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Thread: theorem about family of sets

  1. #1
    Member
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    Mumbai, India
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    theorem about family of sets

    Here's the theorem I am trying prove

    Suppose $\displaystyle A$ is a set, and for every family of sets $\displaystyle \mathcal{F}$
    we have

    $\displaystyle \left[\bigcup \mathcal{F} = A\right]\Rightarrow A \in \mathcal{F}$

    so prove that $\displaystyle A$ has exactly one element. Now there are two parts here. One is the existence part and second is uniqueness part. I have proved the
    existence part using method of contradiction. I chose $\displaystyle \mathcal{F}=\varnothing$ as a particular F to show the contradiction.

    Now coming to the uniqueness part, the given is

    Given ---->

    $\displaystyle \forall \mathcal{F} \left[(\bigcup \mathcal{F} = A )\Rightarrow A \in \mathcal{F}\right]$

    and the Goal is

    $\displaystyle \forall y \forall z \left[(y \in A)\wedge (z \in A)\Rightarrow (y=z)\right]$

    I will assume the negation of the goal and try to find the contradiction. Hence the
    givens are

    $\displaystyle \forall \mathcal{F} \left[(\bigcup \mathcal{F} = A )\Rightarrow A \in \mathcal{F}\right]$

    $\displaystyle (y \in A)\wedge (z \in A)$

    $\displaystyle y \neq z $

    and the goal now is contradiction. Now here I am having some difficulty,
    how do I choose particular $\displaystyle \mathcal{F}$ ? Since $\displaystyle y \in A$
    and $\displaystyle z \in A $ we know that

    $\displaystyle \{y,z\} \subseteq A$

    But there could be more elements in A. Is there any way I can consider different
    cases which will exhaust the possibilities ?

    please comment.
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    410

    Re: theorem about family of sets

    The method of contradiction is fine for existence, but probably not best for uniqueness. Suppose towards a contradiction that $\displaystyle A=\emptyset$. Then $\displaystyle \bigcup\emptyset=\emptyset=A$ and hence $\displaystyle A\in\emptyset$, a contradiction. So there is $\displaystyle a\in A$. Let $\displaystyle \mathcal{F}=\{\{a\},A\setminus\{a\}\}$. Then $\displaystyle \bigcup\mathcal{F}=A$ and hence $\displaystyle A\in\mathcal{F}$. Since $\displaystyle A\neq A\setminus\{a\}$ it follows that $\displaystyle A=\{a\}$.
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  3. #3
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    208

    Re: theorem about family of sets

    hatsoff

    thanks , that is just beautiful...

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