On Categorial Membership

Abstract

We investigate the family of concepts that an agent comes to know through a set of defining features, and examine the role played by these features in the process of categorization. In a qualitative framework, categorial membership is evaluated through an order relation among the objects at hand, which translates the fact that an object may fall more than another under a given concept. For concepts defined by their features, this global membership order depends on the degree with which each feature applies to the objects of the universe. The passage from these individual membership degrees to a global membership order poses a problem analogous to vote aggregation in social choice theory. This similarity leads to an original solution that is particularly well-adapted to the framework of cognitive psychology. The resulting membership order extends to compound concepts, and provides a good description of the guppy paradox and the conjunction effect.

Appendix

Observation 1

Let $\preceq_{\alpha}$ be the relation defined by

$$ \begin{aligned} &x\preceq_{\alpha}y\;iff\;for\;each\;feature\;f\in \Updelta(\alpha)\;such\;thaty\prec_{f}x, there\;exists\;a\;feature \\ &\quad g\in\Updelta(\alpha), g\;more salient\;than\;f, such\;that\;x\prec_{g}y. \end{aligned} $$

Then $\preceq_{\alpha}$ is transitive.

Proof

Let x, y and z be three items such that $x \preceq_{\alpha} y$ and $y \preceq_{\alpha} z$, and suppose that there exists $f \in \Updelta(\alpha)$ such that such that $z\prec_{f}x$. We have to show that there exists a feature $g \in \Updelta(\alpha), g$ more salient than f, such that $x\prec_{g}z$. We make a proof by cases:

• Suppose first that $x\preceq_{f}y$. Then we have $z\prec_{f} y$, and there exists a feature $k \in \Updelta(\alpha), k$ more salient than f, such that $y\prec_{k}z$. We can suppose that k is maximally salient in $\Updelta(\alpha)$ for this property (recall that $\Updelta(\alpha)$ is finite). If $x \preceq_{k}y$, we get $x \prec_{k}z$ and we are done. Otherwise, because of the connectedness of $\preceq_{k}$, we have $y \prec_{k}x$. Since we supposed $x \preceq_{\alpha} y$, this implies that there exists a concept $g \in \Updelta(\alpha), g$ more salient than k, such that $x \prec_{g}y$. We cannot have $z\prec_{g}y$ otherwise there would exist $h \in \Updelta(\alpha), h$ more salient than g, such that $y\prec_{h}z$, contradicting the choice of k. We have therefore $y\preceq_{g}z$, hence $x\prec_{g}z$ as desired.

• Suppose now that $y\prec_{f}x$. Then there exists $k \in \Updelta(\alpha), k$ more salient than f such that $x\prec_{k}y$, and we can again suppose k maximally salient for this property. If $y\preceq_{k}z$, we get $x\prec_{k}z$ and we are through. Otherwise, we have $z\prec_{k}y$ and there exists g more salient than k such that $y\prec_{g}z$. Let us show that $x\preceq_{g}y$: if this were not the case, we would have $y\prec_{g}x$, so that there would exist h more salient than g such that $x\prec_{h}y$. But then h would be more salient than k, which is impossible. We have therefore $x\preceq_{g}y$, hence $x\prec_{g}z$, and the proof is complete.

$\square$

Observation 2

One has x ∼_α y if and only if x∼ _f y for all features f of $\Updelta(\alpha)$.

Proof

It is clear that x ∼_α y holds whenever x∼ _f y for all features f of $\Updelta(\alpha)$. Suppose conversely that we do not have x∼ _f y for some feature f of $\Updelta(\alpha)$, and let f be of maximal salience in $\Updelta(\alpha)$ for this property. Since $\preceq_{f}$ is a total preorder, we necessarily have $x\prec_{f}y$ or $y\prec_{f}x$. In the first case, and by the choice of f, we cannot have $y \preceq_{\alpha} x$; in the second case, we cannot have $x \preceq_{\alpha} y$. This shows that we cannot have x ∼_α y. $\square$

Observation 3

Define the extension Ext α of α as the set of all objects z such that $\delta_{f}(z) = 1\quad \forall f\in\Updelta(\alpha)$. Then it holds $x \prec_{\alpha} z$ and $x\prec_{\alpha}^{*}z$ for all $x \notin Ext\,\alpha$ and $z\in Ext\,\alpha$.

Proof

If z is such that δ _f (z) = 1 for all $f\in\Updelta(\alpha)$, it follows from the definitions of $\preceq_{\alpha}$ and $\preceq_{\alpha}^{*}$ that one has necessarily $x \preceq_{\alpha} z$ and $x \preceq_{\alpha}^{*} z$ for all objects x of the universe. If moreover there exists a defining feature f such that δ _f (x) < 1, these inequalities are strict, so that $x \prec_{\alpha} z$ and $x \prec_{\alpha}^{*}z$ as desired. $\square$

Observation 4

$\preceq_{\alpha}$ and $\preceq_{\alpha}^{*}$ are finite membership orders.

Proof

Concerning $\preceq_{\alpha}$, the proof follows from Observation 2. As for $\preceq_{\alpha}^{*}$, we see, again by Observation 2 that every $\preceq_{\alpha}$—equivalence class is embedded in a $\preceq_{\alpha}^{*}$-equivalence class, whence the result. $\square$

Observation 5

Define the extension $Ext (\beta\,\star\,\alpha)$ of $\beta\,\star\,\alpha$ to be the set of all $\preceq_{\beta \star \alpha}$ -maximal elements, and let z be an arbitrary object. Then the following conditions are equivalent:

• $z \in Ext (\beta\,\star\,\alpha)$

• z is $\prec_{\beta\, \star\, \alpha}^{*}{\text{-}}maximal$

• $z\in Ext\, \alpha\cap Ext\, \beta$

• $x \prec_{\beta\, \star\, \alpha} z \forall x \notin Ext (\beta\, \star\, \alpha)$

• $x\prec_{\beta\, \star\, \alpha}^{*} z \forall x \notin Ext (\beta\, \star\, \alpha)$.

Proof

Since we have $\prec_{\beta\star\alpha}^{*} \subseteq \prec_{\beta\star\alpha}$, any $\prec_{\beta\star\alpha}$-maximal element is $\prec_{\beta\star\alpha}^{*}$-maximal. If x is not $\prec_{\beta\star\alpha}^{*}$-maximal, it holds $x\prec_{\beta\star\alpha}^{*}z$ for any element z of Ext α ∩ Ext β, since $\prec_{\beta\star\alpha}^{*}$ corresponds to the proportional order induced by the fictitious defining features set {α, β}. This shows that any $\prec_{\beta\star \alpha}^{*}$-maximal element must lie in Ext α ∩ Ext β, and is therefore also $\prec_{\beta\star\alpha}$-maximal. The proof of the Observation follows. $\square$