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Thread: Is this correct ??

  1. #1
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    Is this correct ??

    $\displaystyle \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi
    \Longleftrightarrow X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $

    proof:

    $\displaystyle \Longrightarrow \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi $ means :

    $\displaystyle X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $


    $\displaystyle \Longleftarrow $
    Assume $\displaystyle \prod_{\alpha \in \bigtriangleup} X_\alpha \neq \phi $

    $\displaystyle \rightarrow $ $\displaystyle \exists x :\bigtriangleup \to \bigcup_{\alpha \in \bigtriangleup} X_\alpha \ s.t \ x(\alpha) \in X_\alpha, \forall \alpha \in \bigtriangleup \ $ $\displaystyle \ \to X_\alpha \neq \phi $ for all $\displaystyle \alpha \in \bigtriangleup $
    but :
    $\displaystyle X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $ "contradiction"
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  2. #2
    Member mabruka's Avatar
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    Sorry but i need a little context here. What is $\displaystyle X_\alpha$ ? a number? a function? a space?

    What is $\displaystyle \phi$?

    You cant expect an accurate answer on this if you dont give enough information since in general that would not be true:


    $\displaystyle \Delta=\{1,2\}, X_1= \frac{1}{2}, X_2=1 $

    $\displaystyle X_1 \times X_2 = \frac{1}{2} $

    but $\displaystyle X_1\not = X_2 $
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    $\displaystyle \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi
    \Longleftrightarrow X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $

    proof:

    $\displaystyle \Longrightarrow \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi $ means :

    $\displaystyle X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $


    $\displaystyle \Longleftarrow $
    Assume $\displaystyle \prod_{\alpha \in \bigtriangleup} X_\alpha \neq \phi $

    $\displaystyle \rightarrow $ $\displaystyle \exists x :\bigtriangleup \to \bigcup_{\alpha \in \bigtriangleup} X_\alpha \ s.t \ x(\alpha) \in X_\alpha, \forall \alpha \in \bigtriangleup \ $ $\displaystyle \ \to X_\alpha \neq \phi $ for all $\displaystyle \alpha \in \bigtriangleup $
    but :
    $\displaystyle X_\alpha = \phi $ , for some $\displaystyle \alpha \in \bigtriangleup $ "contradiction"
    The fact that the product of non-empty sets is non-empty is, in fact, equivalent to the axiom of choice. Think about it, you need to choose any one element from $\displaystyle X_\alpha$ for every $\displaystyle \alpha$ to form some $\displaystyle \bold{x}\in\prod_{\alpha\in\Delta}X_\alpha$

    The other way's easier. Note that $\displaystyle X=\prod_{\alpha\in \Delta}X_\alpha=\left\{\bold{x}:\Delta\mapsto\bigc up_{\alpha\in \Delta}X_\alpha: \bold{x}(\alpha)\in X_\alpha,\text{ }\forall \alpha\in\Delta\right\}$ now if $\displaystyle X_\beta=\varnothing$ with $\displaystyle \beta\in\Delta$ but $\displaystyle X\ne\varnothing$ then we'd have to have that there is some $\displaystyle \bold{x}\in X\implies \bold{x}(\beta)\in X_\beta$. Oops!
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  4. #4
    Member mabruka's Avatar
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    So $\displaystyle \phi$ meant "emptyset" $\displaystyle \emptyset$


    Sorry about the confusion.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post


    $\displaystyle \Delta=\{1,2\}, X_1= \frac{1}{2}, X_2=1 $

    $\displaystyle X_1 \times X_2 = \frac{1}{2} $

    but $\displaystyle X_1\not = X_2 $
    What did you mean here, out of curiosity?
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