Re: Contradiction of bijections as a measure for infinite sets

On 4/4/24 9:07 AM, WM wrote:

Le 04/04/2024 à 14:19, Richard Damon a écrit :

On 4/4/24 5:33 AM, WM wrote:

Le 03/04/2024 à 15:59, FromTheRafters a écrit :

WM presented the following explanation :

Le 02/04/2024 à 17:51, Jim Burns a écrit :

On 4/2/2024 3:36 AM, WM wrote:

>

If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.

>
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an existing bijection?

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Your 'trick' only fails to demonstrate a bijection. Failing to demonstrate a bijection does not mean that there is no bijection, only that your 'trick' doesn't work to that end.

>
Explain why first bijecting n and n/1 should destroy an existing bijection!
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It doesn't, Bijections are always between two DISTINCT sets, not a set and a piece of itself thought of as a set.
>

"In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is equinumerous to A. Explicitly, this means that there exists a bijective function from A onto some proper subset B of A." Wikipedia.
 Regards, WM
 

Right, but that "Proper Subset" is considered as an independent item, not as just pieces of the original set.
You don't seem to understand what a "Set" is.
Looking at the set of numbers: {1, 2, 3, 4, 5, 6, 7, 8, ...}
Just "marking" every other one as:
{ 1, 2*, 3, 4*. 5, 6*, 7, 8*, ...}
The marked numbers are not a Subset (yet) but just a piece of a set.
You can make them a set, and thus a subset by creating the set as:
{2, 4, 6, 8, ...}
Your sloppiness might work in finite sets, and you only seem to think in finte terms, but it doesn't work for infinte sets, because it causes these sorts of issues.
By your logic, it would be IMPOSSIBLE for the "subset" to be equinumerous, as you have shown.
This doesn't mean that statement is false, just that it uses definitions that differ from what you seem to be willing to use, so you have excluded yourself from those parts of logic.