Re: 2N=E

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.

 

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. Either by Cantor's original setup
{1/1}
{1/1, 1/2,}
{1/1, 1/2, 2/1}
{1/1, 1/2, 2/1, 1/3}
{1/1, 1/2, 2/1, 1/3, 2/2}
...
or in my way
XOOO... XXOO... XXOO... XXXO... ...
XOOO... OOOO... XOOO... XOOO... ...
XOOO... XOOO... OOOO... OOOO... ...
XOOO... XOOO... XOOO... OOOO... ...
....................................
Only if all fractions have been counted, all fractions have been counted. That means that the sequence is complete.
Regards, WM