Is this correct ??

**flower3** · March 8th, 2010, 09:37 AM

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

**mabruka** · March 8th, 2010, 02:10 PM

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 $X_1\not = X_2$

**Drexel28** · March 8th, 2010, 02:26 PM

Originally Posted by flower3

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

**mabruka** · March 8th, 2010, 04:51 PM

So $\phi$ meant "emptyset" $\emptyset$

Sorry about the confusion.

**Drexel28** · March 8th, 2010, 07:04 PM

Originally Posted by mabruka

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

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

but $X_1\not = X_2$

What did you mean here, out of curiosity?

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