Schaum's General Topology, exercise 4.11 - what am I missing? In Schaum's General Topoplogy exercise 11 in chapter 4 we find the following task:

Prove: A set $F$ is closed if and only if its complement $F^c$ is open.

This is all that the exercise says; there are no assumptions on $F$ provided. The book defines closedness in the following way:

A subset $A$ of $\mathbb{R}$ is a closed set iff its complement $A^c$ is an open set.

What I cannot understand is what the exercise is asking of me. If the set $F$ in the exercise is a subset of $\mathbb{R}$ then there is nothing to prove, since it is true by definition, right?
However, if the intention of the author is for $F$ to be a subset of some other set, I cannot see how this could be done, since open and closed sets are only defined for $\mathbb{R}$ and $\mathbb{R}^2$ as per chapter 4. The main material of the book starts from chapter 5, but from what I understand from my "sneak peek" of the later parts of the book, "open" and "closed" need to be defined in terms of some topology, but here all we know that $F$ is a set, but we don't know in what topological space it lives (right?).
I'm sure that there is some thing that I'm overlooking, but I cannot see what it might be. I would be grateful if you could point me in the right direction!
 A: After reading this answer, be sure to read Mark Bennet's subsequent comment + my reactive comment.

For what it is worth, the definition that I was taught is that $F$ is closed $\iff F^c$ is open.
However, based on this definition, consider https://en.wikipedia.org/wiki/Boundary_(topology).
One of the consequences of the definition that I was taught is 
$E_1$ An open set never contains any of its boundary points.
From this, it is easy to see that a closed set contains all of its boundary points (if any).
I have also seen the definition of an open set to be a set that contains none of its boundary points, and the definition of a closed set to be a set that contains all of its boundary points.
Under this alternative set of definitions, it is meaningful
to (for example) 
pronounce $F$ as an open set and ask you to prove that $F^c$ is a closed set.
A: I checked the book, and you are right that they just re-state their definition.
The exercise is:

Prove: A set $F$ is closed if and only if its complement $F^c$ is
open.

And their solution to that exercise is:

Note that $(F^c)^c = F$, so $F$ is the complement of $F^c$. Thus, by
definition, $F$ is closed iff $F^c$ is open.

What I think they were thinking is:
"Ok, so we defined a closed set to be a set whose complement is open. Now we state an exercise that shows exactly what these sets look like."
This would be why their solution involves noting that the complement of a complement of a set is the set itself - but apparently they forgot that they did define a set to be closed in exactly the way stated by the exercise.
