4
$\begingroup$

Tychonoff's theorem:$\phantom{---}$ If $A$ is a non-empty index set and $X_{\alpha}$ is a non-empty compact topological space for every $\alpha\in A$, then $X\equiv\times_{\alpha\in A} X_{\alpha}$ is compact in the product topology.

This theorem is well-known to be equivalent to the axiom of choice.

Converse of Tychonoff's theorem:$\phantom{---}$ If $X$ is not empty and compact in the product topology, then $X_{\alpha}$ is compact for every $\alpha\in A$.

Proof:$\phantom{---}$ Pick any $\alpha\in A$ and $x_{\alpha}\in X_{\alpha}$. For any other $\beta\in A\setminus\{\alpha\}$, pick an arbitrary $x_{\beta}\in X_{\beta}$. Construct $\mathbf x\in X$ such that $\pi_{\alpha}(\mathbf x)=x_{\alpha}$ and $\pi_{\beta}(\mathbf x)=x_{\beta}$ for any $\beta\in A\setminus\{\alpha\}$, where the $\pi_{\cdot}(\cdot)$ denote the coordinate maps. It follows that $x_{\alpha}\in \pi_{\alpha}(X)$, and, in turn, that $X_{\alpha}=\pi_{\alpha}(X)$. Since the coordinate maps are continuous by the very construction of the product topology and $X$ is compact, $X_{\alpha}$ is compact.$\phantom{---}\blacksquare$

Notice that this easy proof makes use of the axiom of choice, which guarantees the existence of $\mathbf x$. What I am wondering about is whether the converse of Tychonoff's theorem can be proved in ZF without AC (or weaker variants of it)?

Thank you.

$\endgroup$
  • 1
    $\begingroup$ Maybe you want to require that $X$ is non-empty. Otherwise there's an obvious dependence on the axiom of choice. $\endgroup$ – Asaf Karagila Aug 13 '14 at 5:03
  • $\begingroup$ @AsafKaragila I tacitly assume that non-emptiness is part of the definition of a topological space. But you're right, this tacitness may well generate misunderstandings, so I'll include it as an explicit assumption. $\endgroup$ – triple_sec Aug 13 '14 at 5:05
2
$\begingroup$

If $X$ is non-empty, then there is no dependence on the axiom of choice. To see this, note that $X_\alpha$ is a continuous map of $X$ with the projection map $\pi=\pi_\alpha(x)=x_\alpha$.

This follows from the fact that if a product $X=\prod_{i\in I}X_i$ is non-empty, then for each $x\in X_i$ there is a function $f\in X$ with $f(i)=x$. Simply pick one element $g\in X$, which exists since $X$ is non-empty, then for each $x\in X_i$ define $f_x=(g\setminus\{\langle i,g(i)\rangle\})\cup\{\langle i,x\rangle\}$.

This $\pi$ is continuous since given $U\subseteq X_\alpha$ which is open, the preimage of $U$ is the product of $U_i$'s where $U_\alpha=U$ and $U_i=X_i$ for $i\neq\alpha$.

And one can easily show that the continuous image of a compact space is compact without using the axiom of choice.


If you allow $X$ to be empty, this is of course equivalent to the axiom of choice. Pick any infinite family of infinite sets whose product is empty, and consider them with discrete topologies.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It is clear that $\pi_{\alpha}(X)\subseteq X_{\alpha}$ for each $\alpha\in A$. What is not (to me) is whether this set inclusion goes actually both ways even without choice. I.e., is every $x_{\alpha}\in X_{\alpha}$ the image of some $\mathbf x\in X$ under $\pi_{\alpha}$? $\endgroup$ – triple_sec Aug 13 '14 at 5:09
  • 1
    $\begingroup$ Pick any element in $X$, then you can modify the $\alpha$-th coordinate as you wish. You only change one coordinate, so there is no appeal to the axiom of choice. $\endgroup$ – Asaf Karagila Aug 13 '14 at 5:10
  • $\begingroup$ Aha, I see! I wanted to construct $\mathbf x$ component-wise (which does require the axiom of choice), while you're saying that we can just pick any element of $X$ (which doesn't require choice given that $X\neq\varnothing$) and modify one coordinate at a time. This makes sense, thank you! As for the non-emptiness of $X$, I don't worry about it—to me, the very concept of an empty topological space makes no sense to me to begin with. $\endgroup$ – triple_sec Aug 13 '14 at 5:16
  • 2
    $\begingroup$ Well, I included this for completeness anyway. One can argue that an empty topological space is necessary since otherwise you can't quite talk about product spaces freely in the absence of AC, and you always have to qualify "If there is a product space bla bla bla". $\endgroup$ – Asaf Karagila Aug 13 '14 at 5:20
  • $\begingroup$ That's a good point. I guess I'm so accustomed to AC making it possible to assume that any topological space is not empty per definitionem without confusion that I never realized that you'd have to be careful about this once you stop, even for a minute, taking AC for granted. $\endgroup$ – triple_sec Aug 13 '14 at 5:23

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.