extension of 0568 Qn 61:

Forum for the GRE subject test in mathematics.
yoyostein
Posts: 36
Joined: Tue Feb 28, 2012 12:14 am

extension of 0568 Qn 61:

Postby 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
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: extension of 0568 Qn 61:

Postby blitzer6266 » Tue Mar 20, 2012 2:08 am


yoyostein
Posts: 36
Joined: Tue Feb 28, 2012 12:14 am

Re: extension of 0568 Qn 61:

Postby 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
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: extension of 0568 Qn 61:

Postby 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




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