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    Prove that
     A \subseteq B \Leftrightarrow A \cap B=A

    To show  A \subseteq B \Rightarrow A \cap B=A

    x in A => x in B. (By definition.)
    Case 1 A=B
    then A \cap B=B=A
    Case 2 A \subset B
    Then there is a element of B that isn't in A so
    A \cap B=A(by definition of a intersection of sets, every element of A is in B)

    This shows that both possibillities leads to A \cap B=A. Only two possibillites by definition, so  A \subseteq B \Rightarrow A \cap B=A

    To show  A \subseteq B \Leftarrow A \cap B=A

    A \cap B=A
    => x in A and x in B= x in A.
    So by definition of intersection A \subseteq B as intersection of both sets equals A, so A must not contain unique element that isn't in B since intersection equals A. Hence, A equals B or A is a subset of B.

    Is this proof okay? Is there a shorter way to prove this?

    P.S. I thought you didn't need to use English to do Maths. Actually, I didn't but lol words.
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    What?

    It's true that A = B \Rightarrow A \subseteq B, but not that A \subseteq B \Rightarrow A = B.
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    (Original post by Simplicity)
    Case 2 A\subset B
    Then there is a unique element of B that isn't in A
    It's not unique. Let A = {1}, B = {all integers}.

    The forward direction could be proven in a much simpler way. Trivially, all x in A\cap B are also in A; conversely, all x in A are also in B (by subset property) and so are in their intersection.

    (Original post by Simplicity)
    Hence, A equals B or A is a subset of B.
    I don't understand. Your initial assumption was that A was a subset of B. Why have you written this as the last line of your proof?
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    (Original post by nuodai)
    What?
    (Original post by generalebriety)
    I don't understand. Your initial assumption was that A was a subset of B. Why have you written this as the last line of your proof?
    Sorry, but I have corrected it so it makes sense now. Latex problems.
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    (Original post by Simplicity)
    Sorry, but I have corrected it so it makes sense now. Latex problems.
    The problem's not stated correctly, so whatever you've proven is wrong.
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    Am I being really thick, or isn't the statement not true, and it should be:

     A \subseteq B \iff A \cap B = A ?

    :confused:
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    (Original post by GHOSH-5)
    Am I being really thick, or isn't the statement not true, and it should be:

     A \subseteq B \iff A \cap B = A ?

    :confused:
    Yeah, it should be.
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    (Original post by GHOSH-5)
    Am I being really thick, or isn't the statement not true, and it should be:
    Oh that explains it. I really should proof read better.
    (Original post by generalebriety)
    Yeah, it should be
    Changed it. Is the proof correct now?
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    (Original post by Simplicity)
    Change it.
    Pardon?
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    (Original post by generalebriety)
    Pardon?
    Is the proof correct now?
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    (Original post by Simplicity)
    Changed it. Is the proof correct now?
    No, the objection GE posted in post #3 still holds.
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    (Original post by DFranklin)
    No, the objection GE posted in post #3 still holds.
    But isn't the definition of A \subset B is that there is something in B that isn't in A, but everything in A is in B. So I don't see my mistake.

    Generally whats with this deduction
     A \subset B \Rightarrow A \cap B=A
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    (Original post by Simplicity)
    But isn't the definition of A \subset B is that there is something in B that isn't in A, but everything in A is in B.
    Well, yes, but that's not what you said. You used the word "unique".

    (Original post by Simplicity)
    Generally whats with this deduction
     A \subset B \Rightarrow A \cap B=A
    I don't understand the problem. I've pointed out what's wrong with your proof and posted a proof of my own.
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    Well, yes, but that's not what you said. You used the word "unique".
    Okay, however I still don't see what having the word unique makes it wrong. Yeah, I could have just said element that isn't in A.

    I don't understand the problem. I've pointed out what's wrong with your proof and posted a proof of my own.
    Oh yeah, that is much easier and shorter. Thanks

    On another subject what is the contrapositon of this
    (A \cap B)=A \Rightarrow A \subseteq B

    I'm thinking you need to get rid of the = sign. But lol.
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    I thought this was going to be a thread about things which Simplicity can't prove
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    Proving that A \subseteq B \Rightarrow (A \cap B) = A isn't too difficult, because a neccessity of A \subseteq B is that \forall x \in A : x \in B and \lnot \exists y \in A : y \not \in B, so the result follows directly that (A \cap B) = A.

    Proving that A \subseteq B \Leftarrow (A \cap B) = A is also fairly straightforward, since A \cap B is the set of all x such that x \in A and x \in B, so if this is equal to A then there cannot exist y \in A such that y \not \in B, because then such y \not \in (A \cap B) leading to a contradiction, so \forall x \in A: x \in B. There is no such requirement on B so it is possible that there are elements in B that are not in A, so the result follows quite simply from there.
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    On a side note is this the correct contraposition.
    A \cap B = A \Rightarrow A \subseteq B
    =>
    (A \cap B \subseteq A or A \subseteq A \cap B) \Rightarrow A \subseteq B
    =>
    \neg (A \subseteq B) \Rightarrow \neg(A \cap B \subseteq A or A \subseteq A \cap B)
    =>
     B \subset A \Rightarrow \neg (A \cap B \subseteq A) and \neg(A \subseteq A \cap B)
    =>
    B \subset A \Rightarrow (A \subset A \cup B) and ( A \cup B \subset A)
    As this looks wrong on many levels. Anyone?

    Its hard to know when to stop negging.
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    Right, taking them step by step...

    1. \newline (A \cap B) = A \Rightarrow A \subseteq B - we proved this above

    2. \newline (A \cap B) \subseteq A is by definition always true, so it does not follow (A \cap B) \subseteq A \Rightarrow A \subseteq B. However, \newline A \subseteq (A \cap B) is only true if A \subseteq B (because A must be wholly included in B), so this is the correct one to use

    3. The opposite to (2) applies here. That is, if A \not \subseteq B, then A \not \subseteq (A \cap B) because that would imply that A is wholly included in B. However, it still remains true that (A \cap B) \subseteq A, so that's of no use here.

    4. Once again, it is always true that (A \cap B) \subseteq A so \lnot((A \cap B) \subseteq A) (equivalently, (A \cap B) \not \subseteq A) is always false, so you can't use that here. However, if B \subset A, then it is true that A \not \subseteq A \cap B, because A contains elements that are not in B, so A \subseteq (A \cap B) cannot be true.

    5. This time, B \ne \{ \} then it is necessarily true that A \subset (A \cup B) (did you mean \cap rather than \cup?). And (A \cup B) \subset A can never be true... so basically this whole statement is wrong.
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    (Original post by Simplicity)
    Okay, however I still don't see what having the word unique makes it wrong. Yeah, I could have just said element that isn't in A.
    Then you clearly don't know what "unique" means. If A is a proper subset of B, then there is an element of B that is not in A, but there is not necessarily a unique element of B that is not in A. "Unique" means "one, and only one".

    (Original post by Simplicity)
    On a side note is this the correct contraposition.
    What's wrong with A \not\subseteq B \implies A\cap B \neq A?
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    (Original post by Totally Tom)


    I thought this was going to be a thread about things which Simplicity can't prove
    Yes, so did I.
 
 
 
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