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!
extension of 0568 Qn 61:

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: extension of 0568 Qn 61:
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.
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.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: extension of 0568 Qn 61:
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