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On 20.10.2024 21:42, Jim Burns wrote:Neither is n+1, what's your point?On 10/20/2024 3:24 PM, WM wrote:>On 20.10.2024 20:20, Jim Burns wrote:On 10/20/2024 3:48 AM, WM wrote:On 20.10.2024 00:54, Jim Burns wrote:On 10/19/2024 2:19 PM, WM wrote:If n is countable.to from 0>>A doubled finite is finite.>
If all finites are doubled,
then not all results can be in that set.
If all finites are doubled,
then all results are in the set of finites.
But not in the mapped or multiplied range.
then ⟨0,1,...,n-1,n⟩ exists
If ⟨0,1,...,n-1,n⟩ exists
then ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n⟩ and ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exist
then ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
If ⟨0,1,...,n-1,n,n+1,...,2⋅n-1,2⋅n⟩ exists
then 2⋅n is countable.to from 0
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If n is countable.to from 0
then 2⋅n is countable.to from 0
If n is countable.to from 0If all finites are doubled,>
then all results are in the set of finites.
But not in the mapped or multiplied range.
then 2⋅n is countable.to from 0
Nevertheless 2n is not in the set {1, ..., n}.
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