Is it possible to prove that -(-A)=A using the concept of set subtraction?

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In summary, the conversation discusses how to prove the equivalence of -(-A) and A, with one person suggesting that x \in -(-A) is the same as x \notin -A and therefore x \in A. The other person agrees with the proof but questions the necessity of the intermediate steps. The conversation concludes with a casual agreement to interpret the proof however one pleases.
  • #1
Kamataat
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I need to prove that [tex]-(-A)=A[/tex]. I guess it's the same as [tex]S-(S-A)=A[/tex], where [tex]S[/tex] is the space. So is it true, that if [tex]x \in S-(S-A)[/tex] then [tex]x \notin S-A[/tex]?

- Kamataat
 
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  • #2
I take it -A means the complement of A (with respect to some universe)? The following are equivalent (~ means "not"):

x [itex]\in[/itex] -(-A)
x [itex]\notin[/itex] -A
~(x [itex]\in[/itex] -A)
~(x [itex]\notin[/itex] A)
~(~(x [itex]\in[/itex] A))
x [itex]\in[/itex] A

That establishes the two inclusions -(-A) [itex]\subseteq[/itex] A and A [itex]\subseteq[/itex] -(-A).
 
  • #3
Thanks, Muzza, I get your proof. Still, why are the NOT steps neccessary? Why not this instead:

[tex]x \in -(-A)[/tex]
[tex]x \notin -A[/tex]
[tex]x \in A[/tex]?

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
*shrug* Do as you please :P
 
  • #5
ok, tnx

- Kamataat
 

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