Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 09. Jan 2025, 20:16:01
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <0132705b-414a-4990-9e4d-0474a7bbd5a8@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Mozilla Thunderbird
On 1/9/2025 11:51 AM, WM wrote:
On 09.01.2025 13:18, Richard Damon wrote:
On 1/9/25 6:39 AM, WM wrote:
Cantor will.
Every set of numbers of
the first and second number class
has a smallest element.
Hence they all are on the ordinal line.
>
Which doesn't prove your claim,
>
It proves that the numbers of
the first and second number class
form a linear system.
Apparently,
'linear' is yet.another term which
means something other than
what you (WM) want it to mean.
Apparently,
what you (WM) think you have concluded is that
the ordinal.line has the Archimedeanⁿᵒᵗᐧᵂᴹ property
(no infinitesimals, no infinites)
(finiteⁿᵒᵗᐧᵂᴹ.sets smaller.than fuller.by.one sets)
The ordinal.line does not have the Archimedeanⁿᵒᵗᐧᵂᴹ property.
It has ordinals not.smaller.than fuller.by.one ordinals.
The ordinal.line is linearⁿᵒᵗᐧᵂᴹ and
its linearityⁿᵒᵗᐧᵂᴹ does follow from its well.order,
but that's not the Archimedeanⁿᵒᵗᐧᵂᴹ property.
⎛ Because ⦃α,β⦄ holds a first ordinal,
⎜ α≠β ⇒ α<β ∨ α>β
⎜ and '<' is connected
⎜
⎜ Because ⦃α,β,γ⦄ holds a first ordinal,
⎜ α<β ∧ β<γ ⇒ α<γ
⎜ and '<' is transitive.
⎜
⎜ By definition, ¬(α<α)
⎜ and '<' is irreflexive.
⎜
⎜ A connected, transitive, irreflexive order
⎝ is a linear order.