Re: Contradiction of bijections as a measure for infinite sets

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Sujet : Re: Contradiction of bijections as a measure for infinite sets
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 26. Mar 2024, 16:40:05
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <44cd3a05-fed8-4604-927a-3ba432440e6b@att.net>
References : 1 2 3 4 5 6 7 8
User-Agent : Mozilla Thunderbird
On 3/26/2024 12:43 AM, Chris M. Thomasson wrote:
On 3/25/2024 5:04 PM, Jim Burns wrote:
On 3/25/2024 6:11 PM, WM wrote:

You cannot add a natural number to ℕ.
>
You cannot add a natural number to ℕ ==
ℕ holds all sizes of sets for which
changing by one element changes the set's size.
>
The set of odd numbers is infinite.
There are an infinite number of natural numbers.
infinity = infinity.
Density is another matter.
I prefer
finity = finity,  and
everything else = everything else.
"Everything else" sounds vague, which is my point.
"Finite" has clear, specific descriptions.
"Other than finite" has what?
It seems to me that
a specific role will have an associated infinity
-- though, even here, other infinities can be
shoehorned into or spackled onto.
It's complicated.

Density is another matter.
>
.01->.001 can represent infinity...
Its denser, so to speak.
Ah, well. It's complicated.
ℕ is discrete. ℚ is dense.
ℕ and ℚ have the same infinity.
ℝ has a larger infinity.
ℝ is denser than dense ℚ? I guess?
For any set S, its power set 𝒫(S) is larger.
Even for infinite sets.
|ℕ| = |ℚ| < |ℝ| ≤ |𝒫(ℚ)| < |𝒫(𝒫(ℚ))| <  ...
𝒫(𝒫(𝒫(ℚ))) is even more denser than dense?
I suspect that this is not a fruitful path
which we're on.
Just spitballing, what if
graduating to a new infinity involves
a semi.arbitrary description inserted somewhere?
Like ??? in
𝒫(ℚ)  ⊇  {S ⊆ ℚ: S is ??? }  ∈  𝒫(𝒫(ℚ))
Perhaps someone else has thought about this
and pointed out its strengths and weaknesses.
I confess that I haven't looked for them, yet.
Spitballing is easier.

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