S(S(f)) = S(f) for reloids

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For a set $X$, let $\mathcal F(X)$ be all filters on $\mathcal P(X)$, including the improper filter $\bot_X=\mathcal P(X)$. A filter is a nonempty family closed under finite intersections and supersets. Order filters by $\mathcal A\sqsubseteq\mathcal B$ when $\mathcal A\supseteq\mathcal B$ as families. Thus $\bigsqcup$ is intersection of filter families and $\bigsqcap$ is the filter generated by their union; the empty join is $\bot_X$ and the empty meet is $\top_X=\{X\}$. For $E\subseteq X$, $\uparrow E=\{U\subseteq X:E\subseteq U\}$ is its principal filter. Write $\mathcal A\mathrel{\#}\mathcal B$ when every $U\in\mathcal A$ and $V\in\mathcal B$ have $U\cap V\ne\varnothing$, equivalently $\mathcal A\sqcap\mathcal B\ne\bot_X$.

A reloid $f:X\to Y$ is a filter on $X\times Y$, ordered by reverse inclusion of filter families. For a relation $R\subseteq X\times Y$, its principal reloid is $\uparrow R$. The reloid $g\circ f:X\to Z$ is generated by $G\circ F$ for $F\in f,G\in g$, where $G\circ F=\{(x,z):\exists y,(x,y)\in F,(y,z)\in G\}$. The reverse $f^{-1}$ is generated by $F^{-1}=\{(y,x):(x,y)\in F\}$ for $F\in f$. The identity $1_X$ is $\uparrow\Delta_X$, where $\Delta_X=\{(x,x):x\in X\}$.For an endoreloid $f:X\to X$, set $f^0=1_X$, $f^{j+1}=f\circ f^j$, and \[ S(f)=\bigsqcup_{j=0}^{\infty}f^j. \] This join is the intersection of the filter families of all powers. It is the endomorphism series $S$, distinct from the source’s separately defined connectivity operator $S^*$.

Statement recorded in the supplied source.

For every set $X$ and endoreloid $f:X\to X$, is \[ S(S(f))=S(f)? \] The inner and outer occurrences use the same endomorphism-series operator. [S2][S3]

Short English statement

Is taking the join of all finite powers, including the identity, idempotent for every endoreloid?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 3407 (OPG-751). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, S(S(f)) = S(f) for reloids, OPG-751. Original problem statement, full mathematical discussion, bibliography and comments. https://www.openproblemgarden.org/op/s_s_f_s_f_for_reloids.
• S3 Victor Porton, General Topology as Ordered Semigroup Actions, author manuscript, version of 21 July 2026. Basic definitions and Composition of reloids. https://github.com/vporton/agt-combined/blob/759e45cced7769cf3a84ce60ed6c30f4ce9ffa8e/chap-reloids.tex.
• S4 Victor Porton, General Topology as Ordered Semigroup Actions, author manuscript, version of 21 July 2026. Endomorphism series; Algebraic properties of S and S*, conjecture S(S(f))=S(f). https://github.com/vporton/agt-combined/blob/759e45cced7769cf3a84ce60ed6c30f4ce9ffa8e/chap-connected.tex.