Let me start by defining some terminology to be sure I made no errors there. Parts of this are translated freely from my mother tongue so feel free to correct terminology or the definitions themselves if needed.
Complete lattice
A partially ordered set in which all subsets have both a supremum and an infimum.Monotonic map
Let $L, \leq$ be a partially ordered set. A map $T : L \mapsto L$ is called monotonic if $x \leq y$ implies that $T(x) \leq T(y)$.Directed set
Let $X$ be an arbitrary subset of a partially ordered set. $X$ is directed if and only if each finite subset of $X$ has an upper bound in X.Continuous map
A map $T : L \mapsto L$ over a complete lattice $L, \leq$ is called a continuous map, if $T(sup(X)) = sup(T(X))$ for every directed subset $X$ of $L$.
$T(X)$ is defined as $T(X) = \{T(x) | x \in X \}$
Now it can be shown that for a map $T : L \mapsto L$ over a complete lattice $L,\leq$
$T$ is continuous $\implies$ $T$ is monotonic
However the other way round is not true. Example:
The map $f : [0,1] \mapsto [0,1] : x \mapsto \begin{cases} \frac{1}{2}x & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}x + \frac{1}{2} & \frac{1}{2} \leq x \leq 1 \end{cases} $
over a complete lattice $[0,1],\leq$ is monotonic, but not continuous.
How can one show that $f$ is not continuous using the definitions above? I figure the discontinuity will be at $x = \frac{1}{2}$ but I cannot find how to prove this. I've tried taking several subsets of $[0,1]$ and tried to find a contradiction, without success.
Anyone wants to shine a light on this? Thanks in advance.
For example: in this case, we can take $X = (0,1/2)$. Verify that this is a directed subset of $[0,1]$.
Note that $$ T(\sup(X)) = T(1/2) = 3/4 \neq \sup(T(x)) = \sup(0,1/4) = 1/4 $$