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Thread: Simple Set Question

  1. #1
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    Simple Set Question

    To prove that there exists a unique set $\displaystyle N \in \mathcal{P}(X) $ such that $\displaystyle A \Delta N = A $ for all $\displaystyle A \in \mathcal{P}(X) $ is it as simple as choosing $\displaystyle N = \emptyset $ and so:

    $\displaystyle (A \cup \emptyset) - (A \cap \emptyset) = A - \emptyset = A $?
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  2. #2
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    I guess its right then?

    Thanks
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  3. #3
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    Not quite. You have shown existence but not uniqueness.
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  4. #4
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    To show uniqueness: We want to find a set $\displaystyle N \in \mathcal{P}(X) $ such that $\displaystyle (A \cup N) - (A \cap N) = A $. This means that $\displaystyle A \cup N = A $ and $\displaystyle A \cap N = \emptyset$ so that $\displaystyle A - \emptyset = A $.

    By definition, $\displaystyle A \cup N = A \Rightarrow N = \emptyset $ (which is unique). And so it follows that $\displaystyle A \cap \emptyset = \emptyset $ and $\displaystyle N = \emptyset $.

    Would this show uniqueness?
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  5. #5
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    This what must be done.
    $\displaystyle A\Delta B = A\quad \Rightarrow \quad B = \emptyset $
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  6. #6
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    Suppose that $\displaystyle x \in A \Delta B $ and $\displaystyle x \in A $. Then $\displaystyle x \in (A \cup B) - (A \cap B) \Rightarrow x \in A $ and $\displaystyle x \in A \Rightarrow x \in (A \cup B) - (A \cap B) $. It follows that $\displaystyle B = \emptyset $.

    Is this correct?
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  7. #7
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    No that does not work.

    I think it is safe to say that most of us use this equivalence:
    $\displaystyle A\Delta B = \left( {A - B} \right) \cup \left( {B - A} \right)$.

    With that we can say if $\displaystyle A\Delta B = A$ then that means $\displaystyle \left( {A - B} \right) \cup \left( {B - A} \right) = A$.
    From which we can conclude that $\displaystyle \left( {B - A} \right) \subseteq A\quad \Rightarrow \quad B = \emptyset $.

    QED
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  8. #8
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    Why doesn't it work?

    $\displaystyle x \in A \Rightarrow x \in A $?

    Oh ok: Its apparent that if $\displaystyle B-A \subseteq A $ then $\displaystyle B = \emptyset $. Sort of like a contradiction if we didn't have the empty set.
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