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This question was inspired by the MathOverflow question "Unnecessary uses of the axiom of choice". I want to know of statements in ZFC that can be proven by assuming the Continuum Hypothesis, but can also be proven by a more elaborate (perhaps even significantly more elaborate) proof without assuming the Continuum Hypothesis.

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    $\begingroup$ The theorem of Malliaris and Shelah that $\frak p=\frak t$ is (like equality between any of the "usual" cardinal characteristics of the continuum) a trivial consequence of CH. But its proof in ZFC is difficult, and indeed the result was quite unexpected. $\endgroup$ Commented Feb 22, 2022 at 20:03
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    $\begingroup$ This is an observation, so not really an answer (hence given as a comment), but I often encounter statements/proofs that some set is uncountable which can automatically be strengthened to (but almost never is) cardinality of the continuum by simply observing that the set is Borel. In fact, this paper has several lengthy proofs that certain known theorems involving countable exceptional sets can be strengthened to cardinality less than the continuum, although this strengthening is automatic from the well known fact that the sets are Borel. $\endgroup$ Commented Feb 22, 2022 at 20:45
  • $\begingroup$ These days, questions become CW only by moderator intervention. I flagged your post for moderator attention in light of your suggestion (and then, since it is now presumably unneeded, edited out the suggestion from the body of the post). $\endgroup$
    – LSpice
    Commented Feb 22, 2022 at 21:18
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    $\begingroup$ Someone who is knowledgeable about such things should post answers about (1) the existence of a hereditarily Lindelof regular space which is not separable (Moore) and (2) coloring the edges of the complete graph on $\aleph_1$ vertices with $\aleph_1$ colors so that every uncountable complete subgraph contains edges of every color (Todorcevic) and (3) every first order sentence which is preserved by reduced products is equivalent to a Horn sentence (Shelah) $\endgroup$
    – bof
    Commented Feb 24, 2022 at 5:22

3 Answers 3

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Theorem: The space $\mathbb N^*$ of non-principal ultrafilters on $\mathbb N$ is not homogeneous.

Using CH, it is fairly straightforward to prove there is a special kind of ultrafilter called a $P$-point. A point $u$ of $\mathbb N^*$ is a $P$-point if any countable intersection of open neighborhoods of $u$ is again a neighborhood of $u$. Not all points of $\mathbb N^*$ are $P$-points (regardless of CH). Walter Rudin proved in 1956 that CH implies that $\mathbb N^*$ contains $P$-points, so this shows the space is non-homogeneous.

But the non-homogeneity of $\mathbb N^*$ is a theorem of ZFC. This was proved years later in 1967 in Frolík - Sums of ultrafilters (building on some unpublished work of Kunen). As I understand it, the non-homogeneity of $\mathbb N^*$ was a hot-topic open question during the intervening years, which demonstrates how much tougher the non-CH proof is.

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    $\begingroup$ @JonasFrey: "Sums of ultrafilters," Bulletin of the AMS 73 (1967), pp. 87-91. projecteuclid.org/journals/… $\endgroup$
    – Will Brian
    Commented Feb 22, 2022 at 20:53
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    $\begingroup$ This is a good example of the fact that English-speaking mathematicians use the word "any" too much. "Anyone can do that" may be synonymous with "Everyone can do that", but "If anyone knows the answer to this question, it's Will Brian" is synonymous with "If someone knows the answer to this question, it's Will Brian." The word "any" is sometimes a universal quantifier and sometimes an existential quantifier, the latter in particular in negative sentences ("There isn't anyone here"), questions ("Is anyone here?"), and conditionals (i.e. after words like "if"). $\qquad$ $\endgroup$ Commented Feb 22, 2022 at 20:58
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    $\begingroup$ Just using "every" would disambiguate this. $\qquad$ $\endgroup$ Commented Feb 22, 2022 at 20:59
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    $\begingroup$ There's another use of "any" besides that: "Anyone on this committee can serve as the chair" doesn't mean the same thing as "Everyone on this committee can serve as the chair." $\endgroup$ Commented Feb 22, 2022 at 21:01
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    $\begingroup$ @YCor: There are two well-known proofs without CH, neither of which assumes the negation of CH. The first (Frolik's, with some help from Kunen) examines a partial order on ultrafilters, now known as the Rudin-Frolik order, to show $\mathbb N^*$ is not homogeneous. The second (due to just Kunen) constructs something called a "weak $P$-point" using just ZFC. This feels much more like a ZFC extension of Rudin's proof. But I should mention that, by a theorem of Shelah, ZFC cannot prove that $\mathbb N^*$ contains any $P$-points. So strictly speaking, Rudin's proof cannot work with just ZFC. $\endgroup$
    – Will Brian
    Commented Feb 23, 2022 at 1:16
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A Dowker space is a normal Hausdorff space whose product with the closed unit interval $I$ is not normal. In 1971, Mary Ellen Rudin constructed the first ZFC Dowker space, which had cardinality $\aleph_\omega^\omega$. This space was considered "large" and for a long time it was an open problem to construct a "small" Dowker space. Various people constructed small Dowker spaces by assuming extra hypotheses. Most relevant to the current question is the construction, by I. Juhász, K. Kunen and M. E. Rudin (Two more hereditarily separable non-Lindelöf spaces), of a small Dowker space assuming CH. Finally, in 1996, Zoltán Balogh constructed A small Dowker space in ZFC.

There might be other examples stemming from the work of M. E. Rudin. Initially, I thought that the normality of ${\mathbb R}^\omega$ in the box topology might be an answer to this question, but apparently it is still open whether it can be proved without CH.

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Let $G_1,G_2,H_1,H_2$ be countable groups. If $G_i$ is elementary equivalent to $H_i$ for $i=1,2$, then $G_1\times G_2$ is elementary equivalent to $H_1\times H_2$.

Proof: (a) assume CH. Fix a nonprincipal ultrafilter $\eta$ on the set of integers. Then the ultrapowers $G_i^\eta$ and $H_i^\eta$ are elementary equivalent, have cardinality $\aleph_1$ (by CH) and are $\aleph_1$-complete. So they are isomorphic. Hence, denoting by $\equiv$ elementary equivalence and $\simeq$ isomorphism, we have $$G_1\times G_2\equiv(G_1\times G_2)^\eta\simeq G_1^\eta\times G_2^\eta \simeq H_1^\eta\times H_2^\eta\simeq (H_1\times H_2)^\eta\equiv H_1\times H_2.$$

(b) So the above result is a theorem of ZFC+CH. By Schoenfield absoluteness, it is a theorem of ZF.

(Note: group axioms play no role: this works with arbitrary countable algebras with countable signature, in the sense of universal algebras. Little further effort should remove the countability assumptions.)

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    $\begingroup$ Doesn't this result also follow (without CH) from the Feferman-Vaught theorem? $\endgroup$ Commented Feb 22, 2022 at 23:57
  • $\begingroup$ @AndreasBlass I was indeed hoping to hear of another direct CH-free approach. $\endgroup$
    – YCor
    Commented Feb 23, 2022 at 0:27
  • $\begingroup$ @LutzLehmann thanks, I've fixed the claim (replacing $G_1\times H_1$ with $G_1\times G_2$, etc.) $\endgroup$
    – YCor
    Commented Feb 24, 2022 at 9:31

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