Help with Proving #F > #R for Set F of Functions f:R --> {0,1}

[maw26]

Let F be the set of all function f:R-->{0,1} . Prove That #F>#R

I'm thinking about relationship between the set F and the power set of R.

but don't know how to continue

any help ?

 

[verty]

Show that the set of functions is no smaller than the powerset of R. How do you do that?

 

[maw26]

Show that the set of functions is no smaller than the powerset of R. How do you do that?

How to proof this?

 

[╔(σ_σ)╝]

I had a similar problem.
This thread would be of some help : https://www.physicsforums.com/showthread.php?t=432668 from post 8 till the end.

 

[HallsofIvy]

Show that the set of functions is no smaller than the powerset of R. How do you do that?

Actually, that is NOT sufficient. That would show #F\ge #R, not #F> #R. maw26, can you verify that the problem is to show #F> #R and NOT #F\ge #R?

 

[Citan Uzuki]

I believe HallsOfIvy is mistaken. The powerset of any set is strictly larger than that set, in particular |P(R)| > |R|. So vertigo's suggestion will work.