Let ($L_1 , \le_1 $) and ($L_2 , \le_2 $) be two posets.
A function $f : L_1 \to L_2 $ is continuous if
$ \forall E_1 \subseteq L_1, E_1 \neq \emptyset, f(LUB(E_1)) = LUB(f(E_1)) $
Prove that if $f$ is continuous then it is monotonic
My idea is -
let $E_1 $ = {$ a, b $}
Then $f(E_1)$ = { $f(a), f(b)$ }
and
$ LUB(f(E_1)) $ = $f(b)$
So, we have:
$LUB(E_1)$ = $b$ means $a\leq b$
and $ LUB(f(E_1)) $ = $f(b)$ which means $ f(a) \leq f(b)$
Hence the function is monotonic as $a\leq b$ $\Rightarrow$ $ f(a) \leq f(b)$
Well, this is what my proof looks like. I'm not sure if this is correct. Could anyone please put some light or take trouble to answer with better proof?