Re: infinity ...

In article <MPG.1dc2f7d4729e142f98a52d@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:
> > In article <MPG.1dc1b69de12df94698a510@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > 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'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.
> >
> > None of this handwaving and doubletalk designates any member of *N which
> > maps to {x in *N: x not in f(x)}. And without it, TO's "bijection " is
> > not even a surjection.
> >
> So, it is required that we determine the very last element in an infinite
> bijection?

Who says anything about "last"?

Only TO!

Order of elements is not relevant in finding an element of *N which f
will map to {x in *N: x not in f(x)}.

TO's continuing obsession with such irrelevances to the issue as order
can only be to support his attempts to avoid facing the issue squarely
and admitting his error.


Why must I name the end of the set in order for the bijection to
> be
> valid? You cannot name anything except some conceptual end of the unending
> set
> as a point where the bijection breaks down. I swear this theory just seems
> more
> and more insane to me. How do you tolerate it? Ugh! Please line up by the
> basketball courts for your head-whacking. Maybe that'll work. At least it'll
> be
> fun.
.

Relevant Pages

  • Re: infinity
    ... >> have a bijection from X to Pthat maps things that aren't ... > the last in the enumeration of the infinite set of subsets in P? ... You have a map. ... Do you agree that a bijection from naturals to evens should ...
    (sci.math)
  • Re: infinity
    ... Therefore I conclude that since this arrangement is ... >> whether there EXISTS another map that is bijective. ... bijective and saying "therefore no bijection exists". ... >> showing me a white dog and saying, "how can you look at this ...
    (sci.math)
  • Re: Cantor Confusion
    ... >> I can't so you can not show a bijection. ... I give a surjection on all existing paths. ... Each part of an edge may map to a single path, ...
    (sci.math)
  • Re: functions in set theory
    ... There is no bijection between S and T, ... The injective map between A to S is one that one needs to find and the ... for which g is not a surjection" ...
    (sci.math)
  • Re: infinity ...
    ... > which they map to through f.... ... So, indeed, it IS a member of w. ... > Neither of those possibilities causes a contradiction. ... >> of f, so f is not an onto function, and it is not a bijection. ...
    (sci.math)
Quantcast