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I'm actually trying to construct a proof to show that there exists an uncountable number of irrationals between $x<y$ for $x,y \in \mathbb{R}$. I'd like to do it by using the statement if E is uncountable and F is countable, then E\F is uncountable.

I started by showing that the interval (x,y) is uncountable, and I'd like to use the fact that I've proved that there are a countably infinite number of rationals between x and y.

Can I express the irrationals $ \mathbb{I}$ as $ \mathbb {I} \in (x,y) \backslash \mathbb{Q}$? And then say since $\mathbb{I} \in (x,y)$ and $\mathbb{I} \notin \mathbb{Q}$ it is uncountable since (x,y) is uncountable and $\mathbb{Q}$ is countable?

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  • $\begingroup$ Sorry, I wanted to say if E is uncountable and F is countable. That's an interesting question you raised though. $\endgroup$ – Nikitau Oct 7 '16 at 4:10
  • $\begingroup$ Yes, if $x\lt y,$ you can show that there are uncountably many irrationals between $x$ and $y$ by showing that there are uncountably many reals in the interval, and only countably many of them are rational. I don't know what $\mathbb I\in(x,y)\setminus\mathbb q$ is supposed to mean. $\endgroup$ – bof Oct 7 '16 at 4:13
  • $\begingroup$ I guess I was trying to show that the irrationals were in the interval of (x,y) but not part of the rationals within that interval. Since I've proved that (x,y) was countable and the rationals were uncountable, then $\mathbb{I}$ would be uncountable. $\endgroup$ – Nikitau Oct 7 '16 at 4:16
  • $\begingroup$ You can in fact show that $|E\setminus F|=|E|$: see this question. $\endgroup$ – Brian M. Scott Oct 7 '16 at 8:25
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I think the easiest way is to suppose that $E \setminus F$ is countable. Then $E = F \cup (E \setminus F)$ is the union of two countable sets and thus must be countable, which is a contradiction to our definition of $E$. Therefore $E\setminus F$ is uncountable.

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You can just use the definitions of countable and uncountable. Biject $F$ with $\Bbb N$, then use the diagonal proof to show that taking $\Bbb N$ elements out of $E$ leaves it uncountable, else $E$ would have cardinality $2|\Bbb N|=\Bbb N$

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  • $\begingroup$ I've not covered the diagonal proof, but I'll look it up. Thanks for the response! $\endgroup$ – Nikitau Oct 7 '16 at 4:20

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