When asked "WHO IS YOUR X?" (X still being unknown to me but is known to the respondents), here are the answers I get:

A answers: "A"

B answers: "C"

C answers: "C"

D answers: "F"

E answers: "F"

F answers: "F"

To sum up, the special phenomenon here is that, everybody has their own X (usually), and if any respondent points at another respondent as the first respondent's X, then the other respondent must point at themself as their X.

I wonder who the unknown X may be, when I only know that X is a natural example from everyday life. I thought about a couple of examples, but none of them are satisfactory, as follows:

X is the leader of one's political party, or X is one's mayor, and the like, but all of these examples attribute some kind of leadership or superiority to X, whereas I'm not interested in this kind of solution - involving any superiority of X.

Here is a second solution I thought about: X is the first (or last) person born in the year/month the respondent was born, and the like. But this solution involves some kind of order (in which there is a "first person" and a "last person"), whereas I'm not interested in this kind of solution - involving any order.

Btw, I've published this question also at the Miscellaneous desk, because this question is about everyday life, but now I decide to publish this question also here, because it's indirectly related to a well known topic in Math. 79.177.151.182 (talk) 13:27, 19 December 2024 (UTC)[reply]

Head of household comes to mind as a fairly natural one. The colours then correspond to different households which can be just one person. One objection is that "head of household" is a fairy traditional concept. With marriage equality now being the norm it's perhaps outdated. --2A04:4A43:909F:F9FF:397E:BBF9:E80B:CB36 (talk) 15:11, 19 December 2024 (UTC)[reply]

I have already referred to this kind of solution, in the example of "my mayor", see above why this solution is not satisfactory. 79.177.151.182 (talk) 15:31, 19 December 2024 (UTC)[reply]

The question has been resolved at the Miscellaneous reference desk.

Resolved

79.177.151.182 (talk) 15:48, 19 December 2024 (UTC)[reply]

X may well be 'the oldest living person of your ancestry'. --CiaPan (talk) 20:46, 19 December 2024 (UTC)[reply]

Resolved or not, let's try to analyze this mathematically. Given is some set ${\displaystyle S}$and some function ${\displaystyle f:S\to S.}$For the example, ${\displaystyle S=\{{\mathsf {A}},{\mathsf {B}},{\mathsf {C}},{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},}$with ${\displaystyle f({\mathsf {A}})={\mathsf {A}},}$${\displaystyle f({\mathsf {B}})=f({\mathsf {C}})={\mathsf {C}},}$${\displaystyle f({\mathsf {D}})=f({\mathsf {E}})=f({\mathsf {F}})={\mathsf {F}}.}$

Knowing that "everybody has their own X (usually)", we can normalize the unusual situation that function ${\displaystyle f}$might not be total in two ways. The first is to restrict the set ${\displaystyle S}$to the domain of ${\displaystyle f,}$that is, the set of elements on which ${\displaystyle f}$is defined. This is possible because of the condition that ${\displaystyle f(u)=v}$implies ${\displaystyle f(v)=v,}$so this does not introduce an undue limitation of the range of ${\displaystyle f.}$The second approach is to postulate that ${\displaystyle f(u)=u}$whenever ${\displaystyle f(u)}$might otherwise be undefined. Which of these two approaches is chosen makes no essential difference.

Let ${\displaystyle T=f(S)}$be the range of ${\displaystyle f}$, given by:

${\displaystyle T=\{v\in S:(\exists u\in S:f(u)=v)\}.}$

Clearly, if ${\displaystyle f(v)=v,}$we have ${\displaystyle v\in T.}$We know, conversely, that ${\displaystyle v\in T}$implies ${\displaystyle f(v)=v.}$

Let us also consider the inverse image of ${\displaystyle f}$, given by:

${\displaystyle f^{-1}(v)=\{u\in S:f(u)=v\}.}$

Suppose that ${\displaystyle f^{-1}(v)\neq \emptyset .}$This means that there exists some ${\displaystyle u\in f^{-1}(v),}$which in turns means that ${\displaystyle f(u)=v.}$But then we know that ${\displaystyle f(v)=v.}$Combining this, we have,

${\displaystyle f(v)=v\Leftrightarrow v\in T\Leftrightarrow f^{-1}(v)\neq \emptyset .}$

The inverse-image function restricted to ${\displaystyle T,}$to which we assign the typing

${\displaystyle f^{-1}:T\to \mathbb {P} (S),}$

now induces a partitioning of ${\displaystyle S}$into non-empty, mutually disjoint subsets, which means they are the classes of an equivalence relation. Each class has its own unique representative, which is the single element of the class that is also a member of ${\displaystyle T}$. The equivalence relation can be expressed formally by

${\displaystyle u\sim v\Leftrightarrow f(u)=f(v),}$

and the representatives are the fixed points of ${\displaystyle f.}$

Applying this to the original example, ${\displaystyle T=\{{\mathsf {A}},{\mathsf {C}},{\mathsf {F}}\},}$and the equivalence classes are:

• ${\displaystyle \{{\mathsf {A}}\},}$with representative ${\displaystyle {\mathsf {A}},}$
• ${\displaystyle \{{\mathsf {B}},{\mathsf {C}}\},}$with representative ${\displaystyle {\mathsf {C}},}$and
• ${\displaystyle \{{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},}$with representative ${\displaystyle {\mathsf {F}}.}$

Conversely, any partitioning of a set defines an equivalence relation; together with the selection of a representative for each equivalence class, this gives an instance of the situation defined in the question.  --Lambiam 20:47, 19 December 2024 (UTC)[reply]

FWIW, the number of such objects on a set of size n is given by OEIS: A000248, and that page has a number of other combinatorial interpretations. If you ignore the selection of a representative for each class, you get the Bell numbers. --RDBury (talk) 00:35, 21 December 2024 (UTC)[reply]