It seems to be known that Tychonoff's Theorem for Hausdorff spaces and the Compactness theorem of first order logic are both equivalent over ZF to the ultrafilter lemma. Does anyone know a slick proof for the implication "Compactness Theorem $\rightarrow$ Tychonoff for Hausdorff spaces" (without using the ultrafilter lemma as an intermediate step)?

  • $\begingroup$ What's wrong with using the ultrafilter lemma as an interpolant? $\endgroup$ – Asaf Karagila Jun 23 '13 at 17:06
  • $\begingroup$ @AsafKaragila: In order to fully appreciate the usefulness of some (simple) argument, it is nice to know a different (and possibly more complicated) approach. Also, if you have a different approach, you can compare with the former approach to better understand the role played, for example, by the ultrafilter lemma. $\endgroup$ – André Caldas Jun 23 '13 at 17:13
  • $\begingroup$ I think a proof of the form "To $\prod X_i$, associate the first order theory $T$, ... , $T$ is obvioiusly finitely satisfiable, satisfiability of T implies that $\prod X_i$ is compact" would be more aesthetically pleasing. $\endgroup$ – Dominik Jun 23 '13 at 17:15
  • $\begingroup$ @Dominik: If anything, then proving that $2^X$ for some set $X$ (probably the disjoint union of the spaces), and then finding a theory describing the space itself being closed, or something. $\endgroup$ – Asaf Karagila Jun 23 '13 at 17:36
  • $\begingroup$ @Andre: I think that in cases of weak choice principles it's usually quite easy to see where they are needed. We use ultrafilters to generate some sort of canonical families, and we use the Hausdorff property for a canonical choice from those families. $\endgroup$ – Asaf Karagila Jun 23 '13 at 18:18

Suppose that we have a family of compact Hausdorff space $(X_i, \tau_i)$ where $\tau_i$ is the topology and $i$ ranges over an index set $I$. Assume that the product $X = \prod_i X_i$ is not compact, so there is an open cover $(W_\alpha: \alpha < \kappa)$ of $X$ which does not have a finite subcover. We define a language and a consistent theory and then apply the compactness theorem to get a point in $X - \bigcup_\alpha W_\alpha$.

W.l.o.g, assume that each $W_\alpha = \prod_i W_{\alpha,i}$ for $W_{\alpha,i} \in \tau_i$ and $spt(W_\alpha) = \{i: W_{\alpha,i} \neq X_i\}$ is finite.

The language consists of

  • constants $c_i$ for $i \in I$;
  • for each $i$ and each $U \in \tau_i$, a unary predicate $P_{i,U}$.

The intended meaning of $(c_i: i \in I)$ is a (non-standard) point in $X - \bigcup_\alpha W_\alpha$; and $P_{i,U}(x)$ means $x \in U$.

The theory $T$ consists of

  • $P_{i,X_i}(c_i)$;
  • for each $\alpha$, $\bigvee_{i \in spt(W_\alpha)} \neg P_{i,W_{\alpha,i}}(c_i)$ (i.e., $(c_i: i \in I) \not\in W_\alpha$);
  • for each finite $F \subset \tau_i$, if $\bigcup_{U \in F} U = X_i$ then the sentence $(\forall x) (P_{i,X_i}(x) \to \bigvee_{U \in F} P_{i,U}(x))$ is in $T$;
  • if $C = X_i - U$ is closed and $U_0, U_1 \in \tau_i$ are s.t. $U_i \cap C \neq \emptyset$ and $U_0 \cap U_1 \cap C = \emptyset$, then the sentence $\neg P_{i,U_0}(c_i) \vee \neg P_{i,U_1}(c_i)$ is in $T$.

By the assumption that $W_\alpha$'s do not have a finite subcover, $T$ is consistent. So take a model $M$ of $T$, let $\mathcal{C}_i$ be the set of $X_i - U$ s.t. $U \in \tau_i$ and $M \models \neg P_{i,U}(c_i)$. Then $\mathcal{C}_i$ has finite intersection property and by compactness $\bigcap \mathcal{C}_i$ is non-empty. By the last set of sentences in $T$, $\bigcap \mathcal{C}_i$ is a singleton $\{p_i\}$ (here we need Hausdorff-ness). The point $(p_i: i \in I)$ is not covered by any $W_\alpha$.

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  • $\begingroup$ How do you deal with the part where you assume $W_\alpha$ to be a basic open set without the axiom of choice ? $\endgroup$ – Maxime Ramzi Apr 25 '17 at 12:16
  • $\begingroup$ @Max By definition, an open set in the product space is a union of basic open sets. But I admit that $(W_\alpha: \alpha < \kappa)$ is somehow misleading and should be replaced by $(W_j: j \in J)$ where $J$ is an index set. $\endgroup$ – Wei Wang Apr 28 '17 at 1:18
  • $\begingroup$ I know that, but it seems to me you have to somehow choose the basic open sets that consistute your union. What I'm saying is I don't see how, without the axiom of choice, for any family $(W_i)_{i\in I}$ there would be a family $(V_j)_{j\in J}$ such that they have the same union. When $I$ is finite, induction proves it, but when it isn't, I don't see how you can prove this without the axiom of choice. You can't just say "for $i\in I$, let $(V_j)_{j\in J_i}$ such that...", because you have to choose those $V_j$'s (there may be many such families) $\endgroup$ – Maxime Ramzi Apr 28 '17 at 4:41
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    $\begingroup$ @Max Let $(W_j: j \in J)$ be given. For each $j$, let $\mathcal{W}_j = \{V: V \text{ is a basic open subset of } W_j\}$. By definition, $W_j = \bigcup \mathcal{W}_j$. $\endgroup$ – Wei Wang Apr 29 '17 at 4:06
  • $\begingroup$ Oh right I'm stupid I always forget this trick ! It's not that $W_j$ is a union of basic open sets, it's that $W_j$ is the union of all the basic open sets it contains. $\endgroup$ – Maxime Ramzi Apr 29 '17 at 7:27

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