I have to show that $\inf(f(A)) \geq f(\sup(A))$ given that $f$ is a decreasing function and that $A$ is a non-empty bounded above subset of $\mathbb{R}$.
In my attempt I showed that for all $a \in A$ you get that $f(a) \geq \inf(f(A))$ (after showing that $f(A)$ is bounded below) and $f(a) \geq f(\sup(A))$ (after using the definition of a supremum). I can't seem to deduce from this that $\inf(f(A)) \geq f(\sup(A))$.
Any help would be appreciated.
Suppose for the purpose of contradiction that $\inf(f(A)) < f(\sup(A))$. By properties of infimum, for every $\epsilon>0$ there exists $y \in f(A)$ such that $y< \inf (f(A))+\epsilon$. If we let $\epsilon = \frac{f(\sup(A))-\inf(f(A))}{2}>0$, then this implies in particular that $y<f(\sup(A))$. Note that since $y \in f(A)$, there exists $x \in A$ such that $y=f(x)$.
Now use the fact that $f$ is decreasing to deduce a contradiction from the inequality $f(x)<f(\sup (A))$. Continue from there.
A proof without the $\epsilon$ definition. Again suppose $\inf(f(A))<f(\sup(A))$. Then by the definition of infimum, $f(\sup(A))$ cannot be a lower bound for $f(A)$. In other words, there exists $y \in f(A)$ such that $y < f(\sup(A))$. As before, since $y \in f(A)$, there exists $x \in A$ with $f(x)=y$. So the inequality becomes $f(x)<f(\sup(A))$ as above.