2

I tried to prove that they are equivalent:

  1. There is no set $A$ such that $\aleph_0<|A|<2^{\aleph_0}$ and

  2. Every non-empty subset $A$ of $\mathbb{R}$ is a surjective image of $\omega$ or $\mathbb{R}$ is a surjective image of $A$. That is, $$\forall A\subseteq \mathbb{R} \exists f : (A\neq\varnothing \to f : \omega \twoheadrightarrow A \>\vee\> f: A\twoheadrightarrow \mathbb{R})$$

I don't think they are equivalent without the axiom of choice, but I don't know how to start to prove the unprovability. Thanks for any help!

Hanul Jeon
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  • https://math.stackexchange.com/questions/404807/how-to-formulate-continuum-hypothesis-without-the-axiom-of-choice/404813#404813 although this requires a slightly more delicate approach, since surjections need not be equivalent to injections. – Asaf Karagila Jun 13 '17 at 17:33
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    It is clear that (1) implies (2). So the real question is whether or not (2) implies (1) without the axiom of choice. Note, also, that a surjection from $\omega$ onto a set means that it is countable. – Asaf Karagila Jun 13 '17 at 17:37
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    The statement of (2) is inaccurate. You really mean non-empty sets. :) – Asaf Karagila Jun 13 '17 at 17:39
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    @Arthur: That is literally the question. – Asaf Karagila Jun 13 '17 at 17:40
  • Is the statement "There exists a surjection of $A$ onto $\Bbb{R}$" equivalent to "There exists an injection of $\Bbb{R} \to A$" when $A\subset \Bbb{R}$ ? It suffices to answer this question (since for $ \omega$ everything is easy as it is well ordered) – Maxime Ramzi Jun 13 '17 at 18:26
  • @Max In the absence of choice, certainly not necessarily - in fact, it's consistent that for every infinite set $X$, there is some set $A$ such that there is a surjection $A\rightarrow X$ but no injection $X\rightarrow A$. In this case we can even have such an $A$ be a set of reals! – Noah Schweber Jun 13 '17 at 18:49
  • @AsafKaragila: How is it clear that (1) implies (2)? What if (1) is true, but there is an infinite Dedekind-finite set of reals? – Eric Wofsey Jun 13 '17 at 18:49
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    @EricWofsey I believe that if there is a IDF set of reals, then (1) fails. Let $A$ be IDF, let $B$ be a countably infinite set disjoint from $A$, and let $C=A\cup B$. Clearly $\vert B\vert< C$ and $\vert C\vert\le 2^{\aleph_0}$. Suppose for contradiction that $f$ was an injection from $\mathbb{R}$ into $C$. Since $B$ is countable, there is a bijection between $\mathbb{R}$ and $\mathbb{R}\setminus f^{-1}(B)$; so we may assume without loss of generality that $im(f)\subseteq A$. But that contradicts the Dedekind-finiteness of $A$. – Noah Schweber Jun 13 '17 at 18:51
  • @Eric: What Noah wrote is correct. If there is a Dedekind finite set of reals, then there is a counterexample to 1. The key point is that every co-countable set of reals has size continuum. – Asaf Karagila Jun 14 '17 at 09:38

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