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.