Hasse diagram

#LaTeX

A Hasse diagram represents a finite partially ordered set as a transitive reduction graph. If an element is smaller than (precedes) another, there is an upwards path between them.

Formally,
elements $x, y$ of a poset $(S, \leq)$ are given vertices, and an edge from $x$ upwards to $y$ exists if $x \leq y$ and there is no distinct $z$ with $x \leq z \leq y$ .

Example: divisors of 60, ordered by "is a divisor of":

\documentclass[tikz,margin={2cm 0cm}]{standalone}
\usetikzlibrary{arrows.meta}
%
\begin{document}
\begin{tikzpicture}[every node/.style={circle,inner sep=1.25ex}]
	%
	\foreach \n/\x/\y in {%
		1/0/0,%
		2/2/1, 3/0/1, 5/-2/1,%
		4/4/2, 6/2/2, 10/0/2, 15/-2/2,%
		12/4/3, 20/2/3, 30/0/3,%
		60/2/4%
		}
		\node[label={center:$\n$}] (\n) at (\x,\y) {};
	%
	\foreach \a/\b in {%
		1/2, 1/3, 1/5,%
		2/4, 2/6, 2/10,%
		3/6, 3/15,%
		5/10, 5/15,%
		4/12, 4/20,%
		6/12, 6/30,%
		10/20, 10/30,%
		15/30,%
		12/60, 20/60, 30/60%
		}
		\draw[-Latex] (\a) -- (\b);
	%
\end{tikzpicture}
\end{document}

\begin{aligned} 60 & = 30 \times 2 = 20 \times 3 = 15 \times 4 = 12 \times 5 = 10 \times 6 \\ 30 & = 15 \times 2 = 10 \times 3 = 6 \times 5 \\ 20 & = 10 \times 2 = 4 \times 5 \\ 12 & = 6 \times 2 = 4 \times 3 \\ 15 & = 5 \times 3 \\ 10 & = 5 \times 2 \\ 6 & = 3 \times 2 \\ 4 & = 2 \times 2 \end{aligned}

A Hasse diagram of a partial order will always be acyclic.
If the set is instead equipped with a preorder, the graph may contain cycles.