# A strictly decreasing function between uncountable subsets of the reals

By a standard technique of inductive killing everything relevant (in this case decreasing homeomorphisms between uncountable $$G_\delta$$-subsets of the real line) it is possible to prove the following fact.

Theorem (CH). Under CH the real line contains an uncountable subset $$X$$ admitting no strictly decreasing function $$f:Z\to X$$, defined on some uncountable subset $$Z$$ of $$X$$.

On the other hand, a known PFA-results of Baumgartner (about the order isomorphness of any $$\aleph_1$$-dense subsets of the real line) implies the following

Theorem (PFA). Under PFA, for any uncountable subset $$X\subset\mathbb R$$ there exists a strictly decreasing function $$f:Z\to X$$, defined on some uncountable subset $$Z\subset X$$.

Now

Question. Can this PFA-theorem be proved under a weaker assumption like OCA or (MA$$+\neg$$ CH)?

• "defined on an": do you mean "defined on any"? the article "a" is ambiguous, especially in such sentences. – YCor Jan 22 at 18:38
• @YCor Thank you for the comment. I changed "an" to "some". – Taras Banakh Jan 22 at 18:45
• OK now the meaning is clear. Anyway the correct English is "any", since the sentence is in the negative. – YCor Jan 22 at 18:58

In Todorcevic's book "Partition Problems in Topology" I have found Proposition 8.4(c) saying that under OCA for any uncountable sets $$X,Y$$ of reals there exists a strictly increasing function $$f:Z\to Y$$ defined on some uncountable subset $$Z$$ of $$X$$.
This proposition implies that for any uncountable set $$X\subset \mathbb R$$ there exists a strictly increasing function $$f:Z\to -X$$ defined on some uncountable subset $$Z\subset X$$. Then the function $$-f:Z\to X$$ is stricly decreasing.
• To complement this answer, I think the answer should be no for $\mathsf{MA}$. Abraham and Shelah proved that $\mathsf{MA}$ doesn't give you Baumgartner's theorem by showing $\mathsf{MA}$ is consistent with the existence of something called an entangled set of reals. I think an entangled set shouldn't admit the sort of map you're describing. (I don't have time to check the details today, which is why I'm just writing a comment -- I hope this helps.) – Will Brian Jan 29 at 14:10