Re: Incompleteness of Cantor's enumeration of the rational numbers

On 18.11.2024 23:40, FromTheRafters wrote:

WM wrote on 11/18/2024 :

On 18.11.2024 22:58, FromTheRafters wrote:

on 11/18/2024, WM supposed :

On 18.11.2024 18:15, FromTheRafters wrote:

WM brought next idea :

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ℕ| - |ℕ| = 0 because if you subtract one element from ℕ then you

have

no longer ℕ and therefore no longer |ℕ| describing it.

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Still wrong.

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If you remove one element from ℕ, then you have still ℵo but no longer all elements of ℕ.

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But you do have now a proper subset of the naturals the same size as before.

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It has one element less, hence the "size" ℵo is a very unsharp measure.

 Comparing the size of sets by bijection. Bijection of finite sets give you a same number of elements, bijection of infinite sets give you same size of set.

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Why? Because only potential infinity is involved. True bijections pr5ove equinumerosity.

If |ℕ| describes the number of elements, then it has changed to |ℕ| - 1.

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Minus one is not defined.

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Subtracting an element is defined. |ℕ| - 1 is defined as the number of elements minus 1.

 Nope!

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The number of ℕ \ {1} is 1 less than ℕ.

 

If you don't like |ℕ| then call this number the number of natural numbers.

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Why would I do that when it is the *SIZE* of the smallest infinite set.

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The set of prime numbers is smaller.

 No, it is not.

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It is, because 4 and 8 are missing.
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 > There is a bijection.
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Only between numbers which have more successors than predecessors, although it is claimed that no successors are remaining.