I am just curious: is there a published proof of the compactness of the [Hilbert cube](http://en.wikipedia.org/wiki/Hilbert_cube) that does not use the Axiom of Choice, or is it well known?



Compactness of the Hilbert cube without the Axiom of Choice
I have found this paper by Peter Loeb:


* Peter A. Loeb, [*A new proof of the Tychonoff Theorem*](http://www.jstor.org/stable/2314411), The American Mathematical Monthly **72** (1965), no. 7, 711--717.


Here is the theorem from this paper that implies that the usual Hilbert cube is compact without using the AC.



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> **Theorem 1.** Let $\{\,X\_\nu\mid\nu\in I\,\}$ be a family of compact spaces which is indexed by a set $I$ on which there is a well-ordering $\ge$. If $I$ is an infinite set, let there also be a choice function $F$ on the collection $\{\,C\mid\text{$C$ is closed},\ C\ne\varnothing,\ \text{$C\subset X\_\nu$ for some $\nu$}\,\}$. Then the product space $\prod\_{\nu\in I}X\_\nu$ is compact in the product topology.
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For the usual Hilbert cube $[0,1]^{\mathbb N}$, the function $F$ can, for example, select the least element in every compact subset of $[0,1]$.


