Sujet : Re: how
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.mathDate : 14. Apr 2024, 20:23:01
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <uvhaem$13m07$2@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 4/14/24 3:07 PM, WM wrote:
Le 13/04/2024 à 15:05, Richard Damon a écrit :
On 4/13/24 8:35 AM, WM wrote:
No, the question is: What elements of {1, 2, 3, ..., ω}*2 fall between ω and ω*2?
>
None.
That is not possible in a linear operation like doubling.
Regards, WM
Why?
Every Natural Number has aleph_0 numbers above it, one of which will be double it. Thus, doubling ALL the Natural Numbers still gives you results that are all Natural Numbers.
You logic blows up, because it can't understand such an infinite system.
Multiplication on Natural Numbers is CLOSED, so ANY Natural Number times any other Natural Number (like 2) is another Natural Number so NEVER gets as big as omega.
You just continue to run into the problem that you finite logic just blows up when attempting to try to handle infinite sets.
Re: Contradiction of bijections as a measure for infinite sets
Re: how