# Thread: set theory different approach

1. ## set theory different approach

if we are to prove
X-(X-A) = A,
what we normally do is for an arbitrary x suppose ,
x E X-(X-A)
--> proof
--> x E A

and then
x E A
--->proof
---> x E X-(X-A)
this tells us X-(X-A) = A,
its perfect and theres no wrong with it.

but my question is what happens if I use biconditional statement as
x E X-(X-A) <---> x E X AND x ~E X -A
<---> some proof here
<---> x E A

then we dont want to split it into two parts as did earlier.

but why dont we normally prove like this???

2. $\displaystyle X\backslash \left( {X\backslash A} \right) = X \cap \left( {X \cap A^c } \right)^c = X \cap \left( {X^c \cup A} \right) = \emptyset \cup A = A$

3. yeah.. but i just wanted to know whether my approach is correct or wrong...

4. Originally Posted by doresa
yeah.. but i just wanted to know whether my approach is correct or wrong...
Frankly, I cannot follow what you are even trying to say in the OP.

5. i have mentioned it in my first post.....

6. Originally Posted by doresa
i have mentioned it in my first post.....
I meant that your first post is impossible to follow.
I have no idea what you are doing with that.

7. suppose i have to prove X-(X-A) = A,

1st step :
for an arbitrary x suppose ,
x E X-(X-A)
--> proof goes here
--> x E A
and 2nd step:
x E A
--->proof goes here
---> x E X-(X-A)
this is the normal structure of proving this sum.ok?

but if i do it like this...using the definition of set difference,
x E X-(X-A) <---> x E X AND x ~E X -A
<---> some proof here
<---> x E A
is it wrong?

8. O.K. That is what is usually called a 'pick-a-point proof'.
Some instructors perfer that all proofs be done that way.

9. x'cuse me
Some instructors perfer that all proofs be done that way.
here that way means what way?

10. and why dont you normally use the latter method?