An additive map that is not a linear transformation over $\mathbb{R}$, when $\mathbb{R}$ is considered as a $\mathbb{Q}$-vector space [duplicate]

Possible Duplicate:
On sort-of-linear functions

I am looking for an example of an additive map that is not a linear transformation over $\mathbb{R}$, when $\mathbb{R}$ is considered as a $\mathbb{Q}$-vector space. I mean, I want to find an example of a map $T:\mathbb{R}\rightarrow\mathbb{R}$ such that $T(u+v)=T(u)+T(v)$ for all $u,v\in \mathbb{R}$, but $T(\alpha v)=\alpha T(u)$ is not true for all $\alpha \in\mathbb{R}$.

Thanks for your kindly help.

-

marked as duplicate by Jonas Meyer, Arturo Magidin, spohreis, Pete L. Clark, Gerry MyersonFeb 8 '12 at 6:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Pick a basis; use the basis. –  Arturo Magidin Feb 7 '12 at 20:45
I assume you mean linear over $\mathbb{Q}$, in which case you should replace "for all $\alpha\in\mathbb{R}$" by "for all $\alpha\in\mathbb{Q}$"... –  M Turgeon Feb 7 '12 at 20:46
@MTurgeon: I believe spohreis wants a map that is $\mathbb{Q}$-linear but not $\mathbb{R}$-linear, so "for all $\alpha\in \mathbb{R}$" is correct, since it is prefaced by "is not true" –  Arturo Magidin Feb 7 '12 at 20:47
@ArturoMagidin Fair enough; I just noticed that additivity implies $\mathbb{Q}$-linearity. –  M Turgeon Feb 7 '12 at 21:03

1 Answer

Let $\{r_\alpha\}$ be a Hamel basis of $\Bbb R$ over $\Bbb Q$. Let $\phi$ map $x$ to $c_{\alpha_1}+\cdots+c_{\alpha_k}$, where the (unique) basis representation of $x$ is $c_{\alpha_1}r_{\alpha_1}+\cdots+c_{\alpha_k}r_{\alpha_k}$. Then $\phi(x+y)=\phi(x)+\phi(y)$, but takes on only rational values.

If $\phi(\alpha v)=\alpha\phi(v)$ for all $\alpha$, $v$ in $\Bbb R$, then $\phi$ would be onto. As this isn't the case, $\phi$ is not $\Bbb R$-linear.

It is $\Bbb Q$-linear, though. In fact, any additive map would automatically be $\Bbb Q$-linear.

As far as I know, you need the axiom of choice to construct a function of this type (?).

-
Andres Caicedo's comment on a closely related question supports your last sentence. –  Jonas Meyer Feb 7 '12 at 21:01