Proof that $-(-A) = A$

Question

I have tried to prove that, given a set $A$, $-(-A)=A$. Are there flaws in my logic? I am very new to proof writing and set theory so any tips on structuring the proof would be greatly appreciated. My proof is:

If $S$ is the space and $A\subset S$, then $-(-A)=S-(-A)$.

Suppose that $x \in S-(-A) \iff x \notin-A \iff x\notin S-A \iff x\in A. $ Therefore, $-(-A)=A$

My main queries with my proof are:

1)Is it necessary to introduce the space $S$ and state $-(-A)=S-(-A)$?

2)Is my use of $\iff$ correct?

Answer 1

I assume that by $-A$ you understand the complement of $A$ in $S$, i.e. we assume that $A\subseteq S$ and $$-A=S-A=\{x\in S\ |\ x\not\in A\}$$

Is it necessary to introduce the space $S$?

Yes, $-A$ is not well defined without the set it is living in. For example if $A=\{1\}$ and $S=A$ then $-A=\emptyset$. But if $S=\{1,2\}$ then $-A=\{2\}$. So as you can see the final result depends on $S$.

However some properties are independent on $S$. And $-(-A)$ is one of them. No matter what $S$ is $-(-A)=A$.

Is my use of $\iff$ correct?

Yes.

Answer 2

An easy way out:

$$x\in A\implies x\notin -A\implies x\in-(-A)\implies A\subset -(-A)$$

and on the other hand

$$x\in-(-A)\implies x\notin -A\implies x\in A\implies -(-A)\subset A$$