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On 17.12.2024 13:34, Richard Damon wrote:Why not? That is your problem, you are trying to use logic that needs to finish, but can only do work at a finite rate, but needs to do infinite work because the set you are working on is infinte.On 12/17/24 5:30 AM, WM wrote:An unbounded number of members can be subtracted individually but not all.On 17.12.2024 00:57, Richard Damon wrote:>On 12/16/24 3:59 AM, WM wrote:>On 15.12.2024 22:14, Richard Damon wrote:On 12/15/24 2:44 PM, WM wrote:>On 15.12.2024 13:54, Richard Damon wrote:>>
You can't "name" your dark numbers,
because they are dark.>|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.Which just shows that the full set in infinte, and any member in it is finite, and not the last member.>
Many members can be subtracted individually but infinitely many members cannot be subtracted individually. They are belonging to the set. They are dark.Sure an infinite number of members can be subtracted individually,>
An unbounded number can be subtracted individually. However, if all are subtracted individually, then a last one is subtracted. That cannot happen.
An unbound number can't be used individually,
Regards, WM
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