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Is there a simple, direct proof that the open coloring axiom contradicts CH (straight from the definitions, no machinery allowed)? The separable metric spaces version of OCA, if that helps.

Edit: probably it wasn't clear what I was asking. Let's say, hypothetically, that I'm writing a book on forcing that is meant to be as elementary and readable as possible. I've just introduced OCA and I have half a page to show the reader that it contradicts CH. Can I do this without quoting external theorems?

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When you say OCA, I suppose you mean the one from Todorcevic's work, and not the one from the Abraham-Rubin-Shelah paper, right? – Asaf Karagila Oct 5 '13 at 15:55
Yes, I mean Todorcevic's axiom. – Nik Weaver Oct 5 '13 at 15:56

2 Answers 2

Here are two suggestions, whether they follow your instructions of "no machinery allowed", I will leave to you.

The following seems fairly direct to me, and I would think is the "standard" answer to your question:

First, partially order $\mathbb N^{\mathbb N}$ by eventual domination: $f<g$ iff $f(n)<g(n)$ for all $n$ large enough.

If $X\subseteq\mathbb N^{\mathbb N}$ is unbounded and countably directed, then there are $f\ne g$ in $X$ such that $f(n)\le g(n)$ for all $n$. This is proved as Lemma 0.7 in Todorcevic's Partition problems in topology.

From this it follows that under $\mathsf{OCA}$, any subset of $\mathbb N^{\mathbb N}$ of size $\aleph_1$ is bounded. (For example, this follows from Proposition 8.4.(c) and the proof of Theorem 3.5 in Todorcevic's book.)

(Refining this argument (eventually) gives us that $\mathsf{OCA}$ implies that ${\mathfrak b}=\aleph_2$.) The only technical tool this uses is the notion of oscillation, and what is needed can be developed directly in the proof of the statements above.

A different approach goes by building on Proposition 8.4.(c) directly: $\mathsf{OCA}$ implies that if $X,Y$ are uncountable sets of reals, then there is an uncountable subset of $X$ and a strictly increasing map of this subset into $Y$. This easily clashes with $\mathsf{CH}$, for example, using the results of Sierpiński on Sur un problème concernant les types de dimensions.

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(I expect there are more elegant/direct ways of verifying the claim in the fifth paragraph than the 8.4-3.5 route I suggest, but I tried to stay as close to Stevo's book as possible, to avoid sneaking in machinery through additional developments.) – Andrés Caicedo Oct 5 '13 at 19:36
I'm afraid this is exactly the sort of thing I didn't have in mind --- quoting a consequence X of OCA and then showing that X contradicts CH. I would only consider this a "direct" proof if you include the proof that OCA implies X --- which in your case makes your answer not so simple ... – Nik Weaver Oct 5 '13 at 21:57
(I guess I do not understand what you mean, since the proofs are direct implications that are spelled out in good amount of detail in the book.) – Andrés Caicedo Oct 5 '13 at 22:03
Yes, well, every complete proof is a "direct implication", isn't it? – Nik Weaver Oct 5 '13 at 22:11
Here is an inclusive version of Andres' answer: – saf Oct 8 '13 at 22:22

Another option is to use the same approach as Kunen in his book "Set Theory" (2011). In section V.6 he work with $SOCA$ (a weakening of $OCA_{T}$ that only ask for the existance of one of the homogeneous subsets, not for a covering).

In approximately two pages (364-366) he presents $SOCA$ and shows that the axiom implies $\neg CH$. The way he do it is with weakly Luzin sets.

His proof work as follows:

1) Define a set in ${R}^{n}$ ($R$ is the reals) as skinny iff the set of directions is not dense in the unit sphere.

2) Defines a weakly Luzin as a set whose all uncountable subsets are not skinny.

3) Define $A$ as an $\epsilon$-directed set iff there is a point in the unit sphere such that all the directions of $A$ have an angle at most of $\epsilon$ to that point.

4) Notice that every $\epsilon$-directed set is skinny.

5) Using induction and compactness show that $SOCA$ implies that every set is contains an $\epsilon$-directed set.

I was a litle vague because I left the reference. I can try to be more precise with definitions and proofs if needed.

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Another option is using a 2-entangled set of size $\aleph_{1}$ (the definition of entangled set contradicts $SOCA$), and it is know that there always exists an entangled set of size continuum (so $CH$ contradicts $SOCA$). I'm not sure if the construction of the set is short or not. – Iván Ongay Valverde Jun 29 at 6:56

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