Exercise 8.2.2

Proof. Consider any arbitrary continuous additive function $f:𝑹\to 𝑹$. Then, by what was shown in the text, there is a real $a$ such that $f(q)=f_a(q)=a\cdot q$ for all $q\in 𝑸$; in particular $a=f(1)$.

Now suppose to the contrary that $f\ne f_a$ so that there is an $x\in 𝑹$ where $f(x)\ne f_c(x)$. So let $𝜀=\left|f(x)-f_a(x)\right|∕2$, noting that clearly $𝜀>0$ since $f(x)\ne f_a(x)$. Since $f$ is continuous there is a real ${\delta }_1>0$ such that $\left|f(y)-f(x)\right|<𝜀$ for all $y\in 𝑹$ where $\left|y-x\right|<{\delta }_1$. Also clearly $f_a$ is also continuous so that there is a real ${\delta }_2>0$ where $\left|f_a(y)-f_a(x)\right|<𝜀$ for all $y\in 𝑹$ where $\left|y-x\right|<{\delta }_2$. So let $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\left\{{\delta }_1,{\delta }_2\right\}$. Then, since $\delta >0$ it follows that $x-\delta <x+\delta$ so that there is a $q\in 𝑸$ where $x-\delta <q<x+\delta$ since $𝑸$ is order dense in $𝑹$. It then clearly follows that $\left|q-x\right|<\delta$ so that $\left|q-x\right|<\delta \le {\delta }_1$ and $\left|q-x\right|<\delta \le {\delta }_2$. Therefore $\left|f(q)-f(x)\right|<𝜀$ and $\left|f_a(q)-f_a(x)\right|<𝜀$.

We then have

which is a contradiction, noting that $\left|f(q)-f_a(q)\right|=0$ since $f(q)=f_a(q)$ since $q\in 𝑸$. So it must be that in fact $f=f_a$ as desired. □