Rel, bicategory of relations, allegory
left and right euclidean;
extensional, well-founded relations.
Given an extensional relation $\prec$ on a set, two elements are equal if they cannot be distinguished by the structure of the set of elements that they are related to, that those elements are related to, and so on forever in one direction. We generally think of $x \prec y$ as saying that $x$ is a ‘member’ of $y$, so that $\prec$ is extensional when “$x$ is determined by its members” as in the axiom of extensionality for material set theory.
We begin with the historically first definition, which is correct for well-founded relations. We then consider ways to extend this to more general relations, where the last version seems to be most correct.
A relation $\prec$ on a set $S$ is weakly extensional if, given any elements $x$ and $y$ of $S$, $x = y$ whenever (for all $t$) $t \prec x$ if and only if $t \prec y$.
Interpreting $\prec$ as membership $\in$, this corresponds to the axiom of extensionality as usually stated for pure sets. However, it is really only appropriate when $\prec$ is well-founded, just as the usual axiom of extensionality must be strengthened in the absence of the axiom of foundation.
However, when $\prec$ is well-founded, weak extensionality is equivalent to all the stronger notions below; thus in that case it is usually called just extensionality.
Let $\prec^*$ be the reflexive-transitive closure of the relation $\prec$ on $S$ (so $\prec^*$ is a preorder). Given an element $y$ of $S$, let $S \downarrow^* y$ be the downset of $y$ under $\prec^*$:
with $n \geq 0$. Note that $y$ itself belongs to $S \downarrow^* y$ (through $n = 0$, if in no other way), so we may take it to be a pointed set. Then $\prec$ is Finsler-extensional if it is weakly extensional and, given any elements $x$ and $y$ of $S$, $x = y$ whenever $S \downarrow^* x$ and $S \downarrow^* y$ are isomorphic as pointed sets equipped with a relation $\prec$.
Note that this definition includes weak extensionality, which won't follow from the other half unless $\prec$ is well-founded (see the examples below). It is possible to get weak extensionality free by using the transitive closure $\prec^+$ instead; that is, define $S \downarrow^+ y$ with $n \gt 0$ only. But then you need another step; define $(S \downarrow^+ y)^\top$ to be a pointed set with a new point $\top$ adjoined to $S \downarrow^+ y$; let $x \prec \top$ if and only if $x \prec y$ in $S$ itself. (For a well-founded relation, $(S \downarrow^+ y)^\top \cong S \downarrow^* y$; in general, however, $y$ may already belong to $S \downarrow^+ y$, yet $y \ne \top$ in $(S \downarrow^+ y)^\top$.) Then $\prec$ is Finsler-extensional on $S$ if and only if $x = y$ whenever $(S \downarrow^+ x)^\top \cong (S \downarrow^+ y)^\top$ as pointed sets equipped with $\prec$.
It is immediate that Finsler extensionality implies weak extensionality; the converse may be proved (using induction) for well-founded relations.
Given an element $y$ of $S$, let a path to $y$ consist of a sequence
for $n \geq 0$. Then let $T_y$ be the set of paths to $y$; given paths $\vec{t}$ and $\vec{u}$ in $N_y$, say that $\vec{t} \to \vec{u}$ if $\vec{u} = (t_1, \ldots, t_m)$ for some $m \leq n$. Then $T_y$ with the relation $\to$ is the tree to $y$.
$S$ is Scott-extensional if the only automorphism of any such tree $T_y$ (as a set equipped with a relation $\to$) is the identity function on $T_y$.
Then Scott extensionality implies Finsler extensionality, and the converse holds for well-founded relations.
The strongest version of extensionality is motivated by the study of terminal coalgebras and coinduction.
Let $S$ be equipped with a binary relation $\prec$. A bisimulation on $(S,\prec)$ is a binary relation $\sim$ such that whenever $x \sim y$, for any $a \prec x$ there is a $b \prec y$ with $a \sim b$, and conversely for every $b \prec y$ there is an $a \prec x$ with again $a \sim b$. We then say that $\prec$ is strongly extensional if every bisimulation is contained in the identity relation; i.e., $x = y$ whenever $x \sim y$ for any bisimulation $\sim$. In general, this is probably the best situation in which to say that $\prec$ is simply extensional.
Finsler and Scott extensionality may be understood as special cases of this for particular bisimulations $\sim$. (So can strong extensionality, since any set equipped with a relation has a weakest bisimulation $\approx$.) This is the approach taken by Aczel to study all the above notions of extensionality together.
In particuar, strong extensionality implies Scott extensionality, and the converse holds for well-founded relations. Thus, all forms of extensionality are equivalent for well-founded relations.
The axiom of extensionality in material set theory states membership is an extensional relation on the class of pure sets. (Note that the axiom of foundation states that membership is a well-founded relation, so one usually doesn't worry about the different notions of extensionality for ill-founded relations.) More generally, the membership relation on the transitive closure or reflexive-transtive closure of a pure set is an extensional relation on a set.
Conversely, one can define pure sets in structural set theory in part as sets equipped with an extensional (and optionally well-founded) relation.
A well-order is precisely a well-founded, transitive, extensional relation. Removing well-foundedness here gives a theory of ill-founded ordinal numbers.
On the set $\mathbf{2} = \{0,1\}$, now let $0 \prec 0$ and $0 \prec 1$ (but no other relationships). This relation is not weakly extensional, although it does satisfy the other half of Finsler extensionality, since $\mathbf{2} \downarrow^* 0 \ncong \mathbf{2} \downarrow^* 1$. However, $(\mathbf{2} \downarrow^+ 0)^\top \cong (\mathbf{2} \downarrow^+ 1)^\top$; this shows the necessity of defining Finsler extensionality as we do.
On the set $\mathbf{2}$ again, let $0 \prec 1$ and $1 \prec 0$ (but $0 \nprec 0$ and $1 \nprec 1$). This relation is weakly extensional but not Finsler-extensional, since $\mathbf{2} \downarrow^* 0 \cong \mathbf{2} \downarrow^* 1$. Yet $0$ and $1$ can hardly be distinguished by $\prec$ when there is an automorphism of $(\mathbf{2},\prec)$ that swaps them; this and the other examples below motivate the stronger notions of extensionality.
On the set $\mathbf{3} = \{0,1,2\}$, let $2 \nprec 0$ and $i \nprec i$ but all other relationships hold. Then this relation is Finsler-extensional but not Scott-extensional, since $T_2 \cong T_1$.
Finally, on the set $\mathbf{2}$ again, let $1 \nprec 0$ but all other relationships hold. Then this relation is Scott-extensional but not strongly extensional, since $0 \approx 1$.
Weak extensionality is a kind of antisymmetry condition: Let $x \leq y$ mean that $t \prec y$ whenever $t \prec x$. Then $\leq$ is clearly a preorder, which is antisymmetric (so a partial order) if and only if $\prec$ is weakly extensional.
Now suppose that if $t \prec x$ if and only if $t \prec y$ for all $t$, then $x \prec z$ if and only if $y \prec z$ for all $z$. Then if we define $x \equiv y$ to mean that $x \leq y$ and $y \leq x$, then $\equiv$ is an equivalence relation such that $\prec$ descends to a weakly extensional relation on the quotient set $S/{\equiv}$.
Strongly extensional quotients are even easier to construct, although they do create additional relationships. It is easy to see that the union of any family of bisimulations is a bisimulation, and therefore there is a largest bisimulation $\approx$ for any binary relation $\prec$ on $S$. Moreover, $\approx$ is an equivalence relation (though not every bisimulation need be so), and $\prec$ descends to a strongly extensional relation on the quotient $S/\approx$.
Given two sets $S$ and $T$, each equipped with a strongly extensional relation $\prec$, a function $f\colon S \to T$ is a simulation of $S$ in $T$ if
Then sets so equipped form a category with simulations as morphisms.
Note that there is at most one simulation from $S$ to $T$; in fact, strong extensionality for $T$ is equivalent to saying that any two simulations $S \rightrightarrows T$ are equal. Also, any simulation from $S$ to $T$ must be an injection; in fact, strong extensionality for $S$ is equivalent to saying that any simulation $S \to T$ is injective.
Thus, we have a (large) poset of sets equipped with extensional relations, and we can consistently interpret the simulations as subset inclusion. This leads to the model of sets equipped with extensional relations as transitive sets.
Last revised on January 29, 2016 at 09:45:22. See the history of this page for a list of all contributions to it.