Function from closed set to closed set - both true and not true? Consider the function $f:\mathbb{R}^2\to \mathbb{R}$ defined by $f((s,t))=2s$. Is it true or not that $f$ maps closed sets to closed sets? I thought that it is not true, because we could consider the set $A:=\{(s,0)\in \mathbb{R}^2:s\in \mathbb{R}\}$, which is closed in $\mathbb{R}^2$. Then $f(A)=(-\infty,\infty)$, which is open in $\mathbb{R}$. However, $f(A)$ is also closed in $\mathbb{R}$.
So, from the logic standpoint, can we say that this is a counterexample, or not? Or can we say that this example can be both true and false?
 A: A counterexample would consist of a closed set whose image is not closed. Open and closed are not opposites (despite what might be inferred from the name). Indeed, you've noted that $\mathbb{R}$ is both open and closed.
Thus this is not a counterexample at all.
A: $\mathbb{R}$ is closed and open. This is not a matter of logic, but a matter of terminology. "Closed" and "open" are logically independent from each other: "closed" doesn't mean "not open", nor does "open" mean "not closed".
For the set you described $A$, you correctly understood that $A$ is closed and that $f(A)$ is open and closed. Since $f(A)$ is closed, $f$ maps this particular closed set to a closed set. So it is not a counterexample. Now, the question you have to answer is: does $f$ map all closed sets to closed sets?
A: As others have said, your $A$ is not a counterexample. But the set $A = \{(x,y) \in \mathbb{R}^2: xy= 1\}$ is closed in the plane (it's a set of the form $q^{-1}[\{1\}]$ for $q(x,y) = xy$). But $f[A] = \mathbb{R}\setminus\{0\}$, which is not closed (as $0$ is its closure but not in the set, or because $\{0\}$ is not open).
To see that $f[A] = \mathbb{R}\setminus\{0\}$: let $f(s,t) \in f[A]$ , so $(s,t) \in A$. Then $s \neq 0$ (otherwise $(s,t) \notin A$), so $f(s,t) = 2s \neq 0$ as well. And if $y \neq 0$ is a real number, $x:= (\frac{y}{2}, \frac{2}{y}) \in A$ and $f(x) = y$.
