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Re: Cardinalities
Posted:
Jun 7, 2000 3:23 PM
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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
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