Re: Contradiction of bijections as a measure for infinite sets

Liste des GroupesRevenir à s physics 
Sujet : Re: Contradiction of bijections as a measure for infinite sets
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.math
Date : 29. Mar 2024, 14:29:42
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <uu6fo6$3dq4t$1@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 3/28/24 4:07 PM, WM wrote:
Le 27/03/2024 à 18:54, Jim Burns a écrit :
On 3/27/2024 9:38 AM, WM wrote:
Le 26/03/2024 à 16:40, Jim Burns a écrit :
>
ℕ and ℚ have the same infinity.
>
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
>
ℕ and ℚᶠʳᵃᶜ are the same size.
 ℕ and ℕ are the same size.
ℚ and ℚ are the same size.
And, it can be shown that ℕ and ℚ are the same size.

 Removing a proper fraction decreases ℚ but leaves it larger than ℕ. When the size changes it cannot remain the same.
Nope, Removing a single element from an infinite set doesn't change its size.

 Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope, showing NO BIJECTION CAN EXIST proves different sizes.
Having just a failed bijection shows nothing for infinite sets.

 Between infinite sets there cannot exist any mapping because most elements are dark. But we can assume that very simple mappings like f(x) = x are true even for dark elements.
Only because you logic doesn't handle the infinite sets.
Your "Darkness" is just your logic system saying "No" to its misuse.

 Therefore between the rational numbers and the natural numbers f(n) = n/1 can be accepted, also f(n) = 1/n, but not f(n) = 2n.
Sure it can, you just don't understand it because you are using imporoper logic for such a system.
Like trying to argue about colors in a black and white photo.

 Regards, WM
  Regards, WM
 

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