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Is there a compact topological space $(X,\tau)$ such that for no cardinal $\kappa$ there is a surjective continuous map $e:\{0,1\}^\kappa \to X$?

(We assume that $\{0,1\}$ is endowed with the discrete topology, and $\{0,1\}^\kappa$ has the product topology.)

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    $\begingroup$ It is worth noting that every compact Hausdorff space is the continuous image of a closed subset of some product of $\{0,1\}$. You can find this in standard books; in particularly, Kelley's "General Topology". $\endgroup$ – Bill Johnson Mar 18 '15 at 15:34
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    $\begingroup$ (To a now-deleted response) that cannot be right. Every subspace of $\{0,1\}^\kappa$ is zero-dimensional. As Bill says, every compact Hausdorff space is a continuous image of a zero-dimensional space and each zero-dimensional space $X$ embeds into $\{0,1\}^\kappa$ (for $\kappa$ equal to the weight of $X$). See Engelking's General topology. $\endgroup$ – Tomasz Kania Mar 18 '15 at 16:51
  • $\begingroup$ Note that any compact metric space is a continuous image of Cantor´s space $2^\omega$. $\endgroup$ – Ramiro de la Vega Mar 20 '15 at 14:44
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It is well known that the space $\{0,1\}^{\kappa}$ satisfies the countable chain condition. Recall that a topological space $X$ satisfies the countable chain condition if and only if every collection $\mathcal{A}$ of pairwise disjoint open sets is countable. However, it is easy to show that the surjective continuous image of a space satisfying the countable chain condition must also satisfy the countable chain condition, so the only possible spaces which are images of some $\{0,1\}^{\kappa}$ satisfy the countable chain condition. However, there are plenty of compact Hausdorff spaces that do not satisfy the countable chain condition such as $\beta\mathbb{N}\setminus\mathbb{N}$ or $[0,1]\times[0,1]$ with the order topology inherited from the lexicographic ordering.

To see that $\{0,1\}^{\kappa}$ satisfies the countable chain condition, endow $\{0,1\}^{\kappa}$ with the infinite product measure $m$ where each $\{0,1\}$ is given the measure $\mu$ such that $\mu(\{0\})=\frac{1}{2}$ and $\mu(\{1\})=\frac{1}{2}$. Then $m$ extends to a Borel measure $\overline{m}$ on $\{0,1\}^{\kappa}$ where $\overline{m}(U)>0$ for each non-empty open set $U$. However, there cannot be an uncountable collection $\mathcal{A}$ of disjoint open subsets of $X$ since the union of any uncountable pairwise disjoint collection of open subsets of $X$ would have infinite measure.

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  • $\begingroup$ Very nice! Small typo: "surjective image" $\to$ "surjective continuous image". $\endgroup$ – Joel David Hamkins Mar 18 '15 at 13:26
  • $\begingroup$ Yes, very good. While you're at it, there's also a strangely placed question mark for what looks like a declarative sentence. $\endgroup$ – Todd Trimble Mar 18 '15 at 13:35
  • $\begingroup$ I corrected the typos. $\endgroup$ – Joseph Van Name Mar 18 '15 at 14:01
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To elaborate on Joseph's answer, the class of continuous images of Cantor cubes has a fancy name, they are the so called dyadic spaces. There is a nice result by Haydon: every Dugundji space is dyadic. (A space $X$ is Dugundji if the conclusion of the Borsuk--Dungundji theorem holds for $X$.)

R. Haydon, On a problem of Pełczyński: Milutin spaces, Dugundji spaces and AE(0-dim), Studia Math. 52 (1974), 23-31.

It is easy to see that the conclusion of the Borsuk--Dugundji theorem fails for $\beta \mathbb{N}$ (it is actually a paradigm counter-example).

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  • $\begingroup$ I´m not sure what your definition of "Dugundji space" is. The Borsuk-Dugundji theorem that I know talks about a metric space $X$ and a closed subspace $A\subseteq X$. Dugundji spaces are compact (so not every metric space is Dugundji) and the Cantor cubes $2^\kappa$ are all Dugundji but for $\kappa>\omega$ they don´t satisfy Borsuk-Dugundji. $\endgroup$ – Ramiro de la Vega Mar 20 '15 at 14:43
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I'd say that the simplest counter-example is the $1$-point compactification of a non-countable discrete space (of course any compactification of any non-countable discrete space would do).

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