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Thread: simple proof

  1. #1
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    simple proof

    How would you prove this, I'm stuck. Thanks

    Let A and B be sets. Then A union B = A if and only if B is subset of A.
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  2. #2
    Senior Member
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    First, you can see from the definition of a subset and from the definition of the union of two sets that for any sets A and B, $\displaystyle A\subseteq{A\cup{B}}$ and $\displaystyle B\subseteq{A\cup{B}}$

    Now, to prove our "if and only if" statement, we break it up into the two direction if-then statements:

    1. If $\displaystyle A\cup{B}=A$ then $\displaystyle B\subseteq{A}$.
    As noted above, $\displaystyle B\subseteq{A\cup{B}}$ for any sets A and B. Thus, if $\displaystyle A\cup{B}=A$, then $\displaystyle B\subseteq{A}$.

    2. If $\displaystyle B\subseteq{A}$ then $\displaystyle A\cup{B}=A$.
    Assume opposite: we have $\displaystyle B\subseteq{A}$, $\displaystyle A\cup{B}\ne{A}$. As $\displaystyle A\subseteq{A\cup{B}}$, if $\displaystyle A\cup{B}\ne{A}$, then there must be at least one element x in $\displaystyle A\cup{B}$ not in A. By definition of the union of sets, every element of $\displaystyle A\cup{B}$ must be an element of A or of B (or both). Thus if $\displaystyle x\in{A\cup{B}}$ and $\displaystyle x\notin{A}$, then $\displaystyle x\in{B}$. Thus there must be at least one element in B not in A. But this contradicts $\displaystyle B\subseteq{A}$. Thus, we show that if $\displaystyle B\subseteq{A}$ then $\displaystyle A\cup{B}=A$.

    With (1) and (2), we have proved that $\displaystyle A\cup{B}=A$ if and only if $\displaystyle B\subseteq{A}$.

    -Kevin C.
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