Attachment 28521

THe problem is attached. Anyone have any ideas?

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- Jun 4th 2013, 06:49 AMnexttime35Proving Isomorphism
Attachment 28521

THe problem is attached. Anyone have any ideas? - Jun 4th 2013, 07:27 AMPlatoRe: Proving Isomorphism
The title is misleading. All that is required is to show a bijection between the sets.

If $\displaystyle \alpha\in X^{\{1,2\}}$ then $\displaystyle \exists x_1\in X~\&~\exists x_2\in X$ such that $\displaystyle \alpha=\{(1,x_1),(2,x_2)\}$.

Now $\displaystyle (x_1,x_2)\in X\times X$. You should be able to complete the question. - Jun 4th 2013, 02:00 PMnexttime35Re: Proving Isomorphism
Now to finish this question, to show a bijection, I will need to show the injection (1-1) and onto-ness (surjection). In order for a function to be 1-1, for all x1,x2 in X x X, f(x1)=f(x2) implies that x1 = x2. And in order for the function to be onto, for all elements, say (x1,x2) in the codomain X x X , there exists (f(1),f(2)) in the domain such that f(f(1),f(2)) = (x1, x2).

Would that be correct? - Jun 4th 2013, 02:26 PMPlatoRe: Proving Isomorphism
Suppose that $\displaystyle \alpha\in X^{\{1,2\}}$ define $\displaystyle \Phi(\alpha)=(\alpha(1),\alpha(2))~.$

Now is it clear that $\displaystyle \Phi:X^{\{1,2\}}\to X\times X~?$

If $\displaystyle \Phi(\alpha)=\Phi(\beta)$ can you show that $\displaystyle \alpha=\beta~?$

If $\displaystyle (a,b)\in X\times X$ can you show $\displaystyle \exists\gamma\in X^{\{1,2\}}$ such that $\displaystyle \Phi(\gamma)=(a,b)~?$ - Jun 4th 2013, 03:05 PMnexttime35Re: Proving Isomorphism
So, Sigma(alpha) = Sigma (Beta) implies that (alpha(1),alpha(2)) = (beta(1), beta(2)), which implies that alpha = beta.

onto: So, let (a,b) in X x X be arbitrary and let (gamma) = sigma(alpha). Then sigma(gamma)= (gamma(1), gamma(2)) = (a,b)?

Does that follow? Thanks for walking this through, I am definitely learning the process here. - Jun 4th 2013, 03:25 PMPlatoRe: Proving Isomorphism