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I am lost on this proof.
Let A be a set such that If S is an element of A, then S is contained in A, and let P(A) be the power set of A. Prove that if S is an element of P(A), then S is contained in P(A).
P(A) is the power set of A.
Please someone know what this stuff is!!!!!!
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Last edited by JaneFairfax (2007-03-05 14:56:07)
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Thank you very much.
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Youre welcome. Can you follow the proof all right?
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Yes, you got me off on the right start. I appreciate it.
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Hi, I am new to the boards. I was wondering how people get their equation symbols into their posts, like Jane did above. You can also just cop and paste the symbols from the top of this page, correct?
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Welcome to the forum!
You can indeed copy and paste the symbols from the top of the page, that's mainly why they're there.
To get maths symbols like Jane did, you need to enclose some LaTeX code in [math] tags.
So, you'd put {math}*your code*{/math}, but with [] instead of {}.
We have a very good LaTeX tutorial here.
Why did the vector cross the road?
It wanted to be normal.
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I submitted a first draft to the teacher yesterday and he had a problem with me using "S" as an element of the set A, when S is just all of the elements of the set A. Tell me if you see any problems before I submit this assignment.
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Im sorry, but I dont really like it.
What your proof needs is a bit more organization of thought. First, you focus on what are you trying prove. What are you trying to prove? Namely, that if S is a member of P(A) then S is a subset of P(A). Then you think of how to go about proving it. For this particular problem, its absolutely straightforward. Namely, you assume S ∊ P(A), and then you try, by a series of deductions, to arrive at S ⊆ P(A).
In general, when you want to prove a statement of the form P ⇒ Q, you can use one of three methods. (i) The most straightforward way is to assume P and then, by a series of deductions, try and arrive at Q. (ii) You can also assume that Q is false and show that this would lead to P being false. This is known as a contrapositive proof. (iii) The third method is proof by contradiction. Assume that P is true but Q is false, then show that this would lead to a contradiction.
For this particular problem, method (i) is the one to use. Whats more Ive already shown you how to do it (see above).
EDIT: Maybe Ill explain part of my proof in more detail. Up to S ⊆ A should be straightforward. Now S ⊆ A means S is a set of members of A. But A has property that each of its members is also a subset of A. Since the members of S are members of A, we see that the members of S are all subsets of A. So S is a set of subsets of A. Now carry on with the proof.
Last edited by JaneFairfax (2007-03-06 18:32:03)
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