Finding counter examples for two statements Are the following  False? I do think they should be false but couldn't find any counter examples to justify them
Let $(X,\preceq_x)$ $(Y\prec_y)$ be partially ordered sets and let $f:X\to Y$ be an order preserving function and $A\subseteq X$
$m$ is a maximal element of A $\Rightarrow$ $f(m)$ is a maximal element of $f(A)$
$l$ =$SupA$ $\Rightarrow$ $f(l) =sup(f(A))$

One way to get a counterexample for the second one is to introduce a gap between $f[A]$ and $f(\sup A)$. Take $X=Y=[0,1]$ with the usual order on each, let $A=[0,1)$, and let
$$f:X\to Y:x\mapsto\begin{cases}
x/2,&\text{if }x\in A\\
1,&\text{if }x=1\;.
\end{cases}$$
Then $f[A]=\left[0,\frac12\right)$, so $\sup f[A]=\frac12$, but $f(\sup A)=f(1)=1$.
