I am completely lost in this proof. Can anyone please help me out?
Statement to prove : Show that any collection of pairwise disjoint, nonempty open intervals in $R$ is at most countable, Hint : each one contains a rational
Proof:
Let {$g_n$} be some list of all the rational numbers and $L$ be a collection of pairwise disjoint, nonempty open intervals in $R$. Then each $I \in L$ contains some rational $\hat{I}$ that is not found in any other open interval.
Let $K=${$r_i \in Q | r_i \in I$} be a collection of rational representaties of the intervals of $L$. Then we can label $I=I_1$ if $r_I=g_{n1}$ where $n_1$ is the smallest integer $n \in N$ such that $g_n=r_i \forall i \in L$. More generally, $I+I_k$ if the rational chosen is $I$, $r_I$ is equal $g_{nk}$ where $nk$ is the kth smallest integer $n \in N $ for which $g_n$ is among the elements of $k$
Thoughts :
The proof states $L$ is a collection of pairwise disjoint open intervals, then each $I \in L$ contains some rational that is not found in any other open interval.
If its pairwise disjoint, doesn't that mean every number on the interval is not contained in any other interval? Why does the proof focus on 1 rational?
what is meant by : Let $K=${$r_i \in Q | r_i \in I$} be a collection of rational representatives of the intervals of $L$. The set of rational representatives ? Is it just the set of all rational numbers in the interval?
"Then we can label $I=I_1$ if $R_I=g_{n1}$, where $n_1$ is the smallest integer $n\in N$ s.t. $g_n=r_I$ for every $I \in L$
If $R_I$ is every rational in the interval $I$, then why would $r_I$ also be a sequence $g_n$? Shouldn't this read $I=I_1$ if $K=g_{n1}$, since K is the set of all $r_I$ in $I$?
Best Answer
This proof is quite elegantly simple:
We know that each interval contains at least 1 rational, so we can create a injective map between these intervals, and rationals.
This implies that the cardinality of the number of intervals is $\leq$ the cardinality of $\mathbb Q = \aleph_0$. So, the number of intervals is at most countable.