I'm familiar with several proofs that the real numbers are uncountable (Cantor's initial proof, a proof by diagonalization, etc.). However, I've never seen a proof that the reals are uncountable that proceeds by showing that the set of Dedekind cuts of the rationals are uncountable. I'm aware that the set of all subsets of the rationals is uncountable, but not all of these sets are Dedekind cuts.
Is there a simple proof of the uncountability of $\mathbb{R}$ that works by showing the set of Dedekind cuts is uncountable?
Thanks!
Best Answer
You can't really show that $\mathbb R$ is uncountable using Dedekind cuts. If you can show that, then I am unaware of any such proof (and having a lot of work around the intro to set theory course in my university for the past few years, I doubt that I would not have known such proof). You can show that $\mathcal P(\mathbb N)$ is uncountable by this method though.
Dedekind cuts provide a method for constructing and defining the real numbers, rather than a direct method of proving their size is $2^{\aleph_0}$.
However in proving that $|\mathbb R|=2^{\aleph_0}=|\mathcal P(\mathbb N)|$ we can use Dedekind cuts to establish the $\leq$ direction by showing that if we fix an enumeration of $\mathbb Q$ as $q_n$, then the sets $A_r = \{n\in\mathbb N\mid q_n<r\}$ for $r\in\mathbb R$ form an injection from $\mathbb R$ into $\mathcal P(\mathbb N)$. Therefore there are at most $2^{\aleph_0}$ many real numbers.
The other direction usually involves the Cantor set construction and shows that there are exactly $2^{\aleph_0}$ many real numbers.