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Math Help - Transitive Set Problem

  1. #1
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    Transitive Set Problem

    Prove that if A is transitive and A \ne \phi, then \phi \in A.

    I can only say that if A is transitive, then \forall a(a \in A \longrightarrow a \subseteq A). How would I have to do in order to continue from here?

    Thanks in advance.
    Last edited by mr fantastic; August 28th 2011 at 04:47 AM. Reason: Title.
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  2. #2
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    Re: Transitive Set Problem (Help)

    If A is a transitive set, then by definition: whenever x \in A, and y \in x, then y \in A. That is it. Now, if A is nonempty, as is the case by your assumption, then there is some element x \in A. If x is the empty set, then the case satisfies our proof. On the other hand, let us suppose x is nonempty. Then there is some y \in x. By the definition of transitivity, y \in x \Rightarrow y \in A. The same reasoning applied to x can be applied to y now--i.e., either y is empty or it contains something that is also in A. It might be fruitful to index these elements as a sequence. Thus, what we end up having to prove is that this member sequence will "bottom out" at some point. Now, how to go about this proof, I am unsure, and maybe I've approached it in an inefficient or less fruitful manner.
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  3. #3
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    Re: Transitive Set Problem (Help)

    Hi

    we can use the axiom of regularity , which is one of the axioms of
    Zermelo–Fraenkel set theory. Look up here

    Axiom of regularity - Wikipedia, the free encyclopedia

    axiom says

    "Every non-empty set A contains an element B which is disjoint from A"

    Since A \neq \phi , using this axiom,

    B \in A \;\;\mbox{and}\; B\cap A =\phi


    \because B \in A and A is transitive , so

     B \subseteq A

     \forall x[x \in B\Rightarrow x \in A ]

    and

    B\cap A =\phi means

    \forall x[x \in B \Rightarrow x \notin A]


    Now lets assume that

    B \neq \phi

    \exists x \in B

    \Rightarrow x \in B

    since

     \forall x[x \in B\Rightarrow x \in A ]

    and

    \forall x[x \in B \Rightarrow x \notin A]

    we have

    x \in A \wedge x \notin A

    which is contradiction. so the assumption that

    B \neq \phi is wrong.

    \Rightarrow  B=\phi

    \because B \in A \Rightarrow \phi \in A

    hope this helps.... \smile \heartsuit
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  4. #4
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    Mumbai, India
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    Re: Transitive Set Problem (Help)

    just curious about the solution given by me..... can somebody confirm my solution here ...... thanks
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