By definition, well-founded relations can never be reflexive. Therefore, for instance, all partial orders are not well-founded. However, the definition of well-founded relations is not intuitive, and we want to give some concrete examples in order to get a better understanding of this definition.# Example: Examples of Well-founded Relations
(related to Definition: Well-founded Relation)
Every well-order in a strictly-ordered set is a well-founded relation. This follows immediately from the definitions of each of these two concepts. In particular, the strict order "$<$" in the well-ordered set of natural numbers $(\mathbb N, < )$ is well-founded. For instance, the subset $\{0\}$ has the minimal element $0,$ since there is no other natural number $m$ with $m < 0.$
Please note that if we take the total order $(\mathbb N, \le)$ then it is a well-order (version for posets), but it is not well-founded! For instance, the set $\{0\}$ contains $0$ as its minimum, but $0$ is not minimal in $\{0\}$ since $0\le 0.$ The reflexivity of "$\le$" prevents the order to be well-founded!
The relation $R\subset\mathbb N\times\mathbb N$ defined by $nRm:\Leftrightarrow n+1=m$ means a relation in which the natural number $m$ is the successor of the natural number $n.$ This relation is well-founded since every non-empty subset of natural numbers contains at least one element $m$ such that it is not the successor of all of its elements. For instance, the number $3$ is not a successor of any of the elements of the subset $\{3,4,5,6\}.$
Let $\mathcal P(X)$ be the power set of a given set $X,$ and let $R\subseteq\mathcal P(X)\times \mathcal P(X)$ be the relation of being a proper subset "$\subset$", i.e. $xRy:\Leftrightarrow x\subset y$ is well-founded, since a given non-empty subset $S\subseteq \mathcal P(X)$ contains at least one set $x\in S$ such that it is not a proper subset of any other set $y\in S.$
The divisibility relation $R$ defined on sets of natural numbers defined by $X:=\{n\in\mathbb N\mid 1 \le n \}$ with $nRm \Leftrightarrow n\mid m\wedge n\neq m.$ For instance, the non-empty subset $S\subset X=\{4,5,6,7,8,9,10\}$ contains $10,9,8,7,6$ as $R$-minimal elements, since there is no $w\in S$ dividing them with $w|m\wedge w\neq m.$
The relation $R$ defined on the set of finite strings over an alphabet with a length $>1$ by $nRm \Leftrightarrow n$ is a non-empty substring of $m$. For instance, take the alphabet $\{a,b,c\}$ and build the string $S=aabc$. This string has a length $>2$ and contains $a,b,c$ as $R$-minimal elements since there is no substring of these strings, which is non-empty and is still in $S.$
The relation $R$ defined on the Cartesian product $X:=\mathbb N × \mathbb N$ of pairs of natural numbers, defined by $(n_1,n_2)R(m_1, m_2)\Leftrightarrow n_1 < m_1\wedge n_2 < m_2.$ For instance, take the non-empty subset of pairs $$\begin{array}{ccccc} \vdots&\vdots&\vdots&\vdots&\\ (3,0)&(3,1)&(3,2)&(3,3)&\ldots\\ (2,0)&(2,1)&(2,2)&(2,3)&\ldots\\ (1,0)&(1,1)&(1,2)&(1,3)&\ldots\\ (0,0)&(0,1)&(0,2)&(0,3)&\ldots \end{array}$$ has the following $R$-minimal elements $$\begin{array}{ccccc} \vdots \\ (3,0) \\ (2,0) \\ (1,0) \\ (0,0)&(0,1)&(0,2)&(0,3)&\ldots \end{array}$$
since only those pairs are both: contained in the subset, and there is no other pair comparable to them using the relation $R$
The contained relation $\in_X$ defined on a set $X$ is well-founded. This follows from the axiom of foundation. As a reminder, the axiom postulates
"Every non-empty set $X$ contains an element that is disjoint from X, formally $\forall X(X\neq\emptyset \Rightarrow\exists (z\in X)X\cap z=\emptyset).$"
The postulate is equivalent to
"There is an element $z\in X$ and for no other element $x\in X$ we have $z\in x.$"
This is exactly the definition of being a well-founded relation, i.e. the axiom of foundation can be interpreted as the following requirement:
"Every non-empty set $X$ contains a $\in$-minimal element."
