Re: Simple enough for every reader?

On 08.07.2025 17:59, joes wrote:

Am Tue, 08 Jul 2025 17:47:12 +0200 schrieb WM:

On 08.07.2025 09:46, Mikko wrote:

On 2025-07-07 15:37:08 +0000, WM said:

On 07.07.2025 10:29, Mikko wrote:
>

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).

Can you refer to some better author?

That is hardly feasible. But you can look up the definition in every
textbook of your choice. You will find the same result. Even Wikipedia
will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.

>
So no requirement of completeness.

"Each element" means that none is missing.

None is *unpaired*.

Each element of either set is in the bijection.

Nothing about "completeness" of (co)domain, whatever
that may look like.

Look up surjection, for instance in Wikipedia: In mathematics, a surjective function (also known as surjection, or onto function is a function f such that, for every element y of the function's codomain, ...
Regards, WM