Re: Well Ordering the Reals

Tony Orlow wrote:
>> When you say 111...111, it is necessary to give reference to the number of
>> bits. Now, I imagine you mean this string to represent the entire set !N, so it
>> has N bits. That means it represents a value that is equal to 2^N-1. Okay, you
>> got me. That value is outside of the range of the original set !N, which goes
>> from 0 through N-1, so it requires a natural outside of the range of the set of
>> naturals in question, in order to map to the entire set, and most assuredly,
>> any power set will require more bits and be bigger than the root set.
>

David R Tribble wrote:
>> So you agree that there cannot be a bijection between *N and P(*N)?
>> Or between any set S and P(S)?
>

Tony Orlow wrote:
> I have always agreed that the power set is larger. What I am trying to get
> across is an appreciation for keeping track of exactly how many bits you have,
> whether it is 2^N, N^2, 2N, or log2(N), when comparing sets represented with
> bits. When you construct a bijection, there is really no consideration given to
> the uneven rates at which one goes through two sets. In the case of the power
> set you acknowledge some difference you can't wish away, but in other cases, as
> the sets go to infinity, there is no end in sight to the bijection, so the sets
> are declared equal in size. Really, in bijecting !N with P(!N), there is also
> no end in sight to the bit strings, no identifiable place where the bijection
> breaks, except for a theoretical last element. So, it still looks like a
> bijection to me. You folks create bijections between LOTS of unequal sets.

You've got it backwards. We define bijections between equal-sized
sets, and you attempt to define bijections between unequal-sized sets.

Our examples are things like:
f(n) = 2n for all n = 0,1,2,3,...
which bijects all the naturals to all the even naturals, which are
equal-sized sets.

Your examples are attempts to biject *N with P(*N), or !N with P(!N),
which are unequal-sized sets. Your "proof" is that there is no end
to the number of "bits" used for the mapping, so there must be
enough bits available to denumerate all the subsets; you keep
saying this in spite of your admission above that there actually are
not enough bits for the bijection.

If you were really trying, you would prove us all wrong by showing
us which elements are not mapped by our bijection f. And if you
were really, really trying, you would show how your mappings
cover all of the elements between the root set and its powerset.

But you never have pulled off either one. And at the rate you keep
repeating over and over your unfounded "intuitive" arguments,
you never will.

.

Relevant Pages

  • Re: infinity ...
    ... >> Tony Orlow wrote: ... >> Therefore for this mapping, ... > any natural which maps to the entire set of naturals. ... We just agreed that it is impossible under this "bijection" ...
    (sci.math)
  • Re: infinity
    ... > Yes, for your proposed mapping between N and P, ... the largest element would not map to the ... so our bijection never runs into any such endpoint and ... what is the last element in the mapping of naturals to evens? ...
    (sci.math)
  • Re: infinity
    ... >> Tony Orlow wrote: ... since it's the same for all evens. ... Don't we have 1/2 as many evens as naturals? ... bijection from the Tonats to P? ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... > Tony Orlow wrote: ... >>> the uneven rates at which one goes through two sets. ... >>> the sets go to infinity, there is no end in sight to the bijection, so the sets ... >>> which bijects all the naturals to all the even naturals, ...
    (sci.math)
  • Re: infinity
    ... In the latter case, there is a bijection, ... >> regarding the evens, and ask how many bits each set's elements has, the evens ... like the bijection between the naturals and the ... > infinite sets, there is no way to use a "running out of" argument, ...
    (sci.math)