Let $f : \small \mathbb{R} → \mathbb{R}$ be a continuous nondecreasing function. Let $A$ be a nonempty, bounded set.
(a) Show that $f(\sup A) = \sup(f(A)).$
(b) If we drop the assumption that $f$ is continuous, what can you say?
Attempt:
$x\le \sup(A)\ \ \forall x \in A$
$\Rightarrow f(x)\le f(\sup(A))\ \ \forall x \in A$
$\Rightarrow \sup(f(A))\le f(\sup(A))\ \ $
also, $f(x)\le \sup(f(A))\ \ \forall x \in A$
$\Rightarrow x\le f^{-1}(\sup(f(A)))\ \ \forall x \in A$
$\Rightarrow \sup(A)\le f^{-1}(\sup(f(A)))\ \ $
$\Rightarrow f(\sup(A))\le \sup(f(A))\ \ $
Thus $f(\sup(A))= \sup(f(A))$
But in the Highlighted line I have used that $f$ is injective, but $f$ may not be injective as $f$ is given to be non decreasing. So how to correct that part.
You are right about your argument: you can't use $f^{-1}$, so you need something else. Let $y = \sup(A)$. Using the definition of the supremum, for each $n \ge 1$ we can pick $x_n \in A$ such that $y-1/n < x_n \le y$. The squeeze lemma tells us that $x_n \to y$ as $n \to \infty$. Since $f$ is continuous we know that $f(x_n) \to f(y)$ as $n\to \infty$. Let $\epsilon >0$ and pick $N$ such that $n \ge N$ implies that $\vert f(x_n) - f(y) \vert < \epsilon$. Then $n \ge N$ implies that $$ f(y) = f(y) - f(x_n) + f(x_n) \le \vert f(y) - f(x_n) \vert + f(x_n) < \epsilon + f(x_n) \\ \le \epsilon + \sup f(A). $$ Since $\epsilon >0$ was arbitrary, we deduce that $$ f(\sup A) = f(y) \le \sup f(A). $$