Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 2D / b

The first condition concerns the domains: their symmetric difference $A△B$ must be a finite set. The original PDF prints this condition and the agreement condition together; the converted TeX omits both. The two existing solution scopes are used here to prove the domain and value parts of the single equivalence relation assertion.

Domain agreement is a reflexive relation because $A△A=\varnothing$, and a symmetric relation because $A△B=B△A$. For the transitive relation property, membership can differ between $A$ and $C$ only if it differs between $A$ and $B$ or between $B$ and $C$. ThusA set union of two finite sets is finite, so the domain condition passes through a chain of related pairs. This is the domain part of equivalence of partial functions modulo finite changes.

For the value condition, a function agrees with itself outside the empty exceptional set, proving the reflexive relation property. Interchanging the two functions preserves equality, so the same exceptional finite set proves the symmetric relation property.

Suppose $(A,f)$ is related to $(B,g)$ and $(B,g)$ to $(C,h)$, with exceptional finite sets $F_{AB}$ and $F_{BC}$. Points of $A\cap C$ might be absent from $B$, so merely taking $F_{AB}\cup F_{BC}$ would miss a real issue. Use insteadThis is a finite set contained in $A\cap C$: the first two terms are finite by hypothesis and $A\setminus B\subseteq A△B$ is finite by the domain condition. If $x\in (A\cap C)\setminus F_{AC}$, then $x\in B$ and neither original exception occurs, so $f(x)=g(x)=h(x)$. The domain condition is also preserved, as proved in the preceding scope. Thus the full relation is a transitive relation, andFinite domains cause no problem: one may take the whole intersection as the exceptional set. Both conditions are needed for equivalence of partial functions modulo finite changes.