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

enter image description here

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.

  • 5
    $\begingroup$ Does $\Bbb Q$ have the Nested Intervals Property? $\endgroup$ 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
  • 2
    $\begingroup$ $[\sqrt2, \sqrt 2+1/n]\cap\Bbb Q$, $n\in \Bbb N$ gives you a nested sequence of closed intervals of $\Bbb Q$. But the sequence has empty intersection. $\endgroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.