The complement of an open set?

Question

The definition of closed set says that the complement of an open set is a closed set.

I found one theorem and its proof, and one step confused me.

We have to prove that one $x_1$ is unique on some interval $[t_0,t]$. So, we take another $x_2$. Then we have two separate sets:

1. Set $A=\left \{ s \in [t_0,t] \mid x_1 (s) \neq x_2(s) \right \}$

2. Set $A^C=\left \{ s \in [t_0,t] \mid x_1 (s) = x_2(s) \right \}$

We want to prove that $A^C=[t_0,t]$.

One thing that confuses me is that they proved that both of $A$ and $A^C$ are open sets. I won't write that proof here, but, they all make perfect sense. How can two complements be both open?

Answer 1

Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.

Answer 2

This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.