Re: abundance of irrationals!)


mueck...@xxxxxxxxxxxxxxxxx wrote:
> Every non-empty set of natural numbers contains at least one number
> which is not smaller han the cardinality of that set. This theorem
does
> not state anything about the cardinality of the set. It is proven for
> any finite set, because we dobt the existence of infinite sets.
Should
> they exist, however, volation of that theorem would show them to not
> exist.

Whaat? Let's take that step by step:

Mueckenheim's Cardinality Theorem: Every set S of natural numbers
contains at least one element >= |S|.

Proof: 1. It is true for finite sets S [rp: I'll accept that,
it's not hard to prove by the bijection definition of finite
set].

2. It would also be true of infinite sets if they exist,
but I (WM) don't know how to prove that.

WM Corrolary: If you find me an infinite set that doesn't obey
the Cardinality Theorem, then it can't exist because I
know it has to be true of all sets.

-------------------------

You don't see anything wrong with that? You "just know"
infinite sets have to have this property, without proof,
so if they don't have this property that PROVES there's
something inconsistent about them?

Only that they're inconsistent with a property you
think they should have but have no way of proving.
Maybe.... just maybe... this is because they don't
have this property. Wouldn't that explain why you don't
have a way to prove it?

> My list:
> 0.1;
> 0.01; 0.11;
> 0.001; 0.101; 0.011; 0.111;
> ...
> contains every positive real number of (0,1).

It does not contain 1/3 = 0.11111....

> Proof: I can tell you the position of first appearance of EVERY digit
> of EVERY real number of (0,1).

How does that prove it contains 1/3?

Your list has the property that the n-th number has
at most M(n) = ceiling(log2(n)) digits which are nonzero.

Let n be the position at which 1/3 is in your list. Then
this number has at most ceiling(log2(n)) nonzero digits.

But 1/3 has no nonzero digits. Therefore 1/3 is not
in your list.

- Randy

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