Defending the Possibility of Knowledge

Abstract

In this paper, I propose a solution to Fitch’s paradox that draws on ideas from Edgington (Mind 94:557–568, 1985), Rabinowicz and Segerberg (1994) and Kvanvig (Noûs 29:481–500, 1995). After examining the solution strategies of these authors, I will defend the view, initially proposed by Kvanvig, according to which the derivation of the paradox violates a crucial constraint on quantifier instantiation. The constraint states that non-rigid expressions cannot be substituted into modal positions. We will introduce a slightly modified syntax and semantics that will help underline this point. Furthermore, we will prove results about the consistency of verificationism and the principle of non-omniscience by model-theoretical means. Namely, we prove there exists a model of these principles, and delineate certain constraints they pose on a structure in which they are true.

Appendix A: Proof of Main Result

• (a) Suppose that the extension of p is $\{w\}$, then $w\Vdash p$. By (Ver), there exists v such that $R(w,v)$ and $v\Vdash Kp$. Since p is only true at w, $v = w$. Moreover, $w\Vdash Kp$ iff there exists $a\in D(w)$ such that $a[w]\subset \{w\}$. Since $w\in a[w]$, by factivity, this means $a[w] = \{w\}$. But then every proposition true at w will be known at w, contradicting (NO). The second part follows immediately.

• (b) Let $W = \{w, x, y, z\}$. Define a as the reflexive, symmetric and transitive closure of $\{(w, x), (y, z)\}$ and b as the reflexive, symmetric and transitive closure of $\{(w, y), (x, z)\}$. Let $D(w) = D(z) = \{a\}$ and $D(x) = D(y) = \{b\}$. Let S be the total binary relation on W, and $\Pi$ the set $\{\{w, x\}, \{y, z\}, \{w, y\}, \{x, z\}\}$.

  $\Pi$ is clearly closed under complementation, and we can easily verify that (Ver) and (NO) are true in each world (consider each possible case separately). For example, only the propositions $\pi _1 = \{w, x\}$ and $\pi _2 = \{w,y\}$ are true at w. $\pi _{1}$ is known and $\pi _{2}$ is unknown at w, but $\pi _{2}$ is known at y, a world accessible from w. The other cases are verified in a similar fashion.

• (c) To show this part we apply (Ver) and (NO) alternatively. Let $w_{0} \in W$ be any world. By (NO), there exists a proposition $\pi _{0} \in \Pi$ that is unknown at $w_{0}$, i.e. for every knower $a\in D(w_{0})$, $a[w_{0}]\not \subset \pi _{0}$. By (Ver), there exists $w_{1}$, accessible from $w_{0}$ via R, such that $\pi _{0}$ is known at $w_{1}$, i.e. there exists $a_{1}\in D(w_{1})$ with $a_{1}[w_{1}]\subset \pi _{0}$. Clearly, $w_{1}$ is distinct from $w_0$. By (NO), there exists $\pi _{1} \in \Pi$ such that $\pi _1$ is unknown at $w_1$. The proposition $\pi _{0} \cap \pi _{1}$ is also true and unknown at $w_{1}$, so in particular $\pi _{0} \cap \pi _{1}$ is distinct from $\pi _{0}$. By (Ver), there therefore exists $w_{2}$ such that $\pi _{0} \cap \pi _{1}$ is known at $w_{2}$. Since both $\pi _{0}$ and $\pi _{1}$ are known at $w_{2}$, $w_{2}$ is distinct from both $w_{0}$ and $w_{1}$. By (NO), there exists $\pi _{2} \in \Pi$ such that is true but unknown at $w_{2}$. And so on and so forth. We can therefore generate an infinite sequence of distinct worlds $w_{n}$ and propositions $\pi _{n}$ such that the propositions $\pi _{0}, \pi _{1}, ..., \pi _{k-1}$ are known at $w_{k}$ but not the proposition $\pi _{k}$. It follows then that W and $\Pi$ are infinite.

• (d) We define a model $\mathcal {M} = \langle W, D, R, \Pi \rangle$ with the desired properties. Let $W = \mathbb {Z}\times \mathbb {N}$. For each $n\in \mathbb {N}$, let $c_{n}$ be the relation of congruence modulo $2^{n}$: $c_{n}(x, y)$ iff $x - y$ is divisible by $2^{n}$. We define $a_{n}$ as the binary relation on W such that: $a_{n}(w,v)$ iff $w_{2} = v_{2}$ and $c_{n}(w_{1}, v_{1})$, where $w = (w_{1},w_{2})$ and $v = (v_{1}, v_{2})$. For $w = (w_{1}, w_{2})$, let $D(w) = \{a_{n} : n\leq w_{2}\}$. We let R be the total binary relation on W (other choices will do also). Finally, $\Pi$ is defined as the set of $\pi \subset W$ such that $\pi = \pi _{0}\times \mathbb {N}$ and $\pi _{0}$ is of the form:

  $$(*) \quad \pi_{0} = c_{n}[k_{1}] \cup c_{n}[k_{2}] \cup ... \cup c_{n}[k_{m}], $$

  where $n>0$ and $k_{1}, k_{2}, ..., k_{m}$ are such that $0 \leq k_{1} < ... < k_{m} < 2^{n}$ (with m possibly zero, in which case $\pi = \varnothing$).¹⁷In other words, $\pi _0$ is a (possibly empty) union of equivalences classes.

$\Pi$ is closed under complementation because the complement of a (possibly empty) union of equivalence classes is also a (possibly empty) union of equivalence classes (of the same relation). $\Pi$ is also closed under intersection. To show this, first observe that if

$$\begin{array}{rll} \pi_{0} &=& c_{n}[k_{1}] \cup c_{n}[k_{2}] \cup \,\, ... \,\, \cup c_{n}[k_{m}] \\ \pi_{0}' &=& c_{n'}[l_{1}] \cup c_{n'}[l_{2}] \cup \,\, ... \,\, \cup c_{n'}[l_{m'}] \end{array} $$

(A1)

with $0 \leq k_{1} < k_{2} \textrm { ... } k_{m} < 2^{n}$ and $0 \leq l_{1} < l_{2} \textrm { ... } l_{m'} < 2^{n'}$ ($m, m'$ possibly zero), then there exists a similar decomposition for $\pi _0$ and $\pi _0'$ with the equivalence classes of the relation $c_{n''}$, for any $n'' \geq \max (n', n)$. This follows from the fact that $2^{n''}$ is a (common) multiple of $2^{n}$ and $2^{n'}$, and therefore every equivalence class of $c_{n}$ or $c_{n'}$ is the union of equivalence classes of $c_{n''}$. If $\pi _{0}$ is a disjoint union of equivalence classes of the relation $c_{n''}$ and the same is true of $\pi _{0'}$, then their intersection is a disjoint union of the classes they have in common. Hence, $\Pi$ is Boolean.

Let us show now that $\Pi$ is epistemically closed. We show this by induction on the number of connectives in $\phi$. If there are none, the result is a consequence of the definition of an the interpretation function. For the induction step, we proceed by considering the main connective of $\phi$. The Boolean cases follow by the induction hypothesis and the arguments of the preceding paragraph. We are left with the case where $\phi$ is of the form $K[w]\psi$, for some $w = (w_{1},w_{2})\in W$. Suppose $v = (v_{1},v_{2}) \in W$. By definition, we have that $v\Vdash K[w]\psi$

$$ \begin{array}{l} iff there exists a\in D(w) such that a[v]\subset \left[\kern-0.15em\left[ \psi \right]\kern-0.15em\right]\\ iff a_{w_{2}}[v]\subset \left[\kern-0.15em\left[ \psi \right]\kern-0.15em\right]\\ iff c_{w_{2}}[v_{1}]\times \{v_{2}\} \subset \left[\kern-0.15em\left[ \psi \right]\kern-0.15em\right]\\ iff c_{w_{2}}[v_{1}] \subset c_{n}[k_{1}] \cup c_{n}[k_{2}] \cup \,\, ... \,\, \cup c_{n}[k_{m}] \end{array}$$

Since $c_{w_{2}}$ is an equivalence relation, $c_{w_{2}}[v_{1}] = c_{w_{2}}[m]$ for any $m\in \mathbb {Z}$ such that $c_{w_{2}}(v_{1},m)$, so that

$$\left[\kern-0.15em\left[ K[w]\psi \right]\kern-0.15em\right] = \left(\bigcup \{c_{w_{2}}[m] : m \in \mathbb{Z} \textrm{ such that } c_{w_{2}}[m] \subset \left[\kern-0.15em\left[ \psi \right]\kern-0.15em\right] \}\right)\times\mathbb{N}.$$

The proposition defined by ‘$K[w]\psi$’ is therefore in $\Pi$.

We must now show that (Ver) and (NO) are true everywhere in the model. Let us start by (Ver). Let $\pi = \pi _{0}\times \mathbb {N}$ be a true proposition at $w = (w_{1},w_{2})$. There exists a minimal n such that $\pi _{0}$ admits a decomposition like (*) above. Since the equivalence classes of (*) are disjoint and $\pi$ is true at w, there exists a unique class containing $w_{1}$, the class $c_{n}[w_{1}]$. Since the proposition $\pi$ is a consequence of the proposition $c_{n}[w_{1}]\times \mathbb {N}$, i.e. $c_{n}[w_{1}] \times \mathbb {N} \subset \pi$, it suffices to know $c_{n}[w_{1}]\times \mathbb {N}$ in order to know $\pi$. By definition of the relations $c_{m}$, it is clear that $c_{m}[w_{1}]\subset c_{n}[w_{1}]$ whenever $m \geq n$. Hence, for any $m \geq n$, $a_{m}[w_{1}]\subset \pi$ so that $\pi$ is known at $(w_{1}, m)$.

To show (NO), for all $w\in W$, we must show there is a proposition that is true but unknown at w. If $w = (w_{1},w_{2})$, then $a_{w_{2}}$ is the most knowledgeable knower at w. By definition of $\Pi$, $\pi = c_{w_{2}+1}[w_{1}]\times \mathbb {N}$ is a proposition and we have that $c_{w_{2}}[w_{1}] \not \subset c_{w_{2}+1}[w_{1}]$, so that $\pi$ is not known by $a_{w_{2}}$. Since $a_{w_{2}}$ is the most knowledgeable knower at w, this proves the result.