# Thread: Is this correct ??

1. ## 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"

2. 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 $\displaystyle X_1\not = X_2$

3. Originally Posted by flower3
$\displaystyle \prod_{\alpha \in \bigtriangleup} X_\alpha = \phi \Longleftrightarrow X_\alpha = \phi$ , 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!

4. So $\displaystyle \phi$ meant "emptyset" $\displaystyle \emptyset$

$\displaystyle \Delta=\{1,2\}, X_1= \frac{1}{2}, X_2=1$
$\displaystyle X_1 \times X_2 = \frac{1}{2}$
but $\displaystyle X_1\not = X_2$