Re: infinity ...

Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
> stephen@xxxxxxxxxx said:
>> Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
>> > David R Tribble said:
>> >> Tony Orlow wrote:
>> >>
>> >>
>> >> > What do you want me to try, anyway, and infinite mapping,
>> >> > element-by-element? A bijection's a bijection, right?
>> >>
>> >> Yes, that would be nice. Please show us your bijection.
>> > f(0) = ...000 = {}
>> > f(1) = ...001 = {0}
>> > f(2) = ...010 = {1}
>> > f(3) = ...011 = {0,1}
>> > f(4) = ...100 = {2}
>> > f(5) = ...101 = {0,2}
>> > f(6) = ...110 = {1,2}
>> > f(7) = ...111 = {0,1,2}
>>
>> If we define
>> w= { x : x not in f(x) }
>> then we get
>> w = {0, 1, 2, 3, ...... }
>>
>> So for what y does f(y) = { 0, 1, 2, 3, ..... }?
>>
>> And is y in f(y)?
>>
>> Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }.
>> So is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in
>> w, and F(N) does not equal w. If it is not, then N is in w,
>> and F(N) does not equal w.
>>
>> <snip>
>>
>> > But I have constructed a bijection between the two using an intermediate binary
>> > representation. What is the specific rule I have broken concerning the
>> > construction of bijections. If I haven't broken any such rule, then is it true
>> > that a bijection between two sets means that have the same size, or even
>> > cardinality?
>>
>> No you have not.
>>
>> Stephen
>>
> Notice above that no natural maps to a subset which contains it, so w is all of
> N. Imagining any completed w leads to a contradiction, since the natural that
> would map to it is always bigger than every natural in that set.

That means that imagining any completed bijection also leads to a
contradiction. w has just as much existence as your bijection.
If the bijection is complete, so is w.


> That's okay
> though, because for every natural, there's a larger one. If x exists, 2^x
> exists. The sets are infinite, so the bijection continues, even though there is
> a difference between the values which are in the subsets and the values which
> denote the subsets.

It is not a bijection. N is a subset of N, and is therefore
in the powerset of N, and if no natural maps to N, then
there is no bijection.

Stephen
.

Relevant Pages

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