Does the fragment of intuitionistic propositional calculus with just $\to$ have a finitely-valued semantics? Intuitionistic propositional calculus (IPC) has a topological semantics. IPC also does not have a sound and complete semantics with finitely many truth values.
I'm curious whether the fragment of IPC with only the connective $\to$ and no truth value constants has a finitely-valued semantics.
I'm also curious what sort of techniques in general are available for establishing whether a logic (given as a set of inference rules) has a finitely-valued semantics or not.

Let $a^o$ represent the interior of $a$. Let $[a]$ be the interpretation of $a$.
IPC has the following topological semantics. The topology in question is the standard topology over the reals $\mathbb{R}$.
$$ [a \land b] = [a] \cap [b] $$
$$ [a \lor b] = [a] \cup [b] $$
$$ [a \to b] = ([a]^c \cup [b])^o $$
$$ [\bot] = \varnothing $$
$$ [\lnot a] = [a \to \bot] $$
I had a silly idea to try replacing $(\mathbb{R}, \tau)$ with the Sierpiński space and see how many connectives I need to remove from the language of IPC to make the Sierpiński space a correct interpretation.
I started off with $\to$ and $\lor$.
However, the statement $(a \to b) \lor (b \to a)$ is "Sierpiński-valid", but is not true for intuitionistic logic. As proof, consider $a = (0, \infty)$ and $b = (-\infty, 0)$. $0$ is not an element of $[(a \to b) \lor (b \to a)]$.
So, then I considered removing $\lor$.
The classical tautology is $((a \to b) \to a) \to a$ is not valid when $a$ is $\{0\}$ and $b$ is $\varnothing$, so this three-valued semantics successfully fails to be classical logic, but I'm not sure whether it's equivalent to the $\to$-fragment of IPC or not.
 A: The set of equations $\varphi = \top$ that hold in the Heyting algebra of the Sierpiński space is a proper superset of the set of tautologies of the $\rightarrow$-fragment of the intuitionistic propositional calculus.
You already know that the Sierpiński space satisfies the equation $(a \rightarrow b) \vee (b \rightarrow a) = \top$, even though $(a \rightarrow b) \vee (b \rightarrow a)$ is not an intuitionistic tautology. You can use the second-order encoding of the intuitionistic connectives to produce the following sentence in the $\rightarrow$-fragment:
$$\varphi \equiv  ((a \rightarrow b) \rightarrow c) \rightarrow ((b \rightarrow a) \rightarrow c) \rightarrow c$$
An algebra satisfies $(a \rightarrow b) \vee (b \rightarrow a) = \top $ for all $a,b$ precisely if it validates $\varphi = \top$ for all $a,b,c$. Hence, the Sierpiński space satisfies $\varphi = \top$.
edit: There is no finite universal Heyting algebra (and hence finite topological space) that satisfies only the tautologies of the $\rightarrow$-fragment. I don't have a reference off-hand (I'll try to find one), but you can take the Boolean algebra $B_n$ with $n$ generators, adjoin a new "top element" to it, and get a Heyting algebra with $n+1$ elements that satisfies an $\rightarrow$-equation (in many variables) that none of the previous algebras do. Regarding your question about "techniques for establishing that a logic has a finite-valued semantics": this usually happens only when every finite model of your algebra is generated by a single finite model. E.g. every finite Boolean algebra is the $n$-fold product of the $2$-element Boolean algebra, and an equation holds in a product algebra $A \times B$ precisely if it holds in both $A$ and $B$ (this is a fact from universal algebra).
