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.

  • $\begingroup$ You say $X$ is directed if and only if each finite subset of $X$ has an upper bound in $X$. Do you mean to specify finite subsets? If not, then what is an example of and $X$ that is not directed? $\endgroup$ Jan 21, 2015 at 22:36
  • $\begingroup$ By a finite subset I mean something like $[0,1] \subset \mathbb{R}$ should I use the term bounded instead? $[0,1] \subset \mathbb{R}$ is directed. $[0,1[ \subset \mathbb{R}$ is not. (Over a complete lattice $\leq$, that is.) $\endgroup$
    – Auberon
    Jan 21, 2015 at 22:48
  • 1
    $\begingroup$ @Auberon No, you don't mean that. If anything you want to abandon the 'complete lattice' from the definition. I think it's standard to define it for posets. $\endgroup$
    – Git Gud
    Jan 21, 2015 at 22:54
  • $\begingroup$ @GitGud I think I get it, thanks $\endgroup$ Jan 21, 2015 at 22:55
  • $\begingroup$ @Omnomnomnom My example was unnecessarily complicated, so I deleted it. The same idea works with $X=\{a,b\}$ in the following diagram. $\begin{array}{ccc} & & c\\\ &\nearrow&\uparrow\\a& & b\end{array}$ $\endgroup$
    – Git Gud
    Jan 22, 2015 at 10:02

1 Answer 1


For example: in this case, we can take $X = (0,1/2)$. Verify that this is a directed subset of $[0,1]$.

Note that $$ T(\sup(X)) = T(1/2) = 3/4 \neq \sup(T(x)) = \sup(0,1/4) = 1/4 $$

  • $\begingroup$ I've tried a similar approach but: why is $T(1/2) = 1/4$ and not $= 3/4$? $\endgroup$
    – Auberon
    Jan 21, 2015 at 22:38
  • $\begingroup$ Check your definition: $1/2 \leq 1/2 \leq 1$, so $T(1/2) = 1/2(1/2) + 1/2$. $\endgroup$ Jan 21, 2015 at 22:39
  • $\begingroup$ I should've clarified that I meant the $T(1/2)$ on the right hand side of the $\neq$. You say $sup(T(X)) = sup(T(0), T(1/2)) = sup(0, 1/4) = 1/4$. Why is $T(1/2) = 1/4$ and not $= 3/4$ like you just implied in your comment? $\endgroup$
    – Auberon
    Jan 21, 2015 at 22:44
  • $\begingroup$ I am not implying that $T(1/2) = 1/4$. I am implying that $$ \sup_{x \in (0,1/2)}T(x) = 1/4 $$ This is quite a different statement. $\endgroup$ Jan 21, 2015 at 22:47
  • $\begingroup$ Why is that true? If you pop in $x = \frac{1}{2}$ in $f$, $\frac{3}{4}$ rolls out, am I not right? Apparently this is the essence of my problem. Sorry if I sound confused. $\endgroup$
    – Auberon
    Jan 21, 2015 at 22:52

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