Condition for Well-Foundedness

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.


Then $\left({S, \preceq}\right)$ is well-founded if and only if there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.


That is, if and only if there is no infinite sequence $\left \langle {a_n}\right \rangle$ such that $a_0 \succ a_1 \succ a_2 \succ \cdots$.


Proof

Reverse Implication

Suppose $\left({S, \preceq}\right)$ is not well-founded.

So by definition there exists a non-empty subset $T$ of $S$ which has no minimal element.

Let $a \in T$.

Since $a$ is not minimal in $T$, we can find $b \in T: b \prec a$.

Since this holds for all $a \in T$, $\prec \restriction_{T \times T}$, the restriction of $\prec$ to $T \times T$, is a right-total endorelation on $T$.

So, by the Axiom of Dependent Choice, it follows that there is an infinite sequence $\left \langle {a_n}\right \rangle$ in $T$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

$\Box$


Forward Implication

Suppose there exists an infinite sequence $\left \langle {a_n}\right \rangle$ in $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$.

We let $T = \left\{{a_0, a_1, a_2, \ldots}\right\}$.

Clearly $T$ has no minimal element.

Thus by definition $S$ is not well-founded.

$\blacksquare$


Axiom of Dependent Choice

This theorem depends on the Axiom of Dependent Choice, by way of Condition for Well-Foundedness/Reverse Implication.

Although not as strong as the Axiom of Choice, the Axiom of Dependent Choice is similarly independent of the Zermelo-Fraenkel axioms.

The consensus in conventional mathematics is that it is true and that it should be accepted.


Sources