I need the following specific version of Tychonoff's theorem:

Suppose $\{X_\alpha\}_\alpha$ is a collection of finite sets endowed with discrete topologies, then $\prod_\alpha X_\alpha$ is compact.

I wonder whether there's any direct proof for this specific version, since it's pervasive in mathematics (other than general topology), say the theory of pro-finite groups where infinite Galois theory is based. Many existence theorem is deduced from compactness.

I wonder whether there are more direct proofs (but Zorn's lemma is allowed) for this special case. I don't like base many existence theorems (especially some existence of a specific morphism between extension fields) on the black box of Tychonoff's theorem.

For example, suppose $A$ is an integrally closed domain with quotient field $F$, and $K/F$ is a normal extension. Let $B$ be the integral closure of $A$ in $K$. Given a prime ideal $P\subsetneq A$, we want to show that $\operatorname{Gal}(K/F)$ acts transitively on prime ideals lying over $P$. We can first prove it for finite extension $K/F$, and deduce the general case from this special case by compactness of $\operatorname{Gal}(K/F)$, see Serre's Local Algebra, say. However, I want to see how this Galois element is constructed without appealing to Tychonoff's theorem.

Any help is welcome. Thanks!

  • 1
    $\begingroup$ What would you consider a (more) direct proof? Note that this statement is equivalent to the Boolean Prime Ideal Theorem. $\endgroup$ – user642796 Dec 17 '15 at 20:20
  • $\begingroup$ @ArthurFischer In some sense, the proof of existence of maximal ideals for commutative rings is very intuitive, and existence of algebraic closure, and existence of extension to splitting fields. I need to use it to extract something. I'll edit my post to include an example. $\endgroup$ – Yai0Phah Dec 17 '15 at 20:35
  • $\begingroup$ This is an exercise in futility. Tychonoff's theorem is equivalent to Zorn's lemma. $\endgroup$ – Matt Samuel Dec 18 '15 at 6:43

It seems possibly interesting to note that the compactness of an arbitrary product of finite sets gives a "direct" proof of the compactness of a product of arbitrary compact Hausdorff spaces (hence, regardless of what we mean by "direct" in the OP, the result for finite sets can't be much easier than the Tychonoff theorem itself.)

Since the Cantor set $K$ is a countable product of $\{0,1\}$, the compactness of $\prod X_\alpha$ for $X_\alpha=\{0,1\}$ implies the compactness of the product for $X_\alpha=K$. Now there is a continuous map from $K$ onto $[0,1]$, so we get the result for $X_\alpha=[0,1]$. And any compact Hausdorff space can be embedded in a (high-dimensional) cube, hence if $X_\alpha$ is compact Hausdorff then $\prod X_\alpha$ is a closed subset of a compact Hausdorff space.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.