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Thread: set theory algebra manipulation

  1. #1
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    set theory algebra manipulation

    A,B, and C are subsets of a set U.

    prove (A-B) - (B-C) = (A-B) - C

    then....

    0= (B intersect C complement) - C.

    Here I am stuck. I cannot think of a useful identity for C.

    Tried doing absorption law for C, but it just ends up expanding. Nothing really simplifies.
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  2. #2
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    Re: set theory algebra manipulation

    Quote Originally Posted by math951 View Post
    A,B, and C are subsets of a set U.
    prove (A-B) - (B-C) = (A-B) - C
    Say $A=\{1,3,5,6,7\},~B=\{0,7,8,9\}~\&~C=\{1,2,3,5,7\} $ where $U=\{0,1,2,3,4,5,6,7,8,9\}$

    What are $(A-B)-(B-C)~\&~(A-B)-C~?$
    Last edited by Plato; Jul 13th 2019 at 09:40 AM.
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    Re: set theory algebra manipulation

    Quote Originally Posted by Plato View Post
    Say $A=\{1,3,5,6,7\},~B=\{0,7,8,9\}~\&~C=\{1,2,3,5,7\} $ where $U=\{0,1,2,3,4,5,6,7,8,9\}$
    What are $(A-B)-(B-C)~\&~(A-B)-C~?$
    I had hoped that you would use the above example to show what you are trying to show is not true.
    Look at the two attached Venn diagrams. They show that $(A\setminus B)\setminus (B\setminus C)\ne (A\setminus B)\setminus C$
    set theory algebra manipulation-b-c.gifset theory algebra manipulation-b-b-c-.gif

    BTW those were produced by WolfFramAlpha: See here. You should learn to use that website.

    Please answer this What are $(A-B)-(B-C)~\&~(A-B)-C~?$
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    Re: set theory algebra manipulation

    sorry... So it is not true? My professor gave this example for set theory to work on algebraic manipulation...I guess he was was wrong.

    (A-B) is A and everything not in B, so essentially, B-C is B and everything not in C. C is well.... C is defined as C= U- C complement.
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    Re: set theory algebra manipulation

    I've been studying nonstop for midterm 1 on tuesday. It is on logical connectives, proofs, induction, and set theory... I have hard part sometimes doing induction hypothesis part, where you have to relate oneside of equation to other.

    Also if we have a power set, let P be denote a power set. Then is this true? P(X x Y)= P(X) x P(Y). Explain.

    I say this: let X= 2 Let Y= 1

    then P(X x Y) = ( 2 x 1). So 2 x 1 = 2. which is a set and element of P(X x Y). Then P(X) = 2, which is a set and element of P(X). Then P(Y)= 1, which is a set and element of P(Y).

    so this is FALSE. Since the left hand side has 1 set, and right hand side has 2 sets. Is this correct reasoning? @Plato
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    Re: set theory algebra manipulation

    I might be wrong though, second guessing myself lol.
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    Re: set theory algebra manipulation

    Quote Originally Posted by math951 View Post
    I've been studying nonstop for midterm 1 on tuesday. It is on logical connectives, proofs, induction, and set theory... I have hard part sometimes doing induction hypothesis part, where you have to relate oneside of equation to other.
    Also if we have a power set, let P be denote a power set. Then is this true? P(X x Y)= P(X) x P(Y). Explain.
    You do not know what $\mathcal{P}(A\times B)$ means!
    It is all about ordered pairs. If $(h,j)\in(A\times B)$ then $h\in A~\&~j\in B$
    So to say $T\in\mathcal{P}(A\times B)$ means that $T$ is a collection of pairs from $A\times B$
    There is almost no relation of $\mathcal{P}(A\times B) ~\&~\mathcal{P}(A)\times\mathcal{P}(B)$
    Consider this: $(\emptyset,\emptyset)\in \mathcal{P}(A)\times\mathcal{P}(B)$ Can you justify this? i.e. Prove it?
    Is it necessary the case that $(\emptyset,\emptyset)\in(A\times B)~?$

    Now do you think that $\mathcal{P}(X \times Y)= \mathcal{P}(X)\times \mathcal{P}(Y)?~?~?$
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    Re: set theory algebra manipulation

    I'll get back to you on this tomorrow, Plato. But essentially, I think P(A x B) is a subset of sets of ordered pairs. where P(a) is a subset of sets of elements.
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    Re: set theory algebra manipulation

    Plato. I know that they are not equal, but am having a tough time proving this..

    SO basically this

    P(X x Y) = {{(X x y)}}. Where as P(X)= {{X}} and P(Y)= {{Y}}

    So if we plug in empty set:

    P(X x Y)= { empty set, {(empty set, empty set)}} and P(X) = {empty set, {empty set}} and P(Y)= {empty set, {empty set}}

    So what are the differences? P(X x Y) is the set of sets of ordered pairs. Where as P(X) is just the set of all subsets of X. Where in P(X) the elements are sets also. Is this true for P( X x Y)? Am I right with my reasoning here?
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    Re: set theory algebra manipulation

    I am overthinking this??? I know P(A) is a set of A means that, say X, such that X is an element of A and also a subset of A. So it is all combinations of subsets in set A.
    Then what is P(A x B)? Well, it is a set of sets of ordered pairs, so basically, the ordered pairs are elements and subsets as well.

    what if I consider another example:

    Is it true that P(AuB) is a subset of P(A) or P(B)?

    Well this is intuitive and I can see it easily that this is not the case! Say A= {1,2,3} B={2,3,5}
    Then (AuB) = {1,2,3,5}

    So {1,2,3,5} is included in P(AuB).

    But {1,2,3,5) is not all included in P(A) or P(B)


    However the reverse inclusion is true,

    P(A) or P(B) is a subset of P(AuB)
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    Re: set theory algebra manipulation

    Quote Originally Posted by math951 View Post
    Is it true that P(AuB) is a subset of P(A) or P(B)?
    NOTATION: $\mathscr{P}(X)$ is the set of all subsets of $X$.
    Now what you posted is exactly backwards.
    This is the case: $\mathscr{P}(A) \cup \mathscr{P}(B)\subseteq\mathscr{P}(A\cup B)$
    That says that every subset of $A$ is a subset of $A\cup B$ OR every subset of $B$ is a subset of $A\cup B$.

    Now what about that is it that you don't understand?
    Are you simply over your preparation necessary to understand this material?
    I don't think so, I hope not. But you must study the logic of these definitions.
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