# Open coloring axiom vs. CH

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

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.) –  Andres 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.) –  Andres 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: blog.assafrinot.com/?p=3243 –  saf Oct 8 '13 at 22:22