Re: Two results of set geometry

David R Tribble wrote:
Now you're saying that you accept more than two infinities.
All of these infinities in your spectrum must be of the "actual"
variety, I take it.


Tony Orlow wrote:
No, those formulaically related to omega are all countable, including
2^omega. Those formulaically related to c are all uncountable, and not
related to 2^omega. This is where I deeply disagree with standard
theory, among other places.

Can you convert your "deep disagreement" into a
mathematically sound argument? Oh, wait, I think I
see something below...


David R Tribble wrote:
Of course, I never was able to make sense of any of your
talk about "omega - 1", anyways. And you realize you'd
make quite a splash if you could ever pin down an infinity
or two lying between Aleph_0 and c, which would certainly
go a long way towards disproving the CH.


Tony Orlow wrote:
If we assert that f(x)>g(x) for all x > (n in N), and that for all
infinite "counts" c, c > (n in N), then f(c)>g(c).

Whoa there, cowboy. How do you deduce that
f(c) > g(c) for c where c > n for all n in N?

Can you show us how ">" is defined for these values
of c? Because it's not the usual arithmetic meaning
of ">", which only operates on reals, is it?


If we further assert
that there exists a bijection between S and T such that for all seS f(s)
e T, then also for all teT, E g s.t. A teT g(t) e S. There is an inverse
function, which shows the measure of the target set, realtive to the
source.

So there is a function f, which being a bijection maps
elements from S to elements of T. And vice versa, which
is the inverse function g, which also maps elements of
T to elements of S.

So how does g "show" the "measure" of the "target" set
relative to the "source" set? Is S the "source" set and
T the "target" set? If so, how does g show the "relative
measure" of sets S and T?

For instance, let's take the set of naturals,
N = {0, 1, 2, 3, ...}
and the set of positive integer squares:
S = {0, 1, 4, 9, ...}
and define a bijection function:
f:N->S, f(n) = n^2 for all n in N
and its inverse function:
g:S->N, g(n) = sqrt(n) for all n in S

How does g show us the "relative measure" between
sets S and N? I know you'd like to claim that S is
"smaller than" N, but how does g show this?


In this respect,

In what respect? I lost you there.

not only are there countable sets greater than
N and less than R,

Such as? How do you know a set is "greater than" another?

but also countably infinite sets less than N, like E,

How is E "less than" N?

and uncountably infinite sets finitely greater than c.

Such as? How is a set "finitely greater than" another set?

Undoubtedly, this is foreign to you, but maybe it starts to form an
image in your mind. One can only hope. :)

It looks like all those undefined terms you're using just
create more questions than they answer.

.

Relevant Pages

  • Re: Well Ordering the Reals
    ... >> Tony Orlow wrote: ... >> Daryl McCullough wrote: ... >> David R Tribble said: ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... > David R Tribble said: ... > Tony Orlow wrote: ... number of terms AND the same sum, and caliming so is patently false. ...
    (sci.math)
  • Re: Two results of set geometry
    ... Tony Orlow wrote: ... especially when you have infinite sets that are ... David R Tribble wrote: ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... > David R Tribble said: ... > Tony Orlow wrote: ... The first 20 reals ...
    (sci.math)
  • Re: infinity
    ... > David R Tribble said: ... > Tony Orlow wrote: ... >>> Nine out of every ten infinitesimals. ...
    (sci.math)
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