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

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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?

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

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

## 2 Answers

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

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.