Re: Two results of set geometry

On 1 Oct, 00:19, Michael Press <rub...@xxxxxxxxxxx> wrote:
In article
<1191066414.782430.37...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Mike Kelly <mikekell...@xxxxxxxxxxxxxx> wrote:



On 29 Sep, 12:00, Mike Kelly <mikekell...@xxxxxxxxxxxxxx> wrote:
step...@xxxxxxxxxx wrote:
Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
It can be proven that c has the same cardinality as the power set of omega.
As cardinality is defined, in conjunction with Cantor's diagonal
argument, which is flawed, yes.
<sniiiip>

A thought struck me while showering. Why are we discussing diagonal
arguments? To prove that c has the same cardinality as the power set
of omega all we need is a bijection between the power set of omega and
the set of real numbers. William Hughes kindly gave the outline in a
previous post.

Let I(x,n) be an indicator function. x in P(N). n in N.
I(x,n) = 1 if n in x.
= 0 if n not in x.

Now a bijection between the power set of omega and the set of real
numbers is given by..

f : P(omega) -> R
f(x) maps to the equivalence class of Cachy sequences containing ( I(x,
0), I(x,0)+I(x,1)/3, I(x,0)+I(x,1)/3+I(x,2)/9, ... )

And I think that's it.. right?

Yes, that is it.

Thanks for confirmation.

.

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