I'm trying to think of a counterexample to the claim if $f: X \to Y$ is continuous and $E \subseteq X$ is closed and bounded $f(E)$ is closed and bounded. Here, $X$ and $Y$ are metric spaces. Obviously, in Euclidean spaces this would be true, since continuity preserves compactness and compactness is equivalent to closed and bounded by Heine-Borel.
Help is appreciated.
Best Answer
Let $X=Y=\Bbb{R}$ and $f$ be the identity. Equip $Y$ with the standard metric and $X$ with the bounded metric $d(x,y)=\min(|x-y|,1)$. Then $f$ is continuous, and $X$ is closed and bounded (as a subset of itself) but its image $Y$ is not bounded.
More examples: $X=(0,1)$, $Y=\Bbb{R}$, $f(x)=1/x$, $E=X$. Here the image is neither closed nor bounded.
$X=(0,1)$, $Y=\Bbb{R}$, $f(x)=x$, $E=X$. Here the image is bounded but not closed.