Past exam of the logic and set theory course of the University of Oxford / 2007 / B / 8 / c / Solution

Irrationality of $\sqrt{2}$ makes the expression $q+r\sqrt{2}$ unique for rational $q,r$: equality of two such expressions with different $r$ would express $\sqrt{2}$ as a rational number. Therefore the rule selecting the coefficient $r$ defines a function on $A$. Its domain is a vector subspace over $\mathbb{Q}$, and direct coefficient addition and rational scaling show it is $\mathbb{Q}$-linear, hence weakly linear. In particularUse Zorn's lemma on the nonempty family of weakly linear extensions of this particular function. Chain unions remain extensions by (a); hence there is a maximal extension $h$. Any further extension in $\mathcal{F}$ would also contain the original function, so this $h$ is maximal in $\mathcal{F}$. Part (b) therefore gives $\mathrm{dom}h=\mathbb{R}$.

A real-linear map $\mathbb{R}\to \mathbb{R}$ is multiplication by its value at $1$. Were $h$ real-linear, it would satisfycontradicting the inherited value $h(\sqrt{2})=1$. Thus a weakly linear function on all of $\mathbb{R}$ need not be real-linear. Its rational homogeneity of an additive function is insufficient for real homogeneity. This particular extension is also discontinuous: rational numbers approaching $\sqrt{2}$ have value zero, while the value at their limit is one.