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Math Help - Proof without use of charactrist function that:

  1. #1
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    Unhappy Proof without use of charactrist function that:

    Proof without use of charactrist function that:
    a) A ∩ B is the largest set containing both A and B
    b) A B is the smallest set containing both A and B

    please help me.
    thank you
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  2. #2
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    Re: Proof without use of charactrist function that:

    The first statement is not true.
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  3. #3
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    Re: Proof without use of charactrist function that:

    Quote Originally Posted by reza95 View Post
    Proof without use of charactrist function that:
    a) A ∩ B is the largest set containing both A and B
    b) A B is the smallest set containing both A and B
    The way you have translated these makes them questionable.

    Usually they are written as:
    a) $A\cap B$ is the largest subset of both $A~\&~B$.
    b) $A\cup B$ is smallest set such $A\subset A\cup B~\&~B\subset A\cup B$ or $A\cup B$ is the smallest superset of $A~\&~B$.

    Proof by contradiction.
    Suppose that $\exists C$ such that $C\subset A~,~C\subset B~\&~C\ne A\cap B~\&~A\cap B\subset C$.

    So it must be that $\exists x\in C$ such that $x\notin A\cap B$. But that means $x\in A~\&~x\in B.$ that is a contradiction.
    Last edited by Plato; March 7th 2014 at 10:25 AM.
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  4. #4
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    Re: Proof without use of charactrist function that:

    I think It is True.
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  5. #5
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    Re: Proof without use of charactrist function that:

    Quote Originally Posted by reza95 View Post
    a) A ∩ B is the largest set containing both A and B
    I think It is True.
    No it is not true as posted. The problem is in translation,
    Had you written a) A ∩ B is the largest set contained by both A and B, then it would have been true.
    In fact that is what I proved in reply #3.
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  6. #6
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    Re: Proof without use of charactrist function that:

    Yes, it *is* true that A∩B is the largest set contained in A *and* contained in B (note that this is not quite what the original post says). Another proof of this:

    Suppose K is a set such that: K⊆A, and K⊆B. We need to show that K⊆A∩B.

    So pick any element k in K. Since K⊆A, we have that k is in A (since this is the DEFINING property of "subset", everything in K is also in A).

    Likewise, k is in B, since K⊆B.

    Since k is in A, and k is in B, we have that k is in A∩B = {x in U:x is in A, and x is in B} (here U can be any set that contains A and B, but it is typical to take U to be the "universe of discourse" that is: some set large enough to contain A,B and whatever "other sets" we may be considering. At the very least, we must have AUB in U).

    Since we have shown that every k in K is also in A∩B, we conclude K⊆A∩B, QED.

    We can construct a "similar" proof that AUB is the SMALLEST set containing A and B. For suppose we have such a set H, with A⊆H, and B⊆H.

    Here, we must show that AUB⊆H.

    To do this, pick any x in AUB.

    Case 1: x is in A. Since x is in A, x is in H, because A⊆H.

    Case 2: x is in B. The same reasoning shows x is in H.

    Since x is in A, or x is in B (or both), in all cases, x lies in H, so AUB⊆H, QED.
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  7. #7
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    Re: Proof without use of charactrist function that:

    By the way, how you prove things depends very strongly on how the words in the statement are defined. I have seen texts in which A\cap B is defined as "the largest set contained in both A and B" and A cup B is defined as "the smallest set that contains both A and B".
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