Sujet : Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 19. Oct 2024, 17:04:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <ed1862ff-3679-4175-bb25-c317be9713b2@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 27
User-Agent : Mozilla Thunderbird
On 10/19/2024 4:16 AM, WM wrote:
On 18.10.2024 00:34, Jim Burns wrote:
On 10/1v7/2024 2:22 PM, WM wrote:
On 17.10.2024 00:39, Jim Burns wrote:
The only set of natural numbers with no first
is the empty set..
>
No, the set of dark numbers is
another set without smallest element.
A nonempty set without a first element
is not a set of only finite ordinals.
>
The set of dark numbers contains
only natural numbers.
There is a general rule not open to further discussion:
Things which aren't natural numbers
shouldn't be called natural numbers.
What you call a "set of finite ordinals" is
not a set
but a potentially infinite collection.
There is a general rule not open to further discussion:
Finite sets aren't potentially infinite collections.
----
Consider nonempty S of only finite ordinals:
only ordinals with only finitely.many priors.
k ∈ S is a finite ordinal
Its set ⦃j∈𝕆:j<k⦄ of priors is finite.
⦃j∈𝕆:j<k⦄∩S ⊆ ⦃j∈𝕆:j<k⦄
⦃j∈𝕆:j<k⦄∩S is a finite set
⦃j∈𝕆:j<k⦄∩S holds its first or is empty.
⎛ If Priors.in.S ⦃j∈𝕆:j<k⦄∩S is empty
⎝then k is first.in.S
⎛ If Priors.in.S ⦃j∈𝕆:j<k⦄∩S is not empty
⎜ then i is first.in.⦃j∈𝕆:j<k⦄∩S
⎜
⎜⎛ For i and m ∈ S, i≠m,
⎜⎜ consider set {i,m} of finite ordinals
⎜⎜ {i,m} holds first.in.{i,m}
⎜⎜ i<m ∨ m<i
⎜⎜
⎜⎜ i<m
⎜⎜⎛ Otherwise, m<i and
⎜⎜⎜ m ∈ ⦃j∈𝕆:j<k⦄∩S and
⎜⎝⎝ i isn't first.in.⦃j∈𝕆:j<k⦄∩S
⎜
⎜ for i and m ∈ S, i≤m
⎝ i is first.in.S
Nonempty S of only finite ordinals
holds first.in.S
No, the set of dark numbers is
another set without smallest element.
A nonempty set without a first element
is not a set of only finite ordinals.
>
The set of dark numbers contains
only natural numbers.
If dark numbers 𝔻 doesn't hold first.in.𝔻
then
either 𝔻 is empty
or 𝔻 isn't only finite ordinals.
Proof:
If you double all your finite ordinals
you obtain only finite ordinals again,
Yes.
although the covered interval is
twice as large as the original interval
covered by "all" your finite ordinals.
No.
The least.upper.bound of finites is ω
The least.upper.bound of doubled finites is ω