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  • Re: Simple enough for every reader?

    From Mikko@mikko.levanto@iki.fi to sci.logic on Sat Jul 5 11:37:06 2025
    From Newsgroup: sci.logic

    On 2025-07-04 12:32:55 +0000, WM said:

    On 04.07.2025 09:51, Mikko wrote:
    On 2025-07-03 13:08:25 +0000, WM said:

    On 03.07.2025 11:35, Mikko wrote:
    On 2025-07-02 13:51:01 +0000, WM said:

    The definition of bijection requires completeness.

    No, it doesn't.

    The function is injective, or one-to-one, if each element of the
    codomain is mapped to by at most one element of the domain,
    The function is surjective, or onto, if each element of the codomain is >>>>> mapped to by at least one element of the domain; Wikipedia

    Bijection = injection and surjection.

    Note that no element must be missing. That means completeness.

    It does not mean that the bijection is completely known.

    It means that every element of the domain and of the codomain is involved. >>
    Being involved is not the same as being known.

    I only said: The definition of bijection requires completeness.

    You: No, it doesn't.

    I also said what is worng in your claim: bijection only requires that
    there is one and only one element of co-domain for each element of
    domain, regardless of completeness.
    --
    Mikko

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  • From WM@wolfgang.mueckenheim@tha.de to sci.logic on Sat Jul 5 15:15:11 2025
    From Newsgroup: sci.logic

    On 05.07.2025 10:37, Mikko wrote:
    On 2025-07-04 12:32:55 +0000, WM said:

    I only said: The definition of bijection requires completeness.
    ;
    You: No, it doesn't.

    I also said what is worng in your claim: bijection only requires that
    there is one and only one element of co-domain for each element of
    domain, regardless of completeness.

    Bijection requires completeness of domain and codomain.

    Regards, WM



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  • From joes@noreply@example.org to sci.logic on Sat Jul 5 13:42:22 2025
    From Newsgroup: sci.logic

    Am Sat, 05 Jul 2025 15:15:11 +0200 schrieb WM:
    On 05.07.2025 10:37, Mikko wrote:
    On 2025-07-04 12:32:55 +0000, WM said:

    I only said: The definition of bijection requires completeness.
    You: No, it doesn't.

    I also said what is worng in your claim: bijection only requires that
    there is one and only one element of co-domain for each element of
    domain, regardless of completeness.

    Bijection requires completeness of domain and codomain.

    How is that not given?
    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.
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  • From Mikko@mikko.levanto@iki.fi to sci.logic on Sun Jul 6 11:34:00 2025
    From Newsgroup: sci.logic

    On 2025-07-05 13:15:11 +0000, WM said:

    On 05.07.2025 10:37, Mikko wrote:
    On 2025-07-04 12:32:55 +0000, WM said:

    I only said: The definition of bijection requires completeness.
    ;
    You: No, it doesn't.

    I also said what is worng in your claim: bijection only requires that
    there is one and only one element of co-domain for each element of
    domain, regardless of completeness.

    Bijection requires completeness of domain and codomain.

    So you say but cannot prove.
    --
    Mikko

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  • From WM@wolfgang.mueckenheim@tha.de to sci.logic on Sun Jul 6 12:04:50 2025
    From Newsgroup: sci.logic

    On 06.07.2025 10:34, Mikko wrote:
    On 2025-07-05 13:15:11 +0000, WM said:

    On 05.07.2025 10:37, Mikko wrote:
    On 2025-07-04 12:32:55 +0000, WM said:

    I only said: The definition of bijection requires completeness.
    ;
    You: No, it doesn't.

    I also said what is worng in your claim: bijection only requires that
    there is one and only one element of co-domain for each element of
    domain, regardless of completeness.

    Bijection requires completeness of domain and codomain.

    So you say but cannot prove.

    It is so by definition. See e.g. W. Mückenheim: "Mathematik für die
    ersten Semester", 4th ed., De Gruyter, Berlin (2015).

    Regards, WM

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