Re: Simple enough for every reader?

Liste des GroupesRevenir à s logic 
Sujet : Re: Simple enough for every reader?
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.logic
Date : 20. Jun 2025, 13:40:41
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <1033ks9$1075$1@dont-email.me>
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 27 28 29 30 31
User-Agent : Mozilla Thunderbird
On 20.06.2025 11:00, Mikko wrote:
On 2025-06-19 14:13:49 +0000, WM said:

But removing all in their natural or any given order (that does not allow two or more to take the same place) implies a last removed one.
 No, it does not. That implication is not acceptable without a proof.
Removing all means that none remains.
Then either more than one have been removed at last or one has been removed at last. When the order is obeyed, more than one are excluded, because among more than one there is always an order.
 
the result is the same anyway.
>
That means that none remains. I case of known order we know the last one subtracted.
>
But that does not prove that your "the last one" denotes anything.
>
How can an ordered set be completely subtracted in its given order without a last one?
>
Euclid did not specify how to draw a circle. He merely postulated that
for any given centre and radius it can be done.
>
The radius is a length that allows to construct every point of the circumference. Further compasses are in use for constructions.
 Euclid's postulates say nothing about that.
Construction are performed with compass and ruler. See for instance the construction of a line perpendicular to a give line.
 
>
Likewise a set theory
does not specify how one set can subtracted from another.
>
But it postulates or accepts that an order exists.
 Cantor originally did. The ZF, which is the most commonly used formal
set theory, doesn't. The nearest it has is the subset relation, which
is a partial order.
Even without AC, ZF has the well-ordered v. Neumann numbers or Zermelo's set Z_0.
 
It merely postulates that it can be done.
>
I ask how it can be done. Am I the first? Is it forbidden?
 It is outside of the scope of the theory. In a particular interpretation
or application of the theory you can and possibly need to answer that
question.
 
All numbers of the v. Neumann set are claimed to be well ordered. Every ne subset has a smallest element. This does not cease when all numbers are considered. When subtracting all, the problem immediately appears as a topic within ZF.
Regards, WM

Date Sujet#  Auteur
17 May 25 * Simple enough for every reader?203WM
18 May 25 +* Re: Simple enough for every reader?113Mikko
18 May 25 i+- Re: Simple enough for every reader?1Ross Finlayson
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25 Jun 25 i                     i                       i   i i              `* Re: Simple enough for every reader?36WM
26 Jun 25 i                     i                       i   i i               `* Re: Simple enough for every reader?35Mikko
26 Jun 25 i                     i                       i   i i                `* Re: Simple enough for every reader?34WM
27 Jun 25 i                     i                       i   i i                 `* Re: Simple enough for every reader?33Mikko
27 Jun 25 i                     i                       i   i i                  `* Re: Simple enough for every reader?32WM
28 Jun 25 i                     i                       i   i i                   `* Re: Simple enough for every reader?31Mikko
28 Jun 25 i                     i                       i   i i                    `* Re: Simple enough for every reader?30WM
29 Jun 25 i                     i                       i   i i                     `* Re: Simple enough for every reader?29Mikko
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