If A is compact then is f(A) closed?

Question

I have the following question:

Let $f: X \rightarrow Y$ be a continuous map between topological spaces and let $A$ be a subset of $X$.
For the following statement, either give a proof or a counterexample.

• If $A$ is compact then $f(A)$ is closed.

I cannot think of a proof that starting from compact set will end up in a closed set so I think the statement is false. I understand the concept of compactness but I cannot think of an counterexample (e.g if I had an open set I would pick something like $(0,1)$ for an open set or similar but I'm a bit confused with what to pick for a compact set).

Any help/suggestions would be appreciated.
Thank you.

Answer 1

Here is a kind of trivial (i.e. boring) counterexample:

Let $X=\mathbb R$ with the standard (metric) topology, $Y=\mathbb R$ with the trivial topology (i.e. only $\varnothing$ and $\mathbb R$ are open) and define $f:X\to Y$ by $f(x)=0$ for all $x\in X$. This is continuous, and the image of any compact set is $\{0\}$, which is not closed.

As pointed out in the comments, the target space $Y$ needs to fail the Hausdorff condition in order for a counterexample to occur.

Answer 2

Let $X$ be $\mathbb{R}$ together with the usual euclidean topology, and $Y$ be $\mathbb{R}$ together with the trivial topology. Let $A := [0,1]$. This is clearly compact. Let $f:X \rightarrow Y$ be such that $f(x) = x$.

This is continuous trivially, and $f(A)$ is not closed in $Y$.