Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Cardinalities
Replies: 17   Last Post: Jun 18, 2000 6:33 AM

 Messages: [ Previous | Next ]
 Steve Leibel Posts: 300 Registered: 12/6/04
Re: Cardinalities
Posted: Jun 7, 2000 3:23 PM

In article <393E4DD2.D4A141D4@gmx.de>, Sebastian Holzmann
<SHolzmann@gmx.de> wrote:

> Robin Chapman wrote:
> > Dafydd y garreg wen <mavnw@csv.warwick.ac.uk> wrote:
> > > What is the cardinality of the power set of N? Is it the same as the
> > > cardinality of R?

> >
> > Yes.

>
> This answer is incorrect. This statement is an incarnation of the famous
> "continuum hypothesis". It is proven to be independent of the standard
> axioms of set theory. So we cannot prove it true or false using them.
>

That's a common misconception. It's easy to prove that card(P(N)) =
card(R) as follows

Given any subset S of N, consider the sequence of 0's and 1's whose n-th
term is

1 if n is an element of S, and

0 if n is not an element of S.

Putting a radix point to the left of the sequence gives the binary
expression of some real between 0 and 1.

Conversely, given a real between 0 and 1, write down its binary expansion,
which looks like .0011101010101 . . . say, and define S to be the subset
of N consisting of those n for which the n-th term of the binary expansion
is 1.

I've just demonstrated a bijection between the set of reals between 0 and
1, and the power set of N.

The continuum hypothesis asks whether there is a set of reals whose
cardinality is strictly between that of N and R.

Steve L