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

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Homework Help Overview

The discussion revolves around proving that the cardinality of the set of functions from the real numbers to the set {0,1} is greater than the cardinality of the real numbers. Participants are exploring the relationship between this set of functions and the power set of the real numbers.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are considering the relationship between the set of functions and the power set of the reals, questioning how to establish that the cardinality of the function set is strictly greater than that of the reals.

Discussion Status

There is an ongoing exploration of the problem, with some participants suggesting that demonstrating the set of functions is no smaller than the power set is insufficient for proving the desired inequality. Others are questioning assumptions about the relationships between these sets.

Contextual Notes

Some participants note that the problem specifically requires showing a strict inequality (#F > #R), rather than a non-strict inequality (#F ≥ #R).

maw26
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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 ?
 
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Show that the set of functions is no smaller than the powerset of R. How do you do that?
 
vertigo said:
Show that the set of functions is no smaller than the powerset of R. How do you do that?


How to proof this?
 
vertigo said:
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?
 
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.
 

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