Real Analysis / Set Theory

[SlipEternal]

I actually was quite confused by all of this. So, I started digging a bit, and I determined that I was wrong. I withdraw my argument. \(\displaystyle (-\infty,x)\cap\mathbb{Q}\) is a Dedekind cut at \(\displaystyle x \in \mathbb{R}\). Therefore, the set \(\displaystyle \left\{(-\infty,x)\cap\mathbb{Q}\mid x \in \mathbb{R}\right\}\ne\left\{(-\infty,r)\cap\mathbb{Q}\mid r\in \mathbb{Q}\right\}\). In fact, the first is uncountable while the latter is countable. As such, my previous argument was invalid.

 

[pka]

I actually was quite confused by all of this. So, I started digging a bit, and I determined that I was wrong. I withdraw my argument. \(\displaystyle (-\infty,x)\cap\mathbb{Q}\) is a Dedekind cut at \(\displaystyle x \in \mathbb{R}\). Therefore, the set \(\displaystyle \left\{(-\infty,x)\cap\mathbb{Q}\mid x \in \mathbb{R}\right\}\ne\left\{(-\infty,r)\cap\mathbb{Q}\mid r\in \mathbb{Q}\right\}\). In fact, the first is uncountable while the latter is countable. As such, my previous argument was invalid.

Thank you for finding out about Cuts or Schnitts as Dedekind himself called them. In fact the way you defined \(\displaystyle A_x\) is the way he did define the lower cut. I used the French way defining the cuts. There is no doubt that such a function exists between \(\displaystyle \mathbb{R}\) and \(\displaystyle \mathcal{P}(\mathbb{Q})\). But the question I had was: because \(\displaystyle \mathcal{P}(\mathbb{Q})\sim\mathcal{P}(\mathbb{N})\) would the composition carry over? And yes it does.