Question

@satellite73 need your help figuring this out.... going to type in details in the comments

Answer 1

remember this question?

http://math.stackexchange.com/questions/751410/prove-f-infty-a-infty-rightarrow-b-infty-is-a-bijection

Answer 2

I had email my prof and he gave me three hints

Answer 3

oh no not again
i was confused enough the first time

Answer 4

yeah but I've email my prof and he gave me these hints and stuff... would that help a bit? O_O

Answer 5

you tax my mind but i guess i can try
but first i have to say i have no idea what \(f_{\infty}\) means

Answer 6

we got @zarkon there too, maybe he can help

Answer 7

1.) Show that \(f_\infty \)is defined, i.e., if \(a \in A_\infty\), then \(f(a) \in B_\infty.\)
2.) Show that \(f_\infty\) is injective, clear because \(f_\infty\) is a restriction of the
injective function f.
3.) Show that \(f_\infty \)is onto, i.e., if\( b \in B_\infty\), then there is an \(a \in A_\infty\)
and \(f(a) = b\). So, show that there is an appropriate element in A that maps to
b, and then show that it even belong to \(X_\infty\).

Answer 8

and each of those 3 hints have 2 line proofs... which I don't get hth is that possible

Answer 9

AHHHHH! DJPON3!!! she's like my fav pony!

Answer 10

the only thing i can think of is that \(A_{\infty}\) is an infinite sequence of elements of \(A\) and that \(f_{\infty}\) is the extension of \(f\) so that \(f(\{a\}_n)=\{b\}_n\) where \(b_i=f(a_i)\)

Answer 11

ah we could try that approach! At least that's something

Answer 12

what i am trying to say is that i don't know what \(f_{\infty}\) means

Answer 13

:/ me either... but how the heck did my classmates score a 9 out of 10 on this thing?

Answer 14

probably knew what it meant!

Answer 15

tag @Zarkon maybe he knows what it means

Answer 16

nugh I ham so stuck!!!!

Answer 17

do you have somewhere a definition of \(f_{\infty}\)? without it, there is no point in thinking about it

Answer 18

I could look it up ...

Answer 19

I can't help unless I know what \(f_\infty\) is.

Answer 20

http://everything2.com/title/Cantor-Schr%25C3%25B6der-Bernstein+Theorem

Answer 21

found this line from the link

Consider the sets Ainf and Binf. Every b in Binf obviously has an inverse under f in Ainf. Therefore f is a bijection from Ainf to Binf.

Answer 22

f infinity the set that doesn't terminate if it has a sequence idk better ask again. x.x

Answer 23

there sent an email ... going to drop this for now...until tomorrow.

Answer 24

@satellite73 @ wio .. I found out what \[f_\infty\] is.. it's the Restriction of f, with domain and codomain restricted

Answer 25

@wio