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Here is mine. It's taken from page 11 of "An Introduction To Abstract Harmonic Analysis", 1953, by Loomis:

https://archive.org/details/introductiontoab031610mbp

https://ia800309.us.archive.org/10/items/introductiontoab031610mbp/introductiontoab031610mbp.pdf

(By the way, I don't know why this book is not more famous.)

To prove that a product $K=\prod K_i$ of compact spaces $K_i$ is compact, let $\mathcal A$ be a set of closed subsets of $K$ having the finite intersection property (FIP) --- viz. the intersection of finitely many members of $\mathcal A$ is nonempty ---, and show $\bigcap\mathcal A\not=\varnothing$ as follows.

By Zorn's Theorem, $\mathcal A$ is contained into some maximal set $\mathcal B$ of (not necessarily closed) subsets of $K$ having the FIP.

The $\pi_i(B)$, $B\in\mathcal B$, having the FIP and $K_i$ being compact, there is, for each $i$, a point $b_i$ belonging to the closure of $\pi_i(B)$ for all $B$ in $\mathcal B$, where $\pi_i$ is the $i$-th canonical projection. It suffices to check that $\mathcal B$ contains the neighborhoods of $b:=(b_i)$. Indeed, this will imply that the neighborhoods of $b$ intersect all $B$ in $\mathcal B$, hence that $b$ is in the closure of $B$ for all $B$ in $\mathcal B$, and thus in $A$ for all $A$ in $\mathcal A$.

For each $i$ pick a neighborhood $N_i$ of $b_i$ in such a way that $N_i=K_i$ for almost all $i$. In particular the product $N$ of the $N_i$ is a neighborhood of $b$, and it is enough to verify that $N$ is in $\mathcal B$. As $N$ is the intersection of finitely many $\pi_i^{-1}(N_i)$, it even suffices, by maximality of $\mathcal B$, to prove that $\pi_i^{-1}(N_i)$ is in $\mathcal B$.

We have $N_i\cap\pi_i(B)\not=\varnothing$ for all $B$ in $\mathcal B$ (because $b_i$ is in the closure of $\pi_i(B)$), hence $\pi_i^{-1}(N_i)\cap B\not=\varnothing$ for all $B$ in $\mathcal B$, and thus $\pi_i^{-1}(N_i)\in\mathcal B$ (by maximality of $\mathcal B$).


Many people credit the general statement of Tychonoff's Theorem to Cech. But, as pointed out below by KP Hart, Tychonoff's Theorem seems to be entirely due to ... Tychonoff. This observation was already made on page 636 of

Chandler, Richard E.; Faulkner, Gary D. Hausdorff compactifications: a retrospective. Handbook of the history of general topology, Vol. 2 (San Antonio, TX, 1993), 631--667, Hist. Topol., 2, Kluwer Acad. Publ., Dordrecht, 1998

https://books.google.com/books?id=O2Hwaj2SqigC&lpg=PA636&ots=xjvA9nwlO5&dq=772%20tychonoff&pg=PA636#v=onepage&q&f=false

The statement is made by Tychonoff on p. 772 of "Ein Fixpunktsatz" (DOI: 10.1007/BF01472256, eudml) where he says that the proof is the same as the one he gave for a product of intervals in "Über die topologische Erweiterung von Räumen" (DOI: 10.1007/BF01782364, eudml).


Screenshot added to answer a comment of ACL:

enter image description here

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    $\begingroup$ The SMBC argument: "Suppose that Tychonoff's theorem is not provable. All those other proofs of Tychonoff's theorem must be wrong. But they are not wrong. Therefore Tychonoff's theorem is proved". $\endgroup$
    – Asaf Karagila
    Commented Feb 15, 2021 at 23:52
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    $\begingroup$ The link to iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Tycho seems to be dead - I guess you're in the best position to say whether the same text is now available somewhere else. (The springerlink links do not work either - but that's basically a non-issue, since the title of the paper is explicitly mentioned.) $\endgroup$ Commented Nov 27, 2021 at 7:26
  • $\begingroup$ @MartinSleziak - Thanks! I removed the link because it was just a repetition of the argument given in the question. $\endgroup$ Commented Nov 27, 2021 at 12:33

19 Answers 19

45
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Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

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    $\begingroup$ After a closer look at Loomis' one, I agree: in the substance it's the same as H.Cartan (it seems to me your're right with the attribution to Cartan, but I'm not certain either). But, I think here's an important issue: to give or not to give a name to the objects one uses? Sometimes there is no need at all (you know those papers with definitions of weird objects that only enter once in a proof). In this case, I think the proof greatly gains in semplicity introducing the notion of filter. I would even say, the notion of filter is the most important byproduct of the compactness theorem. $\endgroup$ Commented May 30, 2010 at 8:28
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    $\begingroup$ Actually, the axiom of choice is used twice in the proof. First, you have to use it to characterize compactness by ultrafilters. For this, you do not need the full axiom of choice, the boolean prime ideal theorem suffices. The second application is in picking for each coordinate a point the projection converges to. Here you need the full axiom of choice. For Hausdorff spaces, you don't need the second part though, because than a filter cannot converge to different points. $\endgroup$ Commented May 30, 2010 at 8:55
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    $\begingroup$ I also like Cartan's ultrafilter proof best. As people have said, all the cleverness is hidden in the definition of ultrafilters. With this in hand, Tychonoff's theorem becomes a consequence of a few very straightforward facts about filters on products. As to whether it's better to hide the ultrafilters or not, I think the test should be whether ultrafilters are of any use outside of this specific context, the answer to which is of course YES! Thus I would recommend this proof even to someone who doesn't already know about ultrafilters -- learning about them is good in and of itself. $\endgroup$ Commented May 30, 2010 at 9:11
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    $\begingroup$ Dear Pietro, Pete, Spencer: In "L'intégration dans les groupes topologiques" Weil explains that his proof of the existence of a Haar measure on locally compact groups was obtained by hiding filters. I think the situations are very similar. (And I find Weil's proof incredibly beautiful - for the ideas and the style.) $\endgroup$ Commented May 30, 2010 at 10:50
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    $\begingroup$ @Pierre-Yves: Do you know what Weil could mean with a proof of the existence of a Haar measure where filters are not hidden? Thank you in advance. $\endgroup$
    – ACL
    Commented Nov 19, 2012 at 15:12
24
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Since all of the answers to this question (except the one involving Alexander's subbase lemma) refer to a usually strange rehashing of the ultrafilter proof (BOO), I decided to give two nice proofs to Tychonoff's theorem here for Hausdorff spaces.

The first proof of Tychonoff's theorem for Hausdorff spaces uses the Stone-Cech compactification. This proof is useful when one constructs the Stone-Cech compactification before Tychonoff's theorem.

Proof: Assume that $X_{i}$ is compact for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ be the product space. Then each projection $\pi_{i}:X\rightarrow X_{i}$ extends to a continuous map $\overline{\pi_{i}}:\beta X\rightarrow X_{i}$ since each $X_{i}$ is compact. Therefore the map $f:\beta X\rightarrow X$ where $f(x_{i})_{i\in I}=(\overline{\pi_{i}}(x))_{i\in I}$ is a continuous surjection, so $X$ is compact being the continuous surjective image of $\beta X$. QED

For the second proof we use the following facts about uniform spaces that every mathematician should be aware of.

i. Every compact Hausdorff space has a unique compatible uniformity and that uniformity is complete and totally bounded.

ii. If a uniform space is complete and totally bounded, then it is compact.

Tychonoff's theorem then immediately follows from the fact that the product of complete uniform spaces is complete and that the product of totally bounded uniform spaces is totally bounded. And this proof is intuitive because it is easier to imagine that the product of complete and totally bounded uniform spaces is complete and totally bounded than to imagine that the product of compact spaces is compact.

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    $\begingroup$ Is there a good reason that this answer was downvoted? Is it because these proofs only work for Hausdorff spaces or is it because it is a sin to construct the Stone-Cech compactification without Tychonoff's theorem? $\endgroup$ Commented Sep 9, 2013 at 14:50
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I like the proof from Alexander's subbase lemma. E.g. A proof here. That lemma also gives the compactness criterion in ordered spaces (completeness implies compactness).

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  • $\begingroup$ Thank you for your answer. I noticed that Walter Rudin gives this proof in his "Functional Analysis". $\endgroup$ Commented May 31, 2010 at 4:51
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My favorite proof is the one from Johnstone's Stone spaces for locales because it works without the axiom of choice.

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    $\begingroup$ Tychonoff and AC are equivalent in ZF. $\endgroup$ Commented May 30, 2010 at 8:07
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    $\begingroup$ @Martin: Tychonoff's theorem for locales does not require any form of the axiom of choice. However, the categories of spatial locales and sober topological spaces are equivalent only if AC is true. $\endgroup$ Commented May 30, 2010 at 8:39
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    $\begingroup$ No, the categories of spatial locales and sober spaces are always equivalent. The reason that Tychonoff for locales doesn't imply Tychonoff for spaces is that the locale product of a family of spatial locales may no longer be spatial. $\endgroup$ Commented May 31, 2010 at 2:41
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    $\begingroup$ @Mike: Yes, you are right. What I actually meant is that equivalence of categories of compact regular locales and compact regular (or Hausdorff) topological spaces requires some form of AC. $\endgroup$ Commented May 31, 2010 at 8:03
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    $\begingroup$ Minimally - one may reduce appeals to AC to the fact that any compact regular locale is spatial. $\endgroup$ Commented May 7, 2017 at 5:39
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Here is Tychonoff's original proof, for powers of the unit interval. He builds a complete accumulation point of a given infinite set by transfinite recursion along the index set. On page 772 of this paper one finds the formulation of the general theorem (in my translation): "The product of compact spaces is again compact. One proves this theorem word for word as in he case of the compactness of the product of intervals". Some authors (Folland, see comment below and Walter Rudin in his `Functional Analysis') credit Čech with proving the general result but Čech's proof is the same as Tychonoff's and, based on a reading of his papers, I think Tychonoff deserves full credit for the theorem and its proof.

@Henno: not Fundamenta but Mathematische Annalen.

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  • $\begingroup$ Dear KP Hart: Thank you for your answer. (It contains at least one typo ("the the proof").) --- Folland (Real Analysis) claims that the general statement is due to Cech (On bicompact spaces, Ann. of Math. 38 (1937), 823-844. MR 1503374). What's your opinion? $\endgroup$ Commented May 30, 2010 at 16:45
  • $\begingroup$ Walter Rudin expressed the same opinion in his `Functional Analysis' but on pae 772 of the second paper that I linked to you'll find (in my translation): The product of compact spaces is again compact. This one proves word for word as in he case of the compactness of the product of intervals. I think Tychonoff deserves full credit for the theorem and its proof. The proof in \v{C}ech's paper is the same as Tychonoff's. $\endgroup$
    – KP Hart
    Commented May 30, 2010 at 19:01
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    $\begingroup$ Very interesting!!! Thank you! --- I was wondering if you couldn't insert the contents of your comment into your answer, to make them more visible. $\endgroup$ Commented May 31, 2010 at 7:01
  • $\begingroup$ Thank you! Two more suggestions: (1) Replace "This one proves word for word as in he case" by "One proves this theorem word for word as in the case". (2) To type Cech with the caron, copy and paste "Čech" (without the quotation marks). (It should work.) $\endgroup$ Commented May 31, 2010 at 10:45
  • $\begingroup$ Here are different links to Tychonoff's articles: Über die topologische Erweiterung von Räumen (springerlink.com/content/l656352441w67612/…). Ein Fixpunktsatz (springerlink.com/content/n61706447r886l58/…). $\endgroup$ Commented May 31, 2010 at 12:28
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I won't swear it's my absolute favorite, but today I learned of a nice proof due to Clementino and Tholen who take as their starting point the closed-projection characterization of compactness, viz. that a space $X$ is compact iff for every space $Y$, the projection $\pi: Y \times X \to Y$ is a closed map.

If this is assumed, then Tychonoff can be proved without much pain as follows.

Lemma: Let $(X_i)_{i: I}$ be a family of spaces. Then for a point $x$ and subset $A$ of $\prod_{i: I} X_i$, we have $x \in Cl(A)$ (the closure of $A$) if, for every finite $F \subseteq I$, we have $\pi_F(x) \in Cl(\pi_F(A))$ under the projection operator $\pi_F: \prod_{i: I} X_i \to \prod_{i: F} X_i$.

The proof is entirely routine and may be left to the reader.

Proof of Tychonoff: Let $(X_\alpha)_{\alpha \lt \kappa}$ be a family of compact spaces indexed by an ordinal $\kappa$. It is enough to show that the projection

$$Y \times \prod_{\alpha \lt \kappa} X_\alpha \to Y$$

is a closed map for any space $Y$. We do this by induction on $\kappa$. The case $\kappa = 0$ is trivial.

It will be convenient to introduce some notation. For $\gamma \leq \kappa$, let $X^\gamma$ denote the product $Y \times \prod_{\alpha \lt \gamma} X_\alpha$ (so $X^0 = Y$ in this notation), and for $\beta \leq \gamma$ let $\pi_\beta^\gamma: X^\gamma \to X^\beta$ be the obvious projection map. Let $K \subseteq X^\kappa$ be closed, and put $K_\beta := Cl(\pi_{\beta}^\kappa(K))$. In particular $K_\kappa = K$ since $K$ is closed, and we are done if we show $\pi_0^\kappa(K) = K_0$.

Assume as inductive hypothesis that starting with any $x_0 \in K_0$ there is $x_\beta \in K_\beta$ for each $\beta < \kappa$ such that whenever $\beta \lt \gamma \lt \kappa$, the compatibility condition $\pi_\beta^\gamma(x_\gamma) = x_\beta$ holds. In particular, $\pi_0^\beta(x_\beta) = x_0$ for all $\beta \lt \kappa$, and we are now trying to extend this up to $\kappa$.

If $\kappa = \beta + 1$ is a successor ordinal, then the projection

$$\pi_\beta^\kappa: X^\beta \times X_\beta \to X^\beta$$

is a closed map since $X_\beta$ is compact. Thus $\pi_\beta^\kappa(K) = Cl(\pi_\beta^\kappa(K)) = K_\beta$ since $K$ is closed, so there exists $x_\kappa \in K$ with $\pi_\beta^\kappa(x_\kappa) = x_\beta$, and then

$$\pi_0^\kappa(x_\kappa) = \pi_0^\beta \pi_\beta^\kappa (x_\kappa) = \pi_0^\beta(x_\beta) = x_0$$

as desired.

If $\kappa$ is a limit ordinal, then we may regard $X^\kappa$ as the inverse limit of spaces $(X^\beta)_{\beta \lt \kappa}$ with the obvious transition maps $\pi_\beta^\gamma$ between them. Hence the tuple $(x_\beta)_{\beta \lt \kappa}$ defines an element $x_\kappa$ of $X^\kappa$, and all that remains is to check that $x_\kappa \in K$. But since $K$ is closed, the lemma indicates it is sufficient to check that for every finite set $F$ of ordinals below $\kappa$, that $\pi_F(x_\kappa) \in Cl(\pi_F(K))$ (as a subspace of $\prod_{\alpha \in F} X_\alpha$). But for every such $F$ there is some $\beta \lt \kappa$ that dominates all the elements of $F$. One then checks

$$\pi_F(x_\kappa) = \pi_F^\beta \pi_\beta^\kappa(x_\kappa) = \pi_F^\beta(x_\beta) \in \pi_F^\beta(K_\beta) = \pi_F^\beta(Cl(\pi_\beta^\kappa(K))) \subseteq Cl(\pi_F^\beta \pi_\beta^\kappa(K)) = Cl(\pi_F(K))$$

where the inclusion indicated as $\subseteq$ just results from continuity of $\pi_F^\beta$. This completes the proof. $\Box$

(More details at the nLab.)

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  • $\begingroup$ And this proof has the nice feature, it seems, of proving the theorem for well-ordered families in situations without Choice. Or are there other appeals to some choice principle in the lemma or the characterisation of compactness? Clearly getting the full theorem for arbitrary families from the above requires the well-ordering principle, as one would expect... $\endgroup$
    – David Roberts
    Commented May 7, 2017 at 11:24
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    $\begingroup$ There is choice in choosing a lift at successor stages. I think the only applications of AC are that and the well-ordering of the indexing. The characterization of compactness in terms of closed projections doesn't require AC and is something you can enact in bounded Zermelo set theory. For example, you don't need to invoke nonprincipal ultrafilters there. $\endgroup$
    – Todd Trimble
    Commented May 7, 2017 at 12:29
  • $\begingroup$ @ToddTrimble - Thanks Todd for this nice answer. I unaccepted it because I accepted it by mistake. I just wanted to upvote it on my iPad and didn't put my finger at right spot. Sorry! Again, I like your answer very much, but there are so many great answers that it would seem somewhat artificial to me say that one of them is better than the others. $\endgroup$ Commented May 7, 2017 at 12:31
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    $\begingroup$ No problem, Pierre-Yves! I agree that it's artificial to have to choose (I'm old-fashioned and really like the ultrafilter-type proofs for their overall conceptual simplicity). "L'embarras du choix" :-) $\endgroup$
    – Todd Trimble
    Commented May 7, 2017 at 12:40
  • $\begingroup$ @ToddTrimble: You say there is choice at successor stages, but that's just picking an element, right? It seems to me the actual choice (apart from the well-ordering) is in the limit step when you go from "there exists a bunch of $x_\beta$" to "there is a tuple $(x_\beta)_{\beta < k}$". Or am I missing something? $\endgroup$ Commented May 20, 2022 at 11:57
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My favorite is the proof via nets by Paul Chernoff. A VERY clever use of generalized convergence in point set topology! https://www.jstor.org/pss/2324485

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  • $\begingroup$ Thank you! Unfortunately I don't have access to JSTOR. Chernoff's proof is also in Folland's Real Analysis. Does anybody know a public link to Chernoff's paper? $\endgroup$ Commented May 30, 2010 at 7:47
  • $\begingroup$ @Pierre-Yves: Google "Tychonoff sets". @AnrewL: I agree, this is really the most intuitive proof. $\endgroup$ Commented May 30, 2010 at 8:18
  • $\begingroup$ @Pierre Yikes,I forgot,that's true,Folland DOES have the proof in his text! I learned the proof from Chernoff's original paper,which was required reading in John Terilla's point set topology course.Interestingly,John learned of the proof in James Stasheff's topology course when he was a graduate student at the University Of North Carolina. $\endgroup$ Commented May 30, 2010 at 15:49
  • $\begingroup$ This proof is also in Volker Runde's "A taste of topology". $\endgroup$ Commented Jul 8, 2011 at 21:55
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I'm surprised that nobody has mentioned the proof using universal nets. (It can be found, e.g., in Pedersen's 'Analysis NOW' and in Bredon's 'Topology and geometry'.)

A universal net in a set X is a net which, for every $Y\subset X$, ultimately lives in $Y$ or $X\backslash Y$. One easily sees that composition of a universal net in X with a function $f:X\rightarrow Y$ gives a universal net in $Y$. Using the ultrafiler lemma, one proves that every net has a universal subnet. All this involves no topology.

Combining the above with standard facts, the proof of Tychonov is extremely short. All one needs is: - a space is compact if and only if every net has a limit point (equiv., a convergent subnet), - a net in $\prod_iX_i$ converges if and only if it converges coordinate-wise.

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    $\begingroup$ This seems to be another way of "hiding" the ultrafilters in the ultrafilter proof. $\endgroup$ Commented Oct 12, 2012 at 15:29
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I have been teaching general topology for several years, but remained unsatisfied by the proofs given in the books that I based the course upon. Finally I wound up writing my own lecture notes, still not quite finished. In those notes, I give four different proofs. Two of them use (ultra)filters, but one of them avoids the terminology. The other two proofs use nets, namely Chernoff's proof without and Kelley's with universal nets.

The notes can be found at https://www.math.ru.nl/~mueger/topology.pdf (updated link; Wayback Machine)

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  • $\begingroup$ the link is dead... $\endgroup$ Commented May 24, 2021 at 14:39
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The non-standard analysis proof is an interesting "application" of the ultrafilter proof: a topological space $A$ is compact if and only if every point in the associated "non-standard topological space" ${}^*A$ is near-standard, that is to say, if and only if each $x \in {}^*A$ is contained in every open neighborhood of some standard point $y \in A$ ($i.e.,$ for all $U \subset A$, $U$ open and $y \in U$ implies $x \in {}^*U \subset {}^*A$).

So let $\mathcal{X}$ be a set of topological spaces indexed by $I$, and $P$, the product of these spaces; write ${}^\*P$ for the "non-standard product" of the set ${}^*\mathcal{X}$ of topological spaces indexed by ${}^*I$, and let $x \in {}^\*P$. It suffices to show that $x$ is near-standard.

For each $\kappa \in I$, let $x_\kappa \in {}^\*X_\kappa \in {}^*\mathcal{X}$ be the $\kappa$th factor of $x$. Then $x_\kappa$ is necessarily near-standard, because $X_\kappa \in \mathcal{X}$ is compact. But this means we can find a point $y \in P$ with factors $y_\kappa \in X_\kappa$ such that $U \subset X_\kappa$ open and $y_\kappa \in U$ implies $x_\kappa \in {}^*U \subset {}^*X_\kappa$, thus $V \subset P$ open and $y \in V$ implies $x \in {}^*V \subset {}^*P$. But this means $x$ is near-standard, so $P$ is compact.

"Under the hood," this is basically the ultrafilter proof (my favorite, to answer the original question), so the axiom of choice is required in more or less the same places: while the non-standard objects exist by the Boolean prime ideal theorem, "finding" the $y_\kappa$ in non-Hausdorff spaces requires the full axiom of choice.

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  • $\begingroup$ How do you define the associated "non standard space" to a standard space? $\endgroup$
    – mathahada
    Commented May 20, 2011 at 8:42
  • $\begingroup$ W.A.J. Luxembourg noted an interesting consequence of doing the proof this way. If the spaces are Hausdorff, then no additional application of choice is needed. Only the use of AC that produces the nonstandard model ... and for that the Boolean Algebra Maximal Ideal Theorem (strictly weaker than AC) suffices. $\endgroup$ Commented May 20, 2011 at 12:17
  • $\begingroup$ Wait a minute, now I'm confused. I have never heard of the Boolean Algebra Maximal Ideal Theorem, but if it implies Tychonoff in this proof without AC and if Tychonoff implies AC then how can the Boolean Theorem be strictly weaker than AC? $\endgroup$ Commented May 20, 2011 at 16:03
  • $\begingroup$ What I'm asking I guess is: Is Tychonoff for Hausdorff spaces equivalent to AC? It seems the answer must be no or else the Boolean Theorem would be equivalent to AC $\endgroup$ Commented May 20, 2011 at 16:05
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    $\begingroup$ Tychonoff for Hausdorff equals Boolean principal ideal Thm. The ultrafilter proof only needs Boolean principal ideal Thm in the Hausdorff case because you dont need to choose a limit point in each factor. $\endgroup$ Commented Feb 27, 2013 at 3:38
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This proof can be considered a variation of the proof using ultrafilters on $X$. I want mainly to point out that we can avoid transferring the ultrafilters through the projections if we use a slightly more general characterization of compactness using ultrafilters.$\newcommand{\FF}{\mathcal F}\newcommand{\UU}{\mathcal U}$

Definition. Let $X$ be a topological space, $x\in X$, $f\colon M\to X$ be a function and $\FF$ be a filter on $X$. Then we say that $x$ is an $\FF$-limit of $f$ iff for every neighborhood $U\ni x$ we have $$f^{-1}[U]\in\FF.$$

Basically, this definition says that $f^{-1}[U]$ has to be a "big set". (You can compare this with the definition of a limit of a sequence $f\colon\mathbb N\to X$ where $f^{-1}[U]$ has to be a cofinite set, i.e., it belongs to the Fréchet filter.)

Some references concerning this notion can be found in this post: Where has this common generalization of nets and filters been written down?

We can now characterize compactness in the following way

Fact. A topological space is compact if and only if for every function $f\colon M\to X$ and every ultrafilter $\UU$ on $M$ there exists an $\UU$-limit in $X$.

A proof of the "easy" implication can be found, for example, here: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. If presented in some introductory course, the proof of the fact that this characterizes compact spaces will probably depend on the facts which were already proven about compact spaces and (ultra)filters at this point.

Proof of Tychonoff theorem. Let $X=\prod\limits_{i\in I} X_i$ be a product of compact spaces. Suppose we have an ultrafilter $\UU$ on $M$ and a function $f\colon M\to X$. Then for each $i\in I$ there exists som $\UU$-limit of $p_i\circ f$ in the compact space $X_i$. Then the point $x$ determined by $p_i(x)=x_i$ is an $\UU$-limit of $f$ in $X$.

(In the proof, we have also used the fact $\FF$-limit in topological product corresponds to pointwise $\FF$-limits for each $i\in I$.)

Proof of Tychonoff's theorem along these lines is given, for example, in Dixmier's _General Topology (zbmath 0545.54001, MR753644) as Theorem 4.3.6. The whole proof is just a few lines - of course, that is related to the fact that it relies on a lot of things proved before that.

Dixmier

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Late to the party, but I think still worth mentioning as it seems to be relatively unknown:

The “simple proof” given by D, G. Wright is particularly nice, because

  1. It uses only the definition of compactness and is nevertheless rather short and simple.
  2. It has the same feature as Todd Trimble's proof above (and is under the hood somewhat similar): If the index set is well-ordered, the proof shows that the axiom of choice can be replaced by a corresponding (for the cardinality of the index set) principle of dependent choices. In particular, the proof shows that ZF+DC suffices to prove Tychonoff's theorem for countable products.
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I first learnt from Munkres' Topology. He gave a different motivation to use the maximal principle (Hausdorff's to be precise, but Zorn's work too) instead of the historic motivation to characterize compact spaces with a generalized version of "sequence"; i.e. filters.

What was Tychonoff's original proof? To me every proof seem to use some maximal principle; Alexander's subbase theorem also uses Zorn's lemma.

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    $\begingroup$ Every proof uses the axiom of choice, because Tychonoff is equivalent to AC. For the converse, that Tychonoff's theorem implies the axiom of choice, it's hard to beat the original proof by Kelley in Fundamenta Mathematicae 37 (1950) 75--76. It may be viewed here: matwbn.icm.edu.pl/ksiazki/fm/fm37/fm3716.pdf $\endgroup$ Commented May 30, 2010 at 6:31
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    $\begingroup$ The original proof used the characterization (of compactness) that every infinite set has a point of complete accumulation, and involved a transfinite recursion, IIRC. It's probably in Fundamenta as well. $\endgroup$ Commented May 30, 2010 at 6:35
  • $\begingroup$ Dear John Stillwell: As a bourbakist, the expression "axiom of choice" makes no sense to me. But I thank you very much for your comment. (Unrelated aside: Thank you very much also for your wonderful translation of Dirichlet! It changed my life!) Thank you to all contributors! $\endgroup$ Commented May 30, 2010 at 7:42
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    $\begingroup$ As Georges Elencwajg points out, there is a minor flaw in the original proof by Kelley that Tychonoff implies AC. Kelley uses the fact that the product of cofinite topologies is compact, a weaker statement equivalent to the theorem that every filter can be extended to an ultrafilter as shown here: math.vanderbilt.edu/~schectex/papers/kelley.pdf One can easily correct the original proof by Kelley by adding {Lambda} to the open sets, so the original proof is easily corrected. $\endgroup$ Commented May 30, 2010 at 13:59
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    $\begingroup$ I thought that Kelley's book (General Topology) had a correct proof of Tychonoff implies AC: let $\{X_i\}_{i \in I}$ be a family of nonempty sets. Put $Y_i = X_i \sqcup \{p\}$. Topologize $Y_i$ by taking the nontrivial open sets to be $X_i$ and $\{p\}$. Then $Y_i$ is compact; by Tychonoff, $Y = \prod_i Y_i$ is compact. For each $i$, put $K_i = \{y \in Y: y_i \in X_i\}$. Then $K_i$ is closed, and any finite intersection of the $K_i$ is nonempty (use $p$ in all but finitely many components). Hence $\prod_i X_i = \bigcap_\alpha K_i$ is nonempty as well, by compactness. Thus AC follows. $\endgroup$
    – Todd Trimble
    Commented May 7, 2017 at 17:52
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Personally, I've always enjoyed the proof given in Topology, by Hocking and Young. It's essentially the basic ultrafilter proof, but its got a nice feel to it. I guess I'm biased because this was the first real Topology book I was ever able to get my hands on.

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    $\begingroup$ In case it is useful for people who want to have a quick look, in the version I found on Google Books, Tychonoff's theorem is stated on page 25 and the proof is given (after some preparation) on page 28. $\endgroup$ Commented Nov 27, 2021 at 7:37
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A very short proof using nonstandard analysis in M. Machover, J.L. Bell, A Course in Mathematical Logic (1977),

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One can actually give a pretty simple direct proof from the open cover definition of compactness (self advert). I like this as it seems more transparent, without having to resort to (perhaps unfamiliar) machinery like ultrafilters

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  • $\begingroup$ The mentioned pretty simple direct proof also uses Zorn's Lemma. $\endgroup$
    – Paul Fabel
    Commented Sep 13, 2017 at 17:53
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    $\begingroup$ Well yes, it has to use the axiom of choice. One could phrase it as a transfinite recursion along the index set of the product otherwise $\endgroup$
    – McDuffin
    Commented Sep 14, 2017 at 9:00
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I guess that the proof from Munkres' book is essentially the same as this one due to Loomis.

For those who prefer the ultrafilter proof, well, indeed this is the very same proof, because "maximal families which satisfy the p.i.f." are ultrafilters, and the argument essentially proves that every ultrafilter on the product has a cluster point, and thus converges.

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Here is another proof using nonstandard analysis, this time using internal set theory (IST).

First, here is the most useful characterization of compactness in IST.

Theorem: a standard topological space $X$ is compact iff for every point $x \in X$, there is a standard point $y \in X$ such that $x$ is in every standard neighborhood of $y$.

Proof:
($\implies$) let $F$ be the standard set whose standard elements are the standard closed sets containing $x$ (this exists by axiom $S$). For every standard finite subset $S$ of $F$, we have $\cap S \neq \emptyset$. Therefore (by compactness and the transfer principle) $\cap F \neq \emptyset$. Let $y$ be a standard element of $\cap F$. For any standard open set $A$ containing $y$, $\bar A \notin F$, then $x \notin \bar A$, then $x \in A$.
($\impliedby$) Let $O$ be a standard open cover of $X$. Let $G$ be a finite subset of $O$ that contains all of $O$'s standard elements (this exists by axiom I). For any point $x \in X$, there is a standard point $y$ whose standard neighborhoods contain $x$. Since $y$ will be in a standard element of $O$ (by the definition of open cover and the transfer principle), so will $x$. But then $x$ is an element of $G$. So $G$ is a finite subcover of $O$. By the transfer principle, every open cover of $X$ (standard or not) has a finite open cover. $\square$


The above theorem is just a standard result (pun unintended) when setting up topology in IST, and is essentially a "definition" of compactness (but note that it only applies to standard spaces; if $\varepsilon$ is a nonzero infinitesimal, than $[-1,1] - \{\varepsilon\}$ satisfies the above condition but is not compact). Now we'll actually prove Tychonoff's theorem.

Tychonoff's theorem: for any index set $I$ and collection of compact topological spaces $X_{i \in I}$, we have that $\prod_{i \in I} X_i$ is compact.

Proof:

Assume that $I$ and $X$ are standard. Let $x$ be any point of $\prod_{i \in I} X_i$. Form a standard set $S$ whose standard elements are pairs $(i,y_i)$ such that $x_i$ is in every standard neighborhood of $y_i$ according to $X_i$'s topology (this exists by axiom $S$). Use the axiom of choice on $S$ to choose just one $y_i$ for $i$, and call the resulting standard set of choices $y$. (Note that if each $X_i$ is Hausdorff, then the $y_i$ are unique and we can skip this use of the axiom of choice.)

Now we'll show that every standard neighborhood $N$ of $y$ contains $x$. By the definition of the product topology and the transfer principle, there is a finite standard subset $I'$ of $I$ and a standard collection of sets $O_{i \in I'}$ that are open in $X_i$ such that $$y \in \{z : \forall i \in I'. z \in O_i\} \subseteq N$$ By axiom S, every element of standard finite set is standard, so every element of $I'$ is standard. (Note that this is where the proof breakdowns in the box topology; if $I'$ is infinite it will contain nonstandard elements and the rest of the proof won't follow.)

For every $i \in I'$, since $i$ is standard, $x_i$ is in each of $y_i$'s standard neighborhoods, including $O_i$. Therefore, $x \in N$.

$\\prod_{i \in I} X_i$ is compact. By the transfer principle, we may drop the assumption that $I$ and $X$ are standard. $\square$


We have shown that Tychonoff's theorem is a theorem of IST.

Corollary: Tychonoff's theorem is a theorem of ZFC.

Proof: IST is conservative over ZFC. $\square$

(Note that IST without the axiom of choice is not conservative over ZF, so even if the spaces are Hausdorff this doesn't prove Tychonoff's theorem in ZF.)

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  • $\begingroup$ When you write \Pi_{i\in I}X_i instead of \prod_{i\in I}X_i then in a displayed (as opposed to inline) context you see $\displaystyle\Pi_{i\in I} X_i$ instead of $\displaystyle\prod_{i\in I}X_i,$ and there are other differences too. $\endgroup$ Commented May 17 at 16:36
  • $\begingroup$ Do you prefer $\epsilon$ to $\varepsilon$? (To me the former looks too similar to $\in.$) $\endgroup$ Commented May 17 at 16:36
  • $\begingroup$ @MichaelHardy oh I didn't know about $\varepsilon$. Fixed! $\endgroup$ Commented May 17 at 17:19
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For logicians, here's a proof using Gödel's compactness theorem (although this is often proved using ultrafilters, it also a quick corollary of Gödel's completeness theorem which doesn't use ultrafilters).

Let $\prod_{i \in I} X_i$ be a product of compact spaces. Let $F$ be a collection of closed sets of $\prod_{i \in I} X_i$ with the finite intersection property (every finite subset of $F$ has non-empty intersection). We will find a point in $\cap F$.

We introduce a propositional symbol for each open set in each factor. The idea is that a symbol $O$ from a space $X_i$ is the proposition that $x_i \in O$ (where $x$ is the point we are trying to find). We represent each basic open set $B$ as a conjunction of such propositions.

For each factor $X_i$ and every finite open cover $O_1, O_2, ..., O_n$ of $X_i$, we add $O_1 \lor O_2 \lor ... \lor O_n$ as an axiom.

For each $C \in F$ we write $\bar C$ as a union of basic open sets of $\prod_{i \in I} X_i$. For each basic open set $B$ in this union, we add $\lnot B$ as an axiom (representing the fact that $x$ will not be found in $B$).

We call this theory $T$.

Every finite subset of $T$ only refers to finitely many $C \in F$. Their intersection has a point, say $x$, in it. We turn this into a model by setting $O$ (an open set of $X_i$) to true iff $x_i \in O$. This model satisfies the finite subset of $T$.

Therefore, by the compactness theorem of propositional logic, there is model satisfying the entirety of $T$. (Or in terms of the completeness theorem, if $T$ was inconsistent there must be a finitely long proof demonstrating this, contradicting the fact that it's finite subsets are consistent.)

For each $i \in I$, we can find a $x_i$ that is outside of every open set assigned "false" by the model. Assume contrary. Then the open sets of $X_i$ assigned false by the model form an open cover of $X_i$. This has a finite subcover, say $O_1, O_2, ..., O_n$. Then the axiom $O_1 \lor O_2 \lor ... \lor O_n$ is false, a contradiction. Thus such a $x_i$ exists. Note contrapositively that $x_i \in O$ implies $O$'s symbol was assigned true.

Gathering these together, we get a point $x \in \prod_{i \in I} X_i$. We will show that $x \in \cap F$. Assume contrary. There is a set $C \in F$ such that $x \notin C$. Then $x \in \bar C$, which we represented as a union of basic open sets, so $x$ is in one of those, say $B$. $B$ is represented by a conjunction, and each conjunct is true by the argument in the previous paragraph. Thus the axiom $\lnot B$ is false, a contradiction! Thus $x \in \cap F$.

$\square$

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