I was reading chapter 4 of the book Functional Equations by B.J. Venkatachala (Page No. $118$) and found a remarkable proof to the fact that there are infinitely many additive functions ($f:R \to R$ and satisfy the condition $f(x+y)=f(x)+f(y)$ where $x,y \in \mathbb{R}$ ) on $\mathbb{R}$ which are not linear (i.e. of the form $f(x)=cx$ for some constant $c$). The proof made use of the concept of Hamel Basis.
My question is that can we prove the same thing without making the use of Hamel Basis ?
A pedantic response would be highly helpful.
thank you,
best regards !
Consider $\Bbb R$ as a vector space over $\Bbb Q$. It has a basis $X$ (here the Axiom of Choice is used), its cardinality is continuum. Then take all permutations of $X$. The cardinality of the set of these permutations is bigger than continuum. Each of these permutations extends to a $\Bbb Q$-linear (hence additive) map from $\Bbb R$ to itself. So the cardinality of the set of additive maps is bigger than continuum. But the cardinality of all linear maps $x\mapsto \alpha x$ is continuum. So there are many additive maps which are not linear.