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On 11.01.2025 11:25, joes wrote:Infinite things don't have an end.Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:Not finite but complete.On 11.01.2025 01:28, Richard Damon wrote:Nope. This is all just your conception of Aleph_0 as finite.On 1/10/25 4:48 PM, WM wrote:The number of elements remains constant. All odd numbers of ℕ areOn 10.01.2025 21:08, Jim Burns wrote:But the set doesn't grow.
>Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",You are inconsistent. You claim that all natural numbers are an
it's you who's saying it doesn't work,
invariable set. But when all elements are doubled then your set
grows,
showing it is not invariable. That is nonsense.
Which element is in the doubled set that wasn't there in the first
place?
deleted. That implies that new even numbers are added.
"Doubling" IS multiplication by 2, turning the naturals into the evens.It does not behave like that. All countable sets are bijective to eachIt appears so. But it is wrong. Construct a bijection between natural
other.
numbers and even natural numbers. f(n) = 2n. Then double the even
numbers by multiplication. Many are not in the bijection.
All even numbers are naturals. Multiplication by any natural numberBut not natural numbers, since *all* natural numbers have been doubled.If Cantor has constructed a sequence containing all even numbers ofWhat? Doubled even numbers are also even numbers.
the original set ℕ, then the doubled even numbers are missing.
When we mark all natural numbers with an astrisk, then none remains.It is not. The sequence 2*n does not go beyond infinity.
When wie double them, then none remains. The odd numbers leave. All even
numbers remain. But the same number of of even numbers is larger than ω.
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., ω2.No. There is no x e N such that 2*x >= omega. You have listed two
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