$f : D \rightarrow \mathbb{R} $ continuous if and only if sup $f \{ x < a\mid x \in D\} = \inf f\{ x > a\mid x \in D\}$ Let $D$ be an interval. Le $ f: D \rightarrow \mathbb{R}$ be a monotone increasing function.
Show that:
$f$ is continuous in a $\in$ $D$ if and only if
$$\sup f\{ x < a\mid x \in D\} = \inf f\{x > a \mid x \in D \}$$
Remark: We don't know what left-hand limit and right-hand limit is.
I know the definition of sup and inf, but unfortunately I don't really have an idea how to solve this. I think that we can use the $\varepsilon$-$\delta$ criterion. 
Edit : $ \text{“} \Longrightarrow \text{''}$ is clear. I only need a hint for  $\text{“} \Longleftarrow \text{''}$.
 A: You asked for a hint.  Here is the hint:

$\sup f\{x<a\mid x \in D\} - \varepsilon$ is not an upper bound for $f\{x<a\mid x \in D\}$ and $\inf f\{x<a\mid x \in D\} + \varepsilon$ is not an lower bound for $f\{x>a\mid x \in D\}$

===== fulll proof below =====
(I'll use that notation $f(x>a) :=f\{x > a\mid x\in D\}$ and the same for $f(x< a)$.  It's easier to type.)
Suppose $\sup f(x< a) = \inf f(x > a)$.  Let $M = \sup f(x< a) = \inf f(x > a)$
It's easy to prove $M = f(a)$.  If $ f(a) < M$ then there is a $c\in f(x < a)$ so that $f(a) < f(c) \le M$ contradicting $f$ is increasing.  Likewise $f(a) > M$ is impossible for similar reasons.
Let $\varepsilon > 0$.  $f(a) - \varepsilon$ is not an upper bound of $f(x< a)$ so there is $c< a$ so that $f(a)- \varepsilon < f(c) \le f(a)$.  As there are $c': c < c' < a$ and $f(c) < f(c') \le f(a)$ because $f$ is strictly monotonically increasing, we know $f(c) < f(a)$. 
Likewise $f(a) + \varepsilon$ is not a lower bound of $f(x>a)$ so there is $d > a$ so that $f(a) < f(d) < M + \varepsilon$.
Let $\delta = \min(a-c, d-a)$.
Then if $|x-a| < \delta$ then $c \le a-\delta < x < a+\delta \le d$.  As $f$ is monotonically increasing $f(a)- \varepsilon < f(c) < f(x) < f(d)< f(a) + \varepsilon$ or in other words $|f(x) - f(a)| < \varepsilon$.
Thus $f$ is continuous. 
