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Let A be any set in which is a subset of some universal set U.
Prove that:
A intersection U = A
Now, I get the logic behind this and understand the statement. My problem is developing a proof for this. Could anyone help?
Rather than giving me the answer I would really appreciate some guidelines to follow please?
These are others that I also understand them but dont know how to generate a proof:
A union U = U
A intersection {} = {}
Last edited by LuisRodg (2008-01-15 05:34:02)
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By the definition of subset, x ∈ A implies x ∈ U.
Obviously x ∈ A also implies x ∈ A, and so x ∈ A implies x ∈ A intersection U.
For the union one, maybe rewrite "A union U" as "U union A\U" and work from there?
A\U denotes the set of elements that are within A but not within U.
Why did the vector cross the road?
It wanted to be normal.
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for union, can you not show:
the union of A and U is the set of all x ∈ A or ∈ U (inclusive or)
since x ∈ A implies x ∈ U. there is no x in U such that x is not in A. thus the union of A and U = U
note that i have made no assumptions about U. U can be any set here as long as A is its subset
Last edited by luca-deltodesco (2008-01-15 07:47:48)
The Beginning Of All Things To End.
The End Of All Things To Come.
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Just start with expanding on what you know, and remember what it is you are trying to show. It most set theory problems, there is really no more work than this than to say, "duh".
A intersect U = A
We want to show first that A intersect U is a subset of A. So we start by assuming that an element x is in A intersect U and we wish to show that x is in A. Remember that "x is in A intersect U" means "x is in A and x is in U".
Now we want to show that A is a subset of A intersect U. So again, we start by assuming that x is in A. Now we want to show that x is in A and x is U. Well of course x is in U (duh!, it's the universal set), and we're assuming that x is in A.
The union problem follows exactly the same pattern as this.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Ok thanks all.
Got another one.
Prove: A union (A intersection B) = A
Now, I can picture this in my head and see thats its true. But again, developing the proof is tedious for me...
Anyways, I got it up to this:
A union (A intersection B) = (x e A) or x e (A intersection B) // by the definition of union
----------------------------- = (x e A) or ((x e A) and (x e B)) // by the definition of intersection
How do I keep going to simplify it to A?
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Show that A intersection B is a subset of A.
You had a previous question that made you prove that C union D = D, where C is a subset of D, and so combining those two results will get you what is required.
Why did the vector cross the road?
It wanted to be normal.
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How do I keep going to simplify it to A?
You don't simplify it to A. All you need to do is show that x is in A. And you've already done that with this statement:
(x e A) or ((x e A) and (x e B))
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Let A be any set in which is a subset of some universal set U.
Prove that:
A intersection U = A
Now, I get the logic behind this and understand the statement. My problem is developing a proof for this. Could anyone help?
Rather than giving me the answer I would really appreciate some guidelines to follow please?
These are others that I also understand them but dont know how to generate a proof:
A union U = U
A intersection {} = {}
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