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I read the following proof as to why the set of real numbers is uncountable.

Assume that $\mathbb{R}$ is countable. Then we can enumerate $\mathbb{R} = \{x_1, x_2, x_3, \cdots\}$ and be sure that every real number appears somewhere on the list. Let $I_1$ be a closed interval that does not contain $x_1$. Next, given a closed interval $I_n$, construct $I_{n+1}$ to satisfy both of the following:

(i) $I_{n+1} \subseteq I_n$

(ii) $x_{n+1} \not\in I_{n+1}$

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Now consider the intersection $\cap_{n=1}^{\infty} I_n$. If $x_{n_0}$ is some real number from the enumerated list, then $x_{n_0} \not \in I_{n_0}$, so $x_{n_0} \not \in \cap_{n=1}^{\infty} I_n$. But we assumed the list contained every single real number, so this implies $\cap_{n=1}^{\infty} I_n = \emptyset$. However, the Nested Interval Property asserts that $\cap_{n=1}^{\infty} I_n \neq \emptyset$, hence the contradiction.


Question: I don't see why this same argument can't be applied onto $\mathbb{Q}$ and show that the set of rationals is uncountable (which is of course nonsense), or even more generally, show that any infinite set $S = \{s_1, s_2, s_3, \cdots \}$ is uncountable, by using the same construction of intervals used above.

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    $\begingroup$ Does $\Bbb Q$ have the Nested Intervals Property? $\endgroup$ – David Mitra Jan 4 '14 at 16:39
  • $\begingroup$ I thought about that, but what does it mean that $\mathbb{Q}$ doesn't have the Nested Interval Property? $\endgroup$ – Trts Jan 4 '14 at 16:42
  • $\begingroup$ That the intersection may perfectly well be empty, @TrueTears ? $\endgroup$ – DonAntonio Jan 4 '14 at 16:44
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    $\begingroup$ $[\sqrt2, \sqrt 2+1/n]\cap\Bbb Q$, $n\in \Bbb N$ gives you a nested sequence of closed intervals of $\Bbb Q$. But the sequence has empty intersection. $\endgroup$ – David Mitra Jan 4 '14 at 16:45
  • $\begingroup$ Ah okay, so basically the Nested Interval Property states that the non-empty intersection must consist of at least a real number, so it may well be possible that the intersection is empty if we were restricted to only the set of rationals. Correct? $\endgroup$ – Trts Jan 4 '14 at 17:02
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A closed interval in $\mathbb R$ is an example of a compact set. Let $C$ be a compact set and let $\mathcal F$ be a set of closed sets in $C$. Suppose that for every finite subset $F\subseteq \mathcal F$, $\bigcap F\ne\varnothing$. Then in fact $\bigcap \mathcal F\ne \varnothing$. A non-degenerate closed interval in $\mathbb Q$ is not compact.

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It has to do with topological properties of $\mathbb{R}$. In particular, we take adventure of the fact that $\mathbb{R}$ is totally ordered.

If we let $X$ be any topological space, that doesn't have a total ordering on it, then $\forall x,y$ we don't have a well defined notion of $x<y$, so the idea of an interval doesn't make sense. A simple example is $\mathbb{C}$ which cannot be made to be totally ordered.

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