Could someone please explain to me the logic of the proof of the following problem?
Exercise 4.17 Let $f$ be a real function defined on $(a,b)$. Prove that the set of points at which $f$ has a simple discontinuity is at most countable.
The start proof that is hinted in the problem is the following.
Let $E$ be the set on which $f(x-)<f(x+)$. With each point $x$ of $E$ associate a triple $(p,q,r)$ of rational numbers such that
(a)$f(x-)<p<f(x+)$
(b)$a<q<t<x$ implies$f(t)<p$
(c)$x<t<r<b$ implies$f(t)<p$
The set of such triples is countable.
How can such set of triples always be countable?
Since we don't know yet that the number of discontinuous points is countable, how can we know anything about the countability of the set?
I am not familiar with any proof about countablilities, so I understand that I am missing something basic. All comments would be enormously appreciated.
If $A$ is a countable set then $A^3$ the set of all triple orders of elements of $A$ is also countable. Since the set of rational numbers is countable the set $(p,q,r)$ of rational numbers is countable.
Thus the set $(p,q,r)$ in theorem is also countable.