# Simple set question

1. Feb 16, 2005

### Kamataat

I need to prove that $$-(-A)=A$$. I guess it's the same as $$S-(S-A)=A$$, where $$S$$ is the space. So is it true, that if $$x \in S-(S-A)$$ then $$x \notin S-A$$?

- Kamataat

2. Feb 16, 2005

### Muzza

I take it -A means the complement of A (with respect to some universe)? The following are equivalent (~ means "not"):

x $\in$ -(-A)
x $\notin$ -A
~(x $\in$ -A)
~(x $\notin$ A)
~(~(x $\in$ A))
x $\in$ A

That establishes the two inclusions -(-A) $\subseteq$ A and A $\subseteq$ -(-A).

3. Feb 16, 2005

### Kamataat

Thanks, Muzza, I get your proof. Still, why are the NOT steps neccessary? Why not this instead:

$$x \in -(-A)$$
$$x \notin -A$$
$$x \in A$$?

If you go (in your post) from step #1 to step #2 directly, then why don't you go from #2 to #6 (e.g. skip #3, #4 and #5)? I mean, if from #1 follows #2, then doesn't #6 follow from #1 and #2 combined (w/o the intermediate steps)?

- Kamataat

4. Feb 16, 2005

### Muzza

*shrug* Do as you please :P

5. Feb 16, 2005

ok, tnx

- Kamataat