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If $f:X\to Y$ is a continuous function and $A\subset X$ is a closed set, then $f[A]$ is also closed.
I know that if $f[A]$ is closed implies $A$ is closed then $f$ is continuous, but I'm not sure that the converse is true, I can't find a counterexample either.
It is NOT true in general.
For example, let $f: \mathbb R\to\mathbb R$, with $$ f(x)=\mathrm{e}^{x}, $$ which is continuous, and let $A=(-\infty,0]$, which is a closed subset of $\mathbb R$, in its usual topology.
However, $\,\,f[A]=(0,1]$, which is not closed.
Another, and perhaps simpler, counterexample is $$ f(x)=\frac{x}{1+x}:\overbrace{[0,\infty)}^{\text{closed}}\mapsto\overbrace{[0,1)}^{\text{not closed}} $$ You'll notice that all the counterexamples consist of unbounded domains. A closed and bounded set in $\mathbb{R}^n$ is compact, and it is true that if $f$ is continuous and $A$ is compact, then $f(A)$ is compact, which is also closed.
$$\arctan:\mathbb R\to(-\pi/2,\pi/2)\text{ is surjective.}$$
