Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/20/2024 3:54 PM, WM wrote:

On 20.10.2024 21:42, Jim Burns wrote:

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:

A doubled finite is finite.

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If all finites are doubled,
then not all results can be in that set.

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If all finites are doubled,
then all results are in the set of finites.

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But not in the mapped or multiplied range.

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If n is countable.to from 0
then ⟨0,1,...,n-1,n⟩ exists
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If ⟨0,1,...,n-1,n⟩ exists
then ⟨n,n+1,...,2⋅n-1,2⋅n⟩ exists
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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
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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 all finites are doubled,
then all results are in the set of finites.

>
But not in the mapped or multiplied range.

>
If n is countable.to from 0
then 2⋅n is countable.to from 0

>
Nevertheless
2n is not in the set {1, ..., n}.

2⋅n is in the set of finites.
{1,...,n} isn't the set of finites.

A doubled finite is finite.

If n is in the set ℕ≠{1,...,n} of finites,
then 2⋅n is in the set ℕ≠{1,...,n} of finites.