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

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.

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

• The answer is indeed no. For a counterexample on $\mathbb{R}$, try e.g. $x\mapsto\arctan x$. – MisterRiemann Feb 13 at 7:50
Counterexample: $$n=1, A = \mathbb R$$ and $$f(x)=e^x$$. We have $$f(A)=(0, \infty)$$, which is not closed.
Now consider, for the sake of simplicity, $$n = 1$$. Then $$\arctan(x)$$ is not closed, since it sends $$\mathbb{R}$$ into (-$$\frac{\pi}{2}, \frac{\pi}{2}$$).