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Ok...the universe in this problem is all real #'s. Let A be the closed interval [0,2] and let B be the closed interval [-1,1].
Find: A \ B ; B \ A ; complement of A (for lack of symbols); B^c (complement of B :) and etc etc...(Worried)
Yes I posted a portion of this problem earlier and I got the 1st two (I hope)... I got..
A \ B = (1,2]
B \ A = [-1,0)
....Are these right? And how do I represent their complement?
That's correct. The complement is everything in R that is not in A or not in B. The complement of an open set is closed and vice versa.
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Originally Posted by terr13
Yes...I have that in my notes...but how do I represent that in symbolic notation?...or is there any?
And I'm not sure what you mean when you say the complement of an open set is closed and vice versa....So the complement of A is neg. infinity (for lack of ability to produce symbol) to zero and two to pos. infinity?
Let's say for fun that 8==infinity...then what I said in words would be...
A^c (complementof A) = (-8,0) ^ (2,8)?
...Yuk!!!...is this right??
It would be the union of the two sets that you listed.
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Originally Posted by terr13
"It" being the complement of A or the complement of B (or both)?
And thanx so much for bearing with me! Sets were never my strong point and now I'm in a college proof class!
Complement of A would be (8,0) U (2,8).
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Originally Posted by terr13
Any good ideas as to how I would rationalize this answer? Cuz I don't really understand. (Crying)
It's just everything that's not the set, so infinity to 0, but not including 0. Then add on 2 to infinity, but not including 2. The Union is everything that they cover, similar two parts together I guess.
So could I say the complement of A = all real #'s but [0,2], since A=[0,2]?Quote:
Originally Posted by terr13
Yes, the complement of A is the same as R\A.
I think I get it now...Thank You!Quote:
Originally Posted by terr13