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Set theory (cartesian products) - What is wrong with this proof? Watch

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    Status: Problem not yet solved, any help appreciated! (+rep)

    (A × B) \cup (C × D) \subseteq (A \cup C) × (B \cup D)

    Notice that above we do not have equality, but containment. What is wrong
    with the following attempt at proving the containment the other way round?
    Incorrect Theorem. For sets A, B, C, and D we have (A \cup C) × (B \cup D) \subseteq (A × B) \cup (C × D).
    “Proof”. Suppose (x, y) \in (A \cup C) × (B \cup D). Then x \in A \cup C and y \in B \cup D,
    so either x \in A or x \in C, and either y \in B or y \in D. We consider these two cases
    separately.
    Case 1. x \in A and y \in B. Then (x, y) \in A × B.
    Case 2. x \in C and y \in D. Then (x, y) \in C × D.
    Thus either (x, y) \in (A×B) or (x, y) \in C ×D, so (x, y) \in (A×B)\cup(C ×D).
    The question prior to this one was to prove the containment the 'right' way round which wasn't too hard, but I can't spot what's wrong with this proof, or how to explain it in words, so any hints or help is welcome, still got plenty of rep to give out today and thanks in advance for any help.
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    ... see the thread you posted with the original question
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    (Original post by DeanK22)
    ... see the thread you posted with the original question
    from looking at the other thread i can see generally why the containment doesn't work the other way round from the explanations given by you and ttony, but i cant see how to identify/word whats wrong with this false proof given
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    Neg?
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    (Original post by Lewk)
    ...
    There aren't 2 cases to consider, there are 4.

    One is:

    Spoiler:
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    x in A, y in D

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    (Original post by DeanK22)
    Neg?
    it wasn't me i've already repped you in a post yesterday :

    & thanks ghostwalker ill go check that out
 
 
 
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