What is the cardinality of an infinite set of closed disjoint intervals? The question is :
Let A be a set of a closed disjoint intervals, what is the cardinality of A ?
I need to prove it using the Density property of the rational numbers in the real numbers.
So i know that the cardinality of every interval is c, and it is closed under union so it has go be c as well, but i'm not sure that it is a countable union, kinda stuck.
Thanks in advance !
 A: There are two different readings of your question (at least I can see two!).
1) If $A = \{[a_i,b_i] \ : \ i \in I\}$ is a set of pairwise disjoint closed intervals. What is the size of $\bigcup A = \bigcup_{i\in I} [a_i,b_i]$.
2) Same set up, what is the range of sizes of $I$.
You question title and the nod to the rationals makes it look like question 2, but your thoughts about the union of the intervals makes it look like 1. I'll answer both. (Assuming that the intervals have $a_i < b_i$.)
1) The size is $2^{\aleph_0}$. The union in question is a subset of $\mathbb{R}$ which has size continuum (which gives an upper-bound) and we know that any closed interval $[a,b]$ with $a<b$ has size continuum, giving a lower bound. (Assuming $i$ is non-empty!)
2) You can easily witness any finite cardinality, or countability by taking any subset you like of $\{[2n,2n+1] \ : \ n\in \mathbb{N}\}$. Can we get any more? No, because each interval we collect contains a rational not contained in any other (because they are disjoint). As there are only countably many rationals, there can be only countably many intervals in our union.
