Showing properties of discontinuous points of a strictly increasing function Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly monotonically increasing.
(i) Is $f$ not continuous at $p \in \mathbb{R}$, there exists a non-empty, open interval $(a_p, b_p) \subset \mathbb{R}$ such that $f(x)\leq a_p$ for all $x < p$ and $f(x) \geq b_p$ for all $x > p$.
(ii) The set of discontinuity points $$ \{ p \in \mathbb{R} | f \; \mbox{is not continuous at} \; p \}$$ is countable.
For (i) I could be way off, but I am picturing a graph with $p$ on the $x$-axis for which the value $f(p)$ on the $y$-axis is undefined. Am I correct to interpret the open interval $(a_p, b_p)$ as an interval on the $y$-axis which should be contained within the distance between two $f(x)$'s (one for $x<p$ and the other for $x>p$)? If that is so far correct, there could be 2 types of discontinuous points $p$, a jump or removable type. For the jump it would be easier to show that somehow the interval $(a_p, b_p)$ is smaller than the vertical jump... For a removable discontinuity $c$, I would think that the interval could contain just the $y$-axis value $\displaystyle \lim_{x\to c}f(x)$ but I don't really know how to express that the upper and lower bounds would be just above and below that...
With (ii) I am currently trying to understand a proof, what exactly does the notation $f(p-)$ or $f(p+)$ mean in this context? Is it simply the value when approached from the left or the right(respectively)?
 A: Revised answer:
(i)$x < p \Rightarrow f(x) < f(p)$ (because $f$ is strictly monotonically increasing).
Because this set of values is bounded from above there exists a supremum, namely $f(p-)$.
Likewise, $x > p \Rightarrow f(x) > f(p)$ (because $f$ is strictly monotonically increasing).
Because this set of values is bounded from below there exists an infinimum, namely $f(p+)$.
Let $a_p = f(p-)$ and $b_p = f(p+)$.
$\Rightarrow f(x) < a_p \forall x < p$ and $f(x) > b_p \forall x >p$.
Furthermore, $p$ discontinuous $\Rightarrow f(p-) \neq f(p+)$, $f$ strictly monotonically increasing $\Rightarrow f(p+) - f(p-) > 0 \Rightarrow (a_p, b_p) \neq \emptyset$.
(ii)For every point of discontinuity $p_i$, there exists (after (i)) an interval $(a_{p_i}, b_{p_i})$. These intervals are by construction disjoint. Because $\mathbb{R}$ is dense, in every interval there exists an element from $\mathbb{Q} \Rightarrow$ the set of discontinuity points has cardinality at most equal to that of $\mathbb{Q} \Rightarrow$ the set of discontinuity points is countable.
