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

Zorn's lemma states that a nonempty partially ordered set in which every chain has an upper bound contains a maximal element. Here a chain means a subset totally ordered by the inherited order; it need not have a greatest member.

View every partial function as its graph, a subset of $\mathbb{R}\times \mathbb{R}$. The family $\mathcal{F}$ is a set by axiom of separation from the power set of this product. Let $\mathcal{C}\subseteq \mathcal{F}$ be a chain and put $h=\bigcup \mathcal{C}$. If two members of the union assign values to the same input, the chain property makes one graph contain the other, so those values agree. Thus $h$ is a partial function.

If $x\in \mathrm{dom}h$ and $q\in \mathbb{Q}$, choose a chain member containing $x$ in its domain. Its rational-scalar closure gives $qx\in \mathrm{dom}h$. If $x,y\in \mathrm{dom}h$, choose two chain members containing them; the larger one contains both, and therefore contains $x+y$ with value $h(x)+h(y)$. Hence $h$ satisfies the required additive identity and domain closure. Consequentlyand it is an upper bound for the chain. For the empty chain its union is the empty function, which satisfies the defining logical implications vacuously, so this case is covered too.

A nonempty weakly linear function has a domain $D$ closed under rational scalars and addition, hence a vector subspace of $\mathbb{R}$ over $\mathbb{Q}$. Additivity first gives $f(0)=0$ and $f(-x)=-f(x)$, then $f(nx)=nf(x)$ for integers $n$. If $q=m/n$ with $n>0$, rational closure puts $qx$ in $D$ andThus $f(qx)=qf(x)$: it is a $\mathbb{Q}$-linear map. The empty function is not maximal because it extends to the function on $\{0\}$ with value zero.

Suppose a maximal $f$ had proper domain $D\ne \mathbb{R}$. Choose $t\notin D$ and any prescribed real value $a$, for instance $a=0$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$: if two expressions coincide and the coefficients differ, then $t$ would belong to $D$. DefineUniqueness makes this well-defined. The displayed logical formula and the $\mathbb{Q}$-linearity of $f$ show directly that $h$ is additive and has a domain closed under rational scalars. Hence it is a weakly linear function and extends $f$. It is a proper extension because $t$ is now in its domain, contradicting maximality. ThereforePart (a) and Zorn's lemma guarantee the existence of maximal extensions as well as this description of their domains.

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.