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Thread: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

  1. #1
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    2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    For all sets X and Y , Y ∩ (X-Y) ≠

    Ok so I'll assume for all sets X and Y , Y ∩ (X-Y) =

    If I say Let set Y be the set of Integers and also let set X then

    if Y = {1,2,3} and X = {4,5,6}

    then Y ∩ (X-Y) = because is a subset of all sets ..including integers

    subbing the Y value into the original leaves {1,2,3} ∩ ({4,5,6}-{1,2,3}) ≠

    I think I have totally stuffed this but I tried.....

    Any help would be appreciated..
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  2. #2
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    I am not sure what you are asking. If Y= {1, 2, 3} and X= {4, 5, 6} then Y- X= {1, 2, 3}- {4, 5, 6} is the empty set so the intersection with Y is also the empty set. Why do you think otherwise?
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Quote Originally Posted by HallsofIvy View Post
    I am not sure what you are asking. If Y= {1, 2, 3} and X= {4, 5, 6} then Y- X= {1, 2, 3}- {4, 5, 6} is the empty set so the intersection with Y is also the empty set. Why do you think otherwise?
    That is not how set minus works. $X-Y$ is guaranteed to be disjoint from $Y$ because the set minus literally removes any elements of $X$ that are also in $Y$. If you have $X-Y$ where $X,Y$ are disjoint sets, then you get all of $X$ (because $X$ did not contain any elements of $Y$, so there was nothing to remove). The intersection of two disjoint sets is always the empty set.

    Here is my attempt.
    To prove: For all sets $X$ and $Y$, $Y \cap (X-Y) = \emptyset$.
    Proof by contradiction:
    Let $X,Y$ be sets and assume that $Y \cap (X-Y) \neq \emptyset$. Let $x \in Y \cap (X-Y)$ be any element (we know one exists because we already assumed that the intersection is nonempty).
    Since $Y \cap (X-Y) \subseteq Y$ and $Y\cap (X-Y) \subseteq (X-Y) \subseteq X$, it must be that $x \in Y$. and $x \in X$. Thus, $X-Y \subseteq X-\{x\}$, which implies that $x \notin X-Y$, contradicting the fact that $Y \cap (X-Y) \subseteq (X-Y)$. Therefore, the initial assumption is false, and $Y \cap (X-Y) = \emptyset$.
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Quote Originally Posted by bee77 View Post
    For all sets X and Y , Y ∩ (X-Y) ≠
    Ok so I'll assume for all sets X and Y , Y ∩ (X-Y) =
    If I say Let set Y be the set of Integers and also let set X then

    if Y = {1,2,3} and X = {4,5,6}
    then Y ∩ (X-Y) = because is a subset of all sets ..including integers
    subbing the Y value into the original leaves {1,2,3} ∩ ({4,5,6}-{1,2,3}) ≠ .
    If $Y=\{1,2,3\}~\&~X=\{4,5,6\}$ then is it clear to you that $Y\cap X=\emptyset~?$
    Then $X\setminus Y= X$ Yes or No? (Note no element is removed from $Y$.)

    For all sets $X~\&~Y , \begin{align*}Y\cap (X\setminus Y)&=Y\cap( X\cap Y^c)\\&=Y\cap(Y^c\cap X)\\&=(Y\cap Y^c)\cap X\\&=\emptyset\cap X\\&=\emptyset. \end{align*}$
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Sorry it asks to prove by contradiction.

    For all sets X and Y , Y ∩ (X-Y) ≠
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Quote Originally Posted by bee77 View Post
    Sorry it asks to prove by contradiction.

    For all sets X and Y , Y ∩ (X-Y) ≠
    You cannot prove a false statement by contradiction or any other way. You can disprove it, though, as both Plato and I just did.
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Thanks Slip Eternal I'll read through your answer and go through it ..It was a bit tough for me
    cheers
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Thanks Plato ..
    sorry that was a bit tough for me ..but yes I could understand the part with the empty set ..just the reasoning etc I struggled with making the contradiction ...or in the case as slipeternal just pointed out disproving it .
    cheers
    Last edited by bee77; Aug 23rd 2017 at 06:16 AM.
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    Re: 2 of 3 Contradiction with sets and intersection ...this one is a bit tough

    Oh they put that part in the question I had to answer ..." prove by contradiction" ..thanks . ("I'll write that in my answer that it is disproved ").
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