# Prove or disprove that if $f$ is continuous function and $A$ is closed, then $\,f[A]$ is closed.

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.

• Think about the complements. What does it imply about images of open sets? Feb 23, 2014 at 18:17
• Any restrictions on the domain/codomain of the function $f$? Feb 23, 2014 at 18:17
• @CarstenSchultz Thank you, I deleted. But am I right in saying that the OP is wrong in saying that 'if $f(A)$ is closed implies $A$ is closed then $f$ is continuous..'? I am still puzzling about a counterexample. Feb 23, 2014 at 18:48
• @drhab, most likely the OP was confused there, but I could not immediately think of a counterexample either. It might be true... Feb 23, 2014 at 19:53
• @CarstenSchultz If you are still interested. For counterexamples see the answer to my question: math.stackexchange.com/q/688320/75923 Feb 24, 2014 at 10:30

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.
$$\arctan:\mathbb R\to(-\pi/2,\pi/2)\text{ is surjective.}$$