2
$\begingroup$

I know that the continuum hypothesis is not decidable, i.e. we can not prove it, nor disprove it.

The question is, is it theoretically possible to find an explicit set $E\subset \mathbb R$ such that we can not prove neither $\vert \mathbb N\vert =\vert E\vert$, nor $\vert \mathbb R\vert =\vert E\vert$?

Have you ever heard of such a set before?

  • 0
    Look $|2^{\mathbb{R}}|$, one has $|\mathbb{N}|<|\mathbb{R}|=|2^{\mathbb{N}}|<|2^{\mathbb{R}}|$ – 2017-01-30
  • 1
    https://en.wikipedia.org/wiki/Perfect_set_property ​ ​ – 2017-01-30
  • 1
    There is a simple example: take $X=\{x\in \Bbb{R} : x\in \Bbb{N}\lor\mathsf{CH}\}.$ Then either $|X|=\aleph_0$ or $|X|=2^{\aleph_0}$ holds but you can prove neither of them. – 2017-01-30
  • 2
    This seems somewhat related: [Are there any constructive axioms which disprove the continuum hypothesis?](http://math.stackexchange.com/q/1823214) – 2017-01-30

1 Answers 1

2

This has nothing to do with the continuum hypothesis, but it is quite possible. Let $S(n)$ be a predicate of natural numbers $n$ such that neither "there is some $n$ such that $S(n)$" nor "there is no $n$ such that $S(n)$" is provable. Let $E = \mathbb Z \cup \{(n, n+1):\; S(n)\}$. Thus if there is some $n$ such that $S(n)$, $E$ has cardinality of the continuum, while if there is no such $n$, $E$ is countable.

  • 0
    Thank you for the answer! Do you have an explicit example of such a predicate $S(n)$? – 2017-01-30
  • 1
    E.g. it is quite possible that "$n$ is an odd perfect number" is such a predicate. If you're talking about provability in a particular formal system, [Gödel's incompleteness theorems](https://en.wikipedia.org/wiki/G%C3%B6del's_incompleteness_theorems) provide examples of such predicates. – 2017-01-30