Re: infinity

Randy Poe said:
>
> Tony Orlow wrote:
> > Virgil said:
> > > The proof does not involve contradiction at all, unless one claims what
> > > cannot exist. It proves by construction that for any f:X -> P(X), one
> > > can always construct a subset of X (and member of P(X)) not in the image
> > > of f.
> > You constructed no such set.
>
> Incorrect. For any map f:X->P(X),
> w = {x in X: x not in f(x)}
> is a set.
>
> It is a well-defined set, it is constructed from any map f,
> and it is a member of P(X), being a subset of X.
>
> It also has the property that f(x) does not equal w for
> any x in X. This is provably true no matter what X is,
> whether it is finite or infinite, and no matter what
> mapping f:X->P(X) is chosen.
>
> > You defined the subset so that it is the set,
>
> Eh? We show how to define a w for any set X and any
> map f:X->P(X). That's the definition.
>
> For different maps, w will be a different set. It is
> constructed based on f so that no matter what f is, it will
> fail to map any element of X to w. Choose a different f,
> you get a different w.
>
> For your f, w happens to be all of X.
>
> > and derive a contradiction from assuming any natural mapping to
> > the entire set, but that is another theme on the largest finite.
>
> "Largest finite" never comes into it.
>
> Your mapping is defined in such a way that there is no
> x in X such that f(x) = X. This has nothing to do with
> "largest finite". Your map, as you defined it, has the
> property that no element x is in f(x). Therefore it's
> pretty obvious without any references to "largest finite",
> that there is no x in X such that f(x) = X.
Not within the range of any given X, but in the overall infinite set, there is.
>
> I don't have to know any more about your map, or even about
> X, than that property. If you tell me that x is never in
> f(x), then I can guarantee you can't find me x in X such
> that f(x) = X. Because then x would have to be in f(x),
> which by your definition is NOT TRUE FOR ANY ELEMENT.
>
> See how reasoning works? No "largest finite", in fact,
> no reference to the properties of x at all. These could
> be sets of mice or library books.
But, in order to define any element mapping to the entire set, the last element
must be identified and the set completed.

Let me ask you this: When do you expect to "finish" this mapping? Are you going
to go through all the numbers and find the set which corresponds to every
single one? Will you ever get to finding the number corresponding with the full
set, or will there always be another natural, and another subset, after any one
you identify? What you are asking for is equivalent to me asking you for the
even that maps to the largest natural, or any natural in the second half of the
naturals. You cannot identify either, but if you could, you would see that 2x
is not in the original value range. How many significant bits do your naturals
have? How many significant bts do your evens have? One fewer bit, 1/2 as many
evens. If the number of bits matter, then they matter.
>
> - Randy
>
>

--
Smiles,

Tony
.

Relevant Pages

  • 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: CANTORs theorem
    ... > map, hence the map is not surjective, and so the cardinality of the ... But the set of non-generators, which depends on the mapping chosen, ... There are enough naturals remaing. ...
    (sci.math)
  • Re: infinity ...
    ... and that therefore your mapping scheme is not a bijection ... since you claim you have a map that maps ... includes all the finite and infinite naturals. ...
    (sci.math)
  • Re: infinity
    ... >> The proof does not involve contradiction at all, ... It is a well-defined set, it is constructed from any map f, ... This is provably true no matter what X is, ... mapping f:X->Pis chosen. ...
    (sci.math)
  • Re: CANTORs theorem
    ... But set theorists would argue that doesn't matter, because even> twice the naturals are not enough to map onto P. ...
    (sci.math)