extension of 0568 Qn 61:

yoyostein · Post by **yoyostein** » Tue Mar 20, 2012 1:22 am

Hi,

From Q61, we know that the set of all functions from R to {0,1} has cardinality equal to that of power set of R

How about

a) the set of all functions from R to (0,1)
b) the set of all functions from (0,1) to R

Thanks for your help!

blitzer6266 · Post by **blitzer6266** » Tue Mar 20, 2012 2:08 am

They're all the same

http://en.wikipedia.org/wiki/Beth_number#Beth_two

yoyostein · Post by **yoyostein** » Tue Mar 20, 2012 7:11 am

Thanks!

Is there an intuitive way to understand the set of all functions from R to {0,1} has cardinality equal to that of power set of R (Beth two)?

I read something about using an ordered pair (x,f(x)), but didn't quite understand it.

blitzer6266 · Post by **blitzer6266** » Tue Mar 20, 2012 12:38 pm

Sure. Given a function from R to {0, 1} pick all of the elements that go to 1. That's a subset of R. That's where the notation 2^/Omega comes from. The power set of /omega is just the set of functions from /omega to the 2 element set

extension of 0568 Qn 61:

extension of 0568 Qn 61:

Re: extension of 0568 Qn 61:

Re: extension of 0568 Qn 61:

Re: extension of 0568 Qn 61: