Re: 2N=E

On 03.11.2024 22:21, Richard Damon wrote:

On 11/3/24 12:00 PM, WM wrote:

On 03.11.2024 16:55, joes wrote:

Am Sun, 03 Nov 2024 12:56:48 +0100 schrieb WM:

On 03.11.2024 09:50, joes wrote:

pparently you do think that there is a natural n such that 2^n is
infinite.

If all naturals are there, then no further one is available. But
doubling all yields a greater number than all.
In actual infinity there is no way to avoid this.

We don't need any further ones because we ALREADY HAVE ALL OF THEM,
even including the doubles.

>
But you have not what is done to all of them afterwards. You must be clairvoyant if you knew in advance whether something is done at all.

The problem is that if you need to do them in "order" you can't complete the infinite task.

Cantor says that all are there and can be paired with all fractions, for instance. That is what I accept for a moment.

 That is the problem with your finite logic, that it can't actualy DO things in actual infinity,

I assume that it is possible.

We don't need to be clairvoyant to understand what WILL happen with a deterministic operation.

Either all numbers are there before - or not. These are the only alternatives. You must switch to and fro.
Regards, WM