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Math Help - Set Theory basics: an element of an element

  1. #1
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    Set Theory basics: an element of an element

    Hi everyone,


    In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

    There is something wrong here, right? What can I infer from knowing the first two statements?

    Thank you!
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Set Theory basics: an element of an element

    You musn't confuse elements and subsets.

    For example if D = {1,2,3}, and A is { {1,2,3}, {2,4}} (A consists of 2 elements {1,2,3} and {2,4} and D has 3 elements 1,2,3)
    2 is an element of D, and D is an element of A, but 2 is not a subset of A.
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  3. #3
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    Re: Set Theory basics: an element of an element

    It is not always true that x\in D and D\in A imply x\subseteq A, but for some x, D and A this may happen.
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Set Theory basics: an element of an element

    Quote Originally Posted by kripkey View Post
    Hi everyone,


    In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

    There is something wrong here, right? What can I infer from knowing the first two statements?

    Thank you!
    Hi kripkey!

    Suppose x={1}.
    And suppose D={{1}, y}. It means that x is an element of D.
    Then let A={{{1}, y}, 1}. We'd have that D is an element of A and also that x is a subset of A.

    Erm... what's wrong? ;P
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  5. #5
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    Re: Set Theory basics: an element of an element

    Quote Originally Posted by kripkey View Post
    In a proof I'm writing I inferred from, x is an element of D, and D is an element of A, that x is a subset of A.

    Are you working with an ultra-filter on a set?
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  6. #6
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    Re: Set Theory basics: an element of an element

    Thanks for the responses guys!

    Not sure what an ultra-filter on a set is, so probably not! Now I'm curious

    I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.
    So I guess I got a little lost on the way.
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Set Theory basics: an element of an element

    Quote Originally Posted by kripkey View Post
    Thanks for the responses guys!

    Not sure what an ultra-filter on a set is, so probably not! Now I'm curious

    I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.
    So I guess I got a little lost on the way.
    The definition of a subset from D is that it contains only elements that are also in D.
    If you take the union of G with 1 or more other sets, the elements of G will also be in that union, so G will be a subset of that union.
    If D is that union, it means that G is a subset of D and not the other way around.
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  8. #8
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    Re: Set Theory basics: an element of an element

    That makes sense, thanks ILikeSerena.
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  9. #9
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    Re: Set Theory basics: an element of an element

    Quote Originally Posted by kripkey View Post
    I was actually trying to somehow justify that if D = arbitrary union of G, then D is a subset of G.

    It seems as if you are asking about generalized unions.
    This is an example of the use of the Union Axiom.
    If G is a collection of sets then x \in \bigcup G \; \Leftrightarrow \,\left( {\exists b \in G} \right)\left[ {x \in b} \right].

    Here is an example given by Herbert Enderton.

    Suppose G=\{\{1,2,8\},\{2,8\},\{4,8\}\} then
    D=\bigcup G=\{1,2,4,8\}.

    Now it is perfectly clear that D\notin G\text{ and }D\not\subset G.

    Moreover, G\notin D\text{ and }G\not\subset D.
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