Theorem 2

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Source. Section 5, printed pp. 61--62 (PDF pp. 3--4).

Statement. Use the definition of thin sets from Theorem 1. Let $G,H$ be additive abelian groups, let $S\subseteq G$ be thin, and let $f:G\to H$. If

$$f(x+y)=f(x)+f(y)$$

for every $x,y\notin S$, then this identity holds for every $x,y\in G$.

Proof. The exceptional pairs lie in

$$(S\times G)\cup(G\times S),$$

a light set. By Theorem 1, there is a homomorphism $h:G\to H$ such that $f=h$ outside a thin set $T$. Put $k=f-h$ and $U=S\cup T$, which is thin.

Fix $a\in G$. Since $U\cup(a-U)$ is thin and is therefore not all of $G$, choose $a_1$ outside that union and set $a_2=a-a_1$. Then $a_1,a_2\notin U$. The assumed equation applies to $(a_1,a_2)$, while $k(a_1)=k(a_2)=0$. Subtracting the additive identity for $h$ from the identity for $f$ gives

$$k(a)=k(a_1)+k(a_2)=0.$$

Thus $f=h$ on all of $G$, so $f$ is a homomorphism. $\square$

Proof coverage. This is a complete deduction from Theorem 1. It is conditional on that theorem; the group-level proof chain underlying Theorem 1 remains sketch-only on its result page.

Dependencies. Theorem 1.

Bears on. #1126