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On 08.01.2025 21:07, Jim Burns wrote:No, |N| is not a member of itself and does not behave like it.On 1/8/2025 9:35 AM, WM wrote:|ℕ| is invariable. |ℕ| = |ℕ|/2 is wrong.On 08.01.2025 12:04, FromTheRafters wrote:WM formulated on Wednesday :Yes.A set like ℕ has a fixed number of elements.If ω exists, then ω-1 exists.Wrong.
They don't cease. They simply aren't in the same league, if you will.If ω-1 does not exist,
what is the fixed border of existence?
Membership in ℕ is determined by ℕ.rule.compliance,
not by position relative to a border.element.
Each object complying with the ℕ.rule is in ℕ Each object not.complying
is not.in ℕ
The rule is for n there is n+1. But the successor is not created but
does exist. How far do successors reach? Why do they not reach to ω-1?
Where do they cease before?
Exactly omega is the "border", but this suggests a wrong mental image,Compliance and non-compliance do not change. Membership does notThen tell me if n = 7 exists and n = ω-1 does not exist where the border
change.
lies.
No, it is a finite expression of an infinity.No element is the border.element because each element is smaller.than"And so on" can only happen, when the elements are created. Potential
another, fuller element, and so, not on the border.
infinity.
--Which elements are in ℕ doesn't change.
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