# Countability of disjoint intervals

According this problem/solution set from an MIT class (http://ocw.mit.edu/courses/mathematics/18-100c-analysis-i-spring-2006/exams/exam1_sol.pdf), the assertion:

"Every collection of disjoint intervals in R is countable."

is True, because "every interval contains a rational number", and the rationals are countable.

It seems to me this should be False, with possible counterexample:

{ [x,x] | x is an element of R}

ie the set of all singelton intervals on R. Why isn't this a valid counterexample?

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This is a typo. The statement should read: "every collection of disjoint open intervals in R is countable." – François G. Dorais Sep 20 '10 at 11:20
You're right. I agree that $[x,x]$ is a singleton interval and I would guess most other mathematicians would too. There's no good reason not to. – Stefan Smith Oct 18 '13 at 22:08
I have a quiestion about that: We have to choose a rational number of a possible infinite subset of them, then, is the result implied by the Axiom of Choice? What happen with people who dont agree with this axiom? – ILikeMath Mar 23 '14 at 2:06

Your thinking is correct; the set of all singleton sets of R is certainly uncountable.

It seems that the question meant something like "Every collection of disjoint open intervals in R is countable." (In this case, the claim that each interval contains a rational number is valid.)

Maybe there was some convention in the course that "interval" meant open interval, or excluded singleton sets; perhaps it's simply a mistake. Either way, it's good that you noticed this detail!

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Or rather non-empty interior. Here is a more general statement, any collection disjoint subsets of $\mathbb{R}$ having non-empty interiors is countable. – AD. Sep 20 '10 at 10:01

Because "singleton interval" is usually not considered to be an interval.

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I generally agree. Some texts might set up their definitions so that singleton sets are intervals, but enough mathematicians use the term to refer only to non-trivial intervals that you have to interpret each author differently. A parallel question: is the empty set an interval? This is also a matter of convention that affects theorems such as "Every continuous function on a closed interval achieves a maximum value on that interval." Different books have different ways of excluding the empty interval from this theorem. It might not be an interval at all, or it might not be a closed interval. – Carl Mummert Sep 20 '10 at 12:27