Question

I am just curious: is there a published proof of the compactness of the Hilbert cube that does not use the Axiom of Choice, or is it well known?

Answer 1

If by the Hilbert cube you mean only $[0,1]^\mathbb N$ then the answer is yes. There is such proof, you can find it in Herrlich's The Axiom of Choice as Theorem 3.13.

If you mean the general case of $[0,1]^I$ then the answer is no, to prove that all Hilbert cubes are compact is equivalent to BPIT/The ultrafilter lemma/Tychonoff for Hausdorff spaces. The proof of that you can find in the same book as Theorem 4.70.

Answer 2

The compactness of the Hilbert cube follows (without AC) from the compactness of $2^\omega$, since $[0,1]$ as well as $[0,1]^\omega$ are continuous images of $2^\omega$.

(Conversely, $2^\omega$ is a closed subset of the Hilbert cube.)

The compactness of $2^\omega$ is just König's lemma for trees of binary sequences, which is easy to prove (hence certainly well-known) without AC. (I think this is called "weak König's lemma", an important principle in reverse mathematics.)

Answer 3

Some of the comments in Goldstern's answer look like they express doubt as to whether choice is required. Here is a proof without choice, in gory detail, just to make sure. The trick is to notice that the construction of an infinite branch $\alpha$ in an infinite binary tree $T$ requires no appeal to the axion of choice because we can specify a concrete choice: go left if you can, otherwise go right.

The Hilbert cube is a continuous image of the Cantor space $2^\omega$ of infinite binary sequences with the product topology. Thus it suffices to show that $2^\omega$ is compact. Given a finite binary sequence $a = [a_1, \ldots, a_n]$, denote by $|a| = n$ its length, and let $B_a = \lbrace \alpha \in 2^\omega \mid a = [\alpha_1, \ldots, \alpha_{|a|}] \rbrace$ be the basic open subset of those sequences that start with $a$.

Consider any cover $(B_{a_i})_{i \in I}$ of $2^\omega$. We build a binary tree $T$ which consists of those finite binary sequences $a$ for which $B_a$ is not contained in any $B_{a_i}$, $$T = \lbrace a \in 2^{*} \mid \forall i \in I . B_a \not\subseteq B_{a_i} \rbrace.$$ In other words, we put in $T$ any finite sequence $a$ such that all of its prefixes are not in $(a_i)_{i \in I}$. Let us show that $T$ has bounded height, i.e., there is $n$ such that every branch in $T$ has lebgth at most $n$.

Suppose on the contrary that the height of $T$ is unbounded. Then we can build an infinite path $\alpha$ in $T$ by recursion as follows. (This is König's lemma, saying that an unbounded binary tree has an infinite path.) We make sure that at each stage $n$ the subtree of $T$ at $[\alpha_1, \ldots, \alpha_n]$ has unbounded height. Start with the empty sequence $[]$. The tree at $[]$ is all of $T$, which has unbounded height by assumption. If $[\alpha_1, \ldots, \alpha_n]$ has been constructed, let $T'$ be the subtree of $T$ at $[\alpha_1, \ldots, \alpha_n]$. One or both of the trees $$T_0' = \lbrace b \in T' \mid b_{n+1} = 0 \rbrace$$ and $$T_1' = \lbrace b \in T' \mid b_{n+1} = 1 \rbrace$$ have unbounded height. If $T_0'$ does, set $\alpha_{n+1} = 0$, otherwise set $\alpha_{n+1} = 1$. (At this point we did not appeal to the axiom of choice, but we did appeal to excluded middle.) This concludes the construction of $\alpha$. Now we have a problem since $\alpha$ is covered by some $B_{a_i}$, and so $a_i$ is a prefix of $\alpha$, but this contradicts the definition of $T$.

Now we know that the height of $T$ is bounded by some $n$. Consider the subset $J \subseteq I$ of those indices $j \in I$ for which $|a_j| \leq n + 1$. As there are only finitely many binary sequences of length at most $n+1$, the set $J$ is finite. But since every sequence of length $n+1$ has some $j \in J$ such that $a_j$ is its prefix, $(B_{a_j})_{j \in J}$ is a finite cover of $2^\omega$.

Answer 4

Thanks for all the answers and sorry about a silly question. I have also figured out that it can be proved using the usual complete metric on the usual (countable product) Hilbert cube and finite $\epsilon$-nets.

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Update. Here is a proof.

Let $X = [-\frac{1}{2},\frac{1}{2}]\times[-\frac{1}{4},\frac{1}{4}]\times[-\frac{1}{8},\frac{1}{8}]\times\dotsb$ be a Hilbert cube endowed with its $\ell_\infty$ metric. For every positive integer $k$, let $N_k$ be the “natural” $\frac{1}{2^k}$-net for $X$.

Let $\mathcal U$ be a given family of open sets such that no finite subfamily of $\mathcal U$ covers $X$. Then let $S_k$ be the set of those elements of $N_k$ which are within the distance of $\frac{1}{2^k}$ from the complement of any finite union of elements of $\mathcal U$. Each $S_k$ is nonempty. For all $m$ and $n$, the distance from any point of $S_m$ to the set $S_n$ is at most $\frac{1}{2^m}+\frac{1}{2^n}$.

Assume that the points of $X$ are ordered by the lexicographic order of their coordinates. Take the “first” $x_1\in S_1$ (i.e. the smallest in the order), then the “first” $x_2\in S_2$ that is within the distance of $\frac{3}{4}$ from $x_1$, then the “first” $x_3\in S_3$ that is within the distance of $\frac{3}{8}$ from $x_2$, and so forth. The obtained sequence $\lbrace x_k \rbrace_{k=1}^\infty$ is Cauchy. Its limit is not in any element of $\mathcal U$.