If $f$ is continuous and $f(A)$ is open, is $A$ necessarily open?

Question

In the context of general topology, a function is said to be continuous if the pre-image of any open subset is also an open subset. Of course, this does not say that the image of an open set must be an open set (and in fact, this is not true in general). But, what about the statement "all sets whose image is open must be open"? This is a different statement, and I don't know what to think of it intuitively. My thoughts:

Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ continuous. Let $A$ be a set such that $f(A)$ is open. I am trying to decide whether or not $A$ is necessarily open. I know from set theory that $A \subseteq f^{-1}(f(A))$, and the fact that it is not an equality is very important (if this was an equality, the statement would be immediately true). With this, I am able to conclude that $A$ is a subset of the open set $f^{-1}(f(A))$, and apparently I can't conclude anything better. With this in mind, it seems to me that the statement is probably false, but I was unable to find any counterexample either.

Answer 1

Counterexample: The function $f:\mathbb R\to\mathbb R, x\mapsto x^2$ obviously is continuous. Be $S=[-2,-1)\cup (2,3)$. This is clearly not an open set, but $f(S)=(1,9)$ which is open.

Answer 2

Let $A=(-2,0)\cup(0,1]$ (isn't open in $\mathbb R$, and $f(x)=x^2$ (continuous on $\mathbb R$). Then $f(A)=(0,4)$ which is open in $\mathbb R$.

Answer 3

Since you ask it in a general context let me add a simple abstract example.

Let $Y$ be a topological space with a unique element. Let $X$ be any topological space. Then the unique $f:X \to Y$ is continuous.

We have $f(A)$ is open for every $A \subset X$. Thus, as soon as in $X$ not every set is an open set, you have a counterexample.