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This question already has an answer here:

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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marked as duplicate by Martin R, Community Feb 13 at 8:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The answer is indeed no. For a counterexample on $\mathbb{R}$, try e.g. $x\mapsto\arctan x$. $\endgroup$ – MisterRiemann Feb 13 at 7:50
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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.

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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.

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