Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o. $\mathbb{P}$ (i.e. $\mathbb{P}$ is a reflexive transitive relation) and for any countable family of dense subsets of $\mathbb{P}$ there is a generic filter which intersects all dense subsets of the countable family. It is well-known that this statement is equivalent to the Baire Category Theorem for Complete Metric Spaces - and thus it is also equivalent to the Principle of Dependent Choices.

A masters student of mine has found in the literature the following statement: "Rasiowa-Sikorski Lemma is equivalent to the Baire Category Theorem for Compact Hausdorff Spaces, modulo the Boolean Prime Ideal Theorem". We understood this as the assertion that the theory ZF + BPI alone is able to prove the equivalence between the Baire Category Theorem for Compact Hausdorff Spaces and the Rasiowa-Sikorski Lemma.

Well, I asked my student to verify such claim, and at first glance I suggested him to follow the results 3.1 to 3.4 of Chapter II of Kunen's book, where there are proofs for some equivalences of Martin's Axiom at $\kappa$, MA($\kappa$): the idea was to discard the hypothesis "c.c.c." and adapt the reasoning, arguing for $\kappa = \omega$. It turns out that it was not a good suggestion, because in 3.1 a kind of Downward-Lowenheim-Skolem argument is done, to show that it is equivalent to work with a restricted form of the forcing axiom, considering only partial orders of bounded cardinality. However, such argument seems to require the Axiom of Choice, or some part of it other than BPI.

Does any of you know if it is indeed possible to prove the equivalence between "Baire Category Theorem for Compact Hausdorff Spaces" and "Rasiowa-Sikorski Lemma for forcing posets" from ZF + BPI alone ? Any suggestions or references would be appreciated.


1 Answer 1


The key observations are that BPI is equivalent to the Stone representation theorem for Boolean algebras, and that for the Rasiowa–Sikorski lemma we can focus on [complete] Boolean algebras, since they are forcing equivalent (so we can restrict the generality of partial orders).

Now, one implication is a consequence of ZF. Since RS is equivalent to Dependent Choice, which itself is equivalent to the downward Löwenheim–Skolem theorem, one can just use that argument.

Alternatively, if $X$ is a compact Hausdorff space and $D_n$ are dense open sets, and without loss of generality $D_{n+1}\subseteq D_n$. take $U$ to be a non-empty open set, and consider the forcing whose conditions are sequences $(x_i,W_i)_{i<n}$ such that $x_i\in D_i\cap U$ and $W_i$ is open such that:

  1. $x_i\in W_i\subseteq \overline W_i\subseteq U\cap D_i$, and
  2. $\overline W_i\subseteq W_j$ if $j<i$.

Now consider $E_n$ to be the dense open set in the forcing whose conditions are sequences of length at least $n$. By the Rasiowa–Sikorski lemma there is a generic meeting all of these $E_n$s which defines a sequence $(x_i,W_i)_{i<\omega}$. Now observe that $\{\overline W_i\mid i<\omega\}$ is a family of compact sets with a finite intersection property, therefore their intersection is non-empty, and it contains a point in $\bigcap D_i\cap U$ as wanted.

In the other direction we need to use BPI, and we use it in the form of Stone's representation theorem. Given a notion of forcing, we may assume without loss of generality that it is a complete Boolean algebra $B$ and we can consider its Stone space, $S(B)$, the space of all the ultrafilters on $B$.

If $D\subseteq B$ is a dense open set, then $D^*=\{F\in S(B)\mid\exists b\in D, b\in F\}$ is a dense open set in $S(B)$. Therefore, by the BCT for compact Hausdorff spaces, if $D_n$ is a sequence of dense open subsets of $B$, $\bigcap D^*_n$ is dense, and in particular not empty. Take any $G\in\bigcap D_n^*$, then for all $n<\omega$, $G\cap D_n$ is non-empty as wanted.

  • $\begingroup$ ... Wow, is the Principle of Dependent Choices equivalent to Downward Lowenheim Skolem ? This is news to me (I thought Downward Lowenheim Skolem was equivalent to the full Axiom of Choice). I will take a look at that... $\endgroup$ May 5, 2021 at 22:38
  • $\begingroup$ Yes. That's an observation due to George Boolos, but also due to several other people (including myself). The one direction is the usual proof, the other is simple: if $T$ is a tree of height $\omega$, $T^*\prec T$ is a countable elementary submodel, then $T^*$ is a countable subtree of height $\omega$, so either $T^*$ has a maximal node which is maximal in $T$ or it has a branch which is a branch in $T$. $\endgroup$
    – Asaf Karagila
    May 5, 2021 at 22:40
  • $\begingroup$ Oh, I see. Thanks. But there is still a problem. The implication where is supposed to be used Boolean algebras (the Stone Space, etc.) is: Baire for Compact Hausdorff spaces --> Rasiowa-Sikorski for posets, so there is no DC to start with in this part (in fact, we want to prove DC at the end of it, since RS and DC are equivalent). $\endgroup$ May 10, 2021 at 17:47
  • $\begingroup$ But, as I said in the beginning, my reference for such an argument is 3.2 of Chapter II of Kunen, and in this part he refers to 3.1, where the Axiom of Choice is required to define certain functions which will be used (in a closure argument) to construct a p.o. of restricted cardinality. That is, precisey, the point I want to avoid the use of the Axiom of Choice: the use appearing at 3.1 of Kunen in the we-can-work-with-smallers-orders stuff, and this happens in an implication at which DC is at the thesis, not at the hypothesis. $\endgroup$ May 10, 2021 at 17:48
  • $\begingroup$ I'm not using DC in that implication. I'm using BCT. (Also, I don't have Kunen on hand, so it's a bit vague when you keep mentioning 3.1 and 3.2 there.) $\endgroup$
    – Asaf Karagila
    May 10, 2021 at 18:17

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