Re: Uncountable sets in CZF?

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 09/08/04 

Date: 8 Sep 2004 12:14:02 -0700

kramsay@aol.com (KRamsay) wrote in message news:<20040908024845.22164.00000641@mb-m05.aol.com>...
> ...
>
> Do not quote people as claiming things that they expressly inform
> you that they are not saying, just because you think the thing they
> refused to say follows from what they did say. Even if the person is
> simply confused, it's not accurate to describe them as having said
> they agree with your conclusion.
> ...
>
> No, even if you were right, you're not entitled to pretend I agree
> with you until I show you what's wrong with it. It's up to you to
> make sure that you're right, not up to other people to correct you
> when you're mistaken.
> ...

Hi Keith,

That helps to clear up that misunderstanding. It also is quite
obfuscatory.

So now I will agree that you definitely claim that no set bijects with
its own powerset.

Instead, you just claim that it's consistent for the naturals to
surject onto the reals in this IZF or some other framework with no
"excluded middle", and that if you assume you can compute those
functions NxN then you claim that there is a surjection from N onto R.

For a surjection from N onto R, that is an injection from R into N.
By your notion, Cantor-Bernstein demands "excluded middle", thus that
while it might be consistent with IZF or what-have-you, as it (C-B) is
probably pretty definitely not inconsistent, you claim that it is not
itself a theorem but would demand something else to make the same
claim that a surjection either way or an injection either way between
two sets implies bijection.

Indeed, I think the point to make clear is that the set of all reals
numbers is the set of all real numbers, and the set of all naturals
numbers or finite ordinals is the set of all natural numbers, and
those sets contain the same elements in each of these "models". The
set R of real numbers is totally ordered and obeys trichotomy, for any
two real numbers x and y it is exactly one of that x<y, x=y, or x>y.
This is where parallel lines do not intersect, etcetera, yet as well
where they do and the principle value of a real number is the value.

So you are claiming that it is not inconsistent in, say, IZF, for
there to be a surjection from N onto R because that doesn't
necessarily imply a bijection because Cantor-Bernstein is not a
theorem without excluded middle.

Now I'm trying to think about each function from N to N vis-a-vis NxN
and NxNxNx.... For each function F from N to N, for y=F(x) (x,y) is
an element of NxN. That's pretty simple, just saying that each
possible 2-tuple of naturals is contained with NxN.

About the set of all functions from N to N, then that would be
Nx{0,1}xN, no? I think it is obvious that it is a subset of NxNxN.
Plainly to represent a single function it is a proper subset of NxN.

How about all the functions from NxN to NxN? What are those?

Anyways we agree to disagree, I wish you luck in unitizing the analog,
figuring out why there could ever or never be another proper class, if
there is one, explaining why N<->R is consistent, why a set containing
only all the real numbers is the set of real numbers, why all real
numbers are computable, etcetera.

Thanks,

Ross F.

Relevant Pages

  • Re: Uncountable sets in CZF?
    ... there can't be a surjection from the whole of the ... | naturals to the reals. ... set is a countable union of countable sets, ...
    (sci.math)
  • Re: Cantors "diagonal argument". My Objection.
    ... is an injective mapping from NxN to N. There ... we could say we have a function J: NxN -> ... to the reals R. This is a little bit tricky. ... Here too, I don't understand, not even the notation: what is a "finit ...
    (sci.logic)
  • Re: set theory with denumerable.
    ... NxN is denumerable. ... Cartesian product is bijectable to the reals in the unit interval ... (consider each sequence .1101010101010... ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... All I know is that what we know about infinite ... > the sets I call finite have larges members. ... The set of all finite naturals is not infinite, ... Sets defined by mapping functions from the naturals to the reals which have ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> less than sqrthas no largest member. ... > The set of all finite naturals is not infinite, ... >> I WILL claim that your incomplete definition of cardinality ... > Sets defined by mapping functions from the naturals to the reals ...
    (sci.math)
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