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If $A$ is closed in $\mathbb{R}^{n}$, is it true that $f(A)$ is closed in $\mathbb{R}^{n}$ as well?

Here, $f : A \rightarrow \mathbb{R}$ is continuous.

Intuitively, I think the answer is no, but I cannot come up with a counterexample. Can someone please help me?

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    The answer is indeed no. For a counterexample on $\mathbb{R}$, try e.g. $x\mapsto\arctan x$. – MSDG Feb 13 '19 at 07:50

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Counterexample: $n=1, A = \mathbb R$ and $f(x)=e^x$. We have $f(A)=(0, \infty)$, which is not closed.

Fred
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A function between two topological spaces with the property that every image of a closed set is closed is indeed called a closed function.

Now consider, for the sake of simplicity, $n = 1$. Then $\arctan(x)$ is not closed, since it sends $\mathbb{R}$ into (-$\frac{\pi}{2}, \frac{\pi}{2}$).

Bonus: since the image of a compact set under a continuous map is still compact, and a compact set in an Hausdorff space is closed, you should look for counterexamples in closed but unlimited sets.

Simone Ramello
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