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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
Date: Sun, 1 Dec 2024 12:17:25 +0200
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On 2024-11-27 11:10:51 +0000, WM said:

> On 27.11.2024 10:33, Mikko wrote:
>> On 2024-11-26 11:07:57 +0000, WM said:
>> 
>>> On 26.11.2024 10:09, Mikko wrote:
>>>> On 2024-11-25 14:38:13 +0000, WM said:
>>> 
>>>>> The simple example contradicts a bijection between the two sets 
>>>>> described above.
>>>> 
>>>> What does "contradicts a bijection" mean?
>>>> 
>>> It shows that the mapping claimed to be a bijection is not a bijection.
>> 
>> If so, no bijection is contradicted.
> 
> The possibility of a bijection between the sets  ℕ = {1, 2, 3, ...} and 
> D = {10n | n ∈ ℕ} is contradicted.

No, it is not. You merely deny it, disregarding obvious facts. The function
f(x) = 10 * f obviously maps every element of ℕ to a different element of
D and there is no element of D that is not 10 * f for some f so this f is
a bijection between ℕ and D.

-- 
Mikko