Show that any collection of pairwise disjoint, nonempty open interval in $\displaystyle R$ is at most countable. [Hint: Each one contains a rational]
The hint basically gives it away. Let $\displaystyle \left\{I_\alpha\right\}_{\alpha\in\mathcal{A}}$ denote the set of your open intervals. For each $\displaystyle I_\alpha$ since $\displaystyle \mathbb{Q}$ is dense, we know there exists some rational number $\displaystyle q_\alpha\in I_\alpha$. So, define $\displaystyle f:\left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} \to\mathbb{Q}:I_\alpha \mapsto q_\alpha$, since the $\displaystyle I_\alpha$'s are disjoint we know that $\displaystyle f$ is an injection...so.
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dense means that between any two real numbers, we can find a rational number.
"at most countable" means that there is an injection from our collection, to a subset of the natural numbers (this subset could be finite, or infinite).
what drexel28 is saying, is that A is an "indexing set" for our collection of intervals. this is mostly a notational convenience, so that we can distinguish different intervals from each other while still using a notation that tells us they all belong "to the same collection".
since our intervals are pair-wise disjoint, one possible choice for A is the set of all real numbers {a: a = inf(I), for some interval I in our collection} (so if one of our intervals was (2,4), we would label that interval $\displaystyle I_2$).
by your hint, each interval $\displaystyle I_\alpha$, contains a rational number, which we can call $\displaystyle q_\alpha$ . since the intervals are pair-wise disjoint, no other interval contains this rational number.
but the rationals can be put into a bijection with the natural numbers, let's call the bijection h.
thus if C is our collection of intervals, we have an injection k from C into the natural numbers given by:
$\displaystyle k(I_\alpha) = h(q_\alpha)$