A proof of Tychonoff Theorem implies AC

Topology Atlas Document # paca-01

A proof of Tychonoff Theorem implies AC

Henno Brandsma

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It is well known that AC¹ is used in all proofs of the Tychonoff theorem "A product of compact spaces is compact". See any textbook. Most will use ultrafilters of closed sets, but others use nets or the Alexander Subbase lemma followed by full AC. (Alexander Subbase Lemma itself is equivalent to Tychonoff Theorem for finite spaces, maybe more about this later). But is this necessary? Kelly first proved that it is: The full Tychonoff theorem implies full AC, and the argument is clean and simple, and nice to know.

Here it goes:

I'll prove the following standard formulation of AC: If S_i (i in I) is a (non-empty) family of non-empty sets then S : = P_{i in I} S_i is non-empty.

We'll add a default element to each set S_i, which we call p_i. Do it such that p_i is not already in S_i (for each i in I) (find one element p not in the union of the sets S_i or S_i ×{i}, and put p_i = p ×{i}, for the formalists). Let X_i = S_i union {p_i}, and give X_i the topology {X_i, S_i, {p_i}, {}}. This topology is finite, so X_i is certainly compact. (note: so even Tychonoff's theorem for spaces with finitely many open sets, a very small subclass of compact spaces, already implies AC!) So we know that X : = P_{i in I} X_i is compact.

Now define U_j = {(x_i) in X : x_j in p_j (or x_j = p_j)} for each j in I. This is a (sub)basic open subset of X: we have specified an open set in coordinate j and nothing in other coordinates.

Claim: no finite subfamily of the (U_j)_{j in
I} covers X.

Why: let U_j₁,...,U_{j_n} be a finite subfamily. Define a point x in X as follows: for each of the j₁,...,j_n, pick a point x_{j_i} in S_{j_i} (i=1,...,n). This can be done because we are making finitely many choices from non-empty sets. This does not need choice (only infinitely many choices need AC). For j not in {j₁,...,j_n} let x_j = p_j (use the default choice that we provided ...). Then this is a well-defined point of X such that x in U_{j_i} (for i=1,...,n) because x_{j_i} is not p_{j_i} there. So x is not in the union of the U_{j_i}.

But this means that {U_j: j in I} is not a cover of X. (X is compact, so if it were a cover it would have a finite subcover, but the claim says that this is not the case.) So there is a point q that is not in the union of all the U_i. So for this q: q_i is not equal to p_i, or for all i in I: q_i in S_i (as X_i = S_i union {p_i}) or q in P_{i in I} S_i, as required: this product of non-empty sets is non-empty, and AC is proved.

¹AC is the Axiom of Choice

Date: March 1, 2003