Images of $\{0,1\}^\kappa$ 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.)
 A: 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).
A: 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).
A: 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.
