Sujet : Re: Contradiction of bijections as a measure for infinite sets
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.mathDate : 06. Apr 2024, 14:58:55
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <uurkev$8bgp$3@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 4/6/24 9:55 AM, WM wrote:
Le 06/04/2024 à 15:40, Richard Damon a écrit :
On 4/6/24 9:26 AM, WM wrote:
Le 05/04/2024 à 12:57, FromTheRafters a écrit :
WM explained on 4/4/2024 :
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Explain why first bijecting n and n/1 should destroy an existing bijection!
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You still seem to think that sets change. If you mean 'n' is an element of the naturals then of course N bijects with the naturals as embedded in Q.
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Of course. But if someone doubts it, I could directly map the naturals n/1 to the fractions with the result that there is no bijection.
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No, not "No Bijection", but that mapping isn't a bijection.
That mapping is Cantor's proposal. But for every other mapping, the O's would also remain. All O's! It is th lossless exchange which proves it.
Cantor's proposal is between members of two distinct sets.
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No, that is disproved by the remaining Os.
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Which only shows that this one mapping doesn't work.
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It is Cantor's famaous mapping, more than a century believed to be a bijection.
But HIS does work, when you do it right.
And, when you try it within one set, as opposed to between two sets,
If it operates, it must operate within one set too.
Why?
IT is a mapping betweens elements of two defined sets.
If you don't have those two sets, you are following the mapping.
You are just proving your stupidity.
Regards, WM