Sujet : Re: Does the number of nines increase?
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 05. Jul 2024, 19:08:34
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <9aac1c05-e35c-4b94-97cd-2b66a6877ca8@att.net>
References : 1 2 3 4 5 6 7 8
User-Agent : Mozilla Thunderbird
On 7/5/2024 6:42 AM, Peter Fairbrother wrote:
On 05/07/2024 10:59, FromTheRafters wrote:
A difference between the idea of 'same' and 'equal size'.
'Same' if each can be a subset/superset of the other (matching elements)
and 'equal size' in terms of cardinality (pairing elements).
>
Maybe. but I'm not concerned about "same" here. only size.
And the elements are pairable (if an orange can be paired to a number),
but one set has an orange in it and the other doesn't.
>
So, pairing (and cardinality) don't really work for sizing these sets;
in everyday terms, the second is bigger.
The other way around.
It's set.inclusion which stops working as a guide to size.
Some sets (the everyday sets) with an orange inserted
cannot fit in that set un.oranged.
These sets have finite cardinals.
Consider the set ℕ of all and only finite cardinals.
There is no cardinal ℵ₀ in ℕ which is the cardinal |ℕ|
⎛ Assume otherwise.
⎜ Assume |ℕ| = ℵ₀ is in ℕ
⎜ ℵ₀ is finite.
⎜ ℕ⁺ᵒʳᵃⁿᵍᵉ doesn't fit in ℕ
⎜ |ℕ⁺ᵒʳᵃⁿᵍᵉ| = |{|α|<ℵ₀⁺¹}|
⎜ {|α|<ℵ₀⁺¹} doesn't fit in ℕ
⎜
⎜ However,
⎜ {|α|<ℵ₀⁺¹} ⊆ ℕ the finite cardinals
⎜ {|α|<ℵ₀⁺¹} fits in ℕ
⎝ Contradiction.
Therefore,
there is no cardinal ℵ₀ in ℕ which is the cardinal |ℕ|
and
ℕ⁺ᵒʳᵃⁿᵍᵉ is not bigger than its proper subset ℕ