[0,1]$ by any function.

  • $\begingroup$ Thanks for this idea. I was thinking of building such an example by induction, starting with a function $f:\omega\to[0,1]$ with a dense image, extend it to all of $\beta\omega$, pick one guy in the preimage of each point. This defines $X_0$. To go from $X_\alpha$ to $X_{\alpha+1}$, choose one point in the closure of each countable set of $X_\alpha$ (if it does not already have one in $X_\alpha$). If $\alpha$ is limit, take the union of the $X_\beta$ below it. Then $X_{\omega_1}$ is countably compact and of cardinality $\mathfrak{c}$. But your example is even better. $\endgroup$ – Mathieu Baillif May 6 '18 at 15:04

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