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Math Help - Is this correct ??

  1. #1
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    Is this correct ??

     \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi  <br />
\Longleftrightarrow X_\alpha = \phi , for some  \alpha \in \bigtriangleup

    proof:

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

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


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

     \rightarrow  \exists x :\bigtriangleup \to  \bigcup_{\alpha \in \bigtriangleup} X_\alpha  \ s.t \ x(\alpha) \in X_\alpha, \forall \alpha \in \bigtriangleup \   \  \to X_\alpha \neq \phi for all   \alpha \in \bigtriangleup
    but :
    X_\alpha = \phi , for some  \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 X_\alpha ? a number? a function? a space?

    What is \phi?

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


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

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

    but  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
     \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi  <br />
\Longleftrightarrow X_\alpha = \phi , for some  \alpha \in \bigtriangleup

    proof:

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

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


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

     \rightarrow  \exists x :\bigtriangleup \to  \bigcup_{\alpha \in \bigtriangleup} X_\alpha  \ s.t \ x(\alpha) \in X_\alpha, \forall \alpha \in \bigtriangleup \   \  \to X_\alpha \neq \phi for all   \alpha \in \bigtriangleup
    but :
    X_\alpha = \phi , for some  \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 X_\alpha for every \alpha to form some \bold{x}\in\prod_{\alpha\in\Delta}X_\alpha

    The other way's easier. Note that 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 X_\beta=\varnothing with \beta\in\Delta but X\ne\varnothing then we'd have to have that there is some \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 \phi meant "emptyset"  \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


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

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

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