Yesterday I learnt that if we have a function $ f : D \rightarrow \mathbb{R} $ , $ D $ $ \subset \mathbb{R} $ an interval, monotone increasing. Then the following is true: $f$ is continuous in $ a \in D $ if and only if $ \sup f\{ x < a\mid x \in D\} = \inf f\{x > a \mid x \in D \} $.
Now I have found an interesting fact for a monotone function $ f : D \rightarrow \mathbb{R} $: We can show that the set of discontiuous points is at most countable.
Remark : we don't know what left-hand and rights limits are.
Attempt: Define $ D_a $ set of discontinuous points. Notice that for a discontinuous point $a$ the following is true: $ \sup f\{ x < a\mid x \in D\} < \inf f\{x > a \mid x \in D \} $. So far so good. We can now associate every discontinuous point $a$ with an open interval $I_a = ( \sup f\{ x < a\mid x \in D\} , \inf f\{x > a \mid x \in D \}) $. Now I need that two different intervals are disjoint. Can somebody proof that fact? If I have this I can say:
Let $ g : D_a \rightarrow \mathbb{Q} $ be a function. $g(a) \in I_a $. g is injective. $g(D_a)$ is countable because $\mathbb{Q}$ is countable. $g$ is injective. Hence $D_a$ is at most countable.
If $a, b$ are two points of discontinuity for $f$ (it is $f$ that is discontinuous, not the points) with $a < b$, then choose some $c$ with $a < c < b$ (for example $c = (a + b)/2$).
Let $I_a = (m_a, M_a), I_b = (m_b, M_b)$.
Since $c > a$ we have $f(c) \in f(\{x > a\})$, so $M_a \le f(c)$. Similarly, since $c < b, f(c) \le m_b$, and thus $M_a \le m_b$.
Therefore $I_a$ and $I_b$ can have no point in common.