Re: 2N=E

Am Thu, 24 Oct 2024 16:17:38 +0200 schrieb WM:

On 24.10.2024 15:54, joes wrote:

Am Thu, 24 Oct 2024 15:42:55 +0200 schrieb WM:

On 24.10.2024 14:24, FromTheRafters wrote:

WM has brought this to us :

>

The infinite must adhere to correct mathematics. Otherwise it is
only matheology, to be believed by believers who despise
mathematics.

Sets don't change. If you want to double each element of the
naturally ordered set of natural numbers you do it by creating
another set with the naturally ordered doubled elements.
{1,2,3,4,...}
{2,4,6,8,...}
These sets are the same |size|

In fact, here "the same size" applies. The change of size between both
sets is +/- 0. But since the first set contains all natural numbers,
the second set contains larger numbers because 2n > n.

Please explain how you can count past omega.

I don't. The required steps are dark. But if the set ℕ is complete, then
it covers a domain at the ordinal axis. If its density is reduced, then
its extension is increased.

If you don't even reach omega, you can forget about larger numbers.
The naturals, evens, and powers of 2 all only go up to omega.

your swapped Bob is *always* in the 'next' room since there is no
'last' room at the infinite hotel.

But if all fractions can be counted, then there is a last state,
namely when this is counting is complete.

No, "countable" means "bijective to (possibly a subset of) the
naturals",
not that there is a last one. That would imply finiteness, but N is
countably infinite.

We can set up a sequence of sets where in every set one one fractions is
indexed.

No element of the sequence indexes all.

Only if all fractions have been counted, all fractions have been
counted. That means that the sequence is complete.

And infinite.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.