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Math Help - Set theory proof?

  1. #1
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    Set theory proof?

    IS my proof correct?


    Prove: For all sets A, A U (empty set)= A

    This statement is true.

    Proof:
    Suppose that A is a set where x is an element of A U (empty set).

    So, x is an element of A OR x is an element of (empty set).

    If x is an element of A then A=A because every set is a subset of itself.

    If x is an element of (empty set) then x is STILL an element of A b/c (empty set) is an element of all sets, namely, set A.

    There fore if x is an element of A or x is an element of empty set it is always an ELEMENT of A. Therefore the original statement is TRUE and for all sets A U (empty set)= A.
    Last edited by matthayzon89; April 10th 2010 at 10:07 AM.
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  2. #2
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    Hello matthayzon89
    Quote Originally Posted by matthayzon89 View Post
    IS my proof correct?


    Prove: For all sets A, A U (empty set)= A

    This statement is true.

    Proof:
    Suppose that A is a set where x is an element of A U (empty set).

    So, x is an element of A OR x is an element of (empty set).

    If x is an element of A then A=A because every set is a subset of itself. Grandad says: this doesn't really make sense.

    If x is an element of (empty set) then x is STILL an element of A b/c (empty set) is an element of all sets, namely, set A. Grandad says: this doesn't either. And note that the empty set is a subset of all sets, not an element.

    There fore if x is an element of A or x is an element of empty set it is always an ELEMENT of A. Therefore the original statement is TRUE and for all sets A U (empty set)= A.
    First, let's define first what we mean by the union of two sets A and B.
    The union of two sets A and B is the set of elements that are in A or in B. Using Set and Logic notation, this means that
    (x \in A \cup B) \Leftrightarrow (x \in A \lor x \in B)
    Next, note that to prove two sets A and B are equal, we usually show that A \subseteq B and B \subseteq A.

    So here you'll need to show that
    (i) A \cup \oslash \subseteq A
    and
    (ii) A\subseteq A \cup \oslash
    Now we prove that A \subseteq B by proving that x \in A \Rightarrow x \in B. So we start the proof of (i) with:
    x \in A \cup \oslash
    \Rightarrow (x \in A) \lor (x \in \oslash), from the definition of union, above

    \Rightarrow x \in A, since x \in \oslash is False for all x

    So we have now proved (i) that:
    A \cup \oslash \subseteq A
    For (ii), for all sets B:
    x \in A \Rightarrow (x \in A) \lor (x \in B)
    \Rightarrow x \in A \cup B, again using the definition of union, above
    \Rightarrow A \subseteq A \cup B
    In particular, when B = \oslash:
    A \subseteq A \cup \oslash
    And that completes the proof.

    Grandad
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