Dense subset of the plane that intersects every rational line at precisely one point?

Question

It seems there should exist a non-measurable bijection $f: \mathbb{R}\rightarrow \mathbb{R}$. And thus we can obtain a non-measurable graph on $\mathbb{R}^2$ which intersects every horizontal or vertical line at precisely one point. Does anyone know how to "construct" such a map? Can it be further made into a automorphsim (w.r.t the addtive group ~~or field~~ structure of $\mathbb{R}$)?

EDIT: Thanks to Piotr and Mark we can choose an addive (actually $\mathbb{Q}$-linear) automorphism $f$ of $\mathbb{R}$, and now we can see an amazing picture:

The graph of $f$ forms a dense subset of $\mathbb{R}^2$ yet it intersects every horizontal or vertical line at precisely one point. What's more amazing, thanks to Hanning's answer to this post, we can choose $f$ so that every rational line intersects the graph at at most one point!

Now my new question (sorry for adding it) is:

Can we choose $f$ so that the graph of $f$ intersects every rational line at precisely one point?

I am not sure that $\mathbb R$ has any field automorphisms. The additive group do have automorphisms, namely $\alpha\mapsto c\cdot\alpha$ for some $c>0$. — Asaf Karagila, Oct 09 '11 at 19:56
@AsafKaragila: Plus all those other weird ones that depend on the Axiom of Choice. — Arturo Magidin, Oct 09 '11 at 20:21
@Arturo: Oh yeah, sure... if you have the freedom to assume the axiom of choice! Ah, such a wonderful world it becomes... where countable unions of countable sets remain countable... :-) — Asaf Karagila, Oct 09 '11 at 20:29
Both answers say that it is not hard to show (or an exercise) that a measurable homomorphism $\mathbb{R} \to \mathbb{R}$ is continuous. I don't consider it that easy: It's a theorem of Steinhaus and has been discussed here, here and on MO and yet another time. — t.b., Oct 09 '11 at 20:42
It is true that it is not "obvious", but it is a medium exercise for anyone who started to learn Lebesgue theory — Piotr Pstrągowski, Oct 10 '11 at 20:46

score 3 · Accepted Answer · edited Oct 13 '11 at 15:05

Any bijection that exchanges some non-measurable subset of $\mathbb{R}$ of cardinality continuum with $(0,1)$ interval will do.

You can make it into an additive isomorphism. Indeed, it's not this hard to show that for real additive function the following holds: if it is measurable, it is actually continuous. So any $\mathbb{R}$ automorphism (as a vector space over $\mathbb{Q}$) that is not continuous will be ok.

On the other hand, $\mathbb{R}$ has no field automorphisms other than identity.

score 2 · Answer 2 · answered Oct 09 '11 at 20:24

Choose a Hamel basis for the vector space $\mathbb{R}$ over the field $\mathbb{Q}$. Define a linear map $\mathbb{R} \to \mathbb{R}$ by permuting some two elements of the basis, acting as an identity on the rest of the basis elements, and then extending it linearly to all of $\mathbb{R}$. Obviously you get a permutation, actually an automorphism of this vector space (and in particular of the underlying additive group).

Why is it not measurable? Well, here's a set of exercises for you:

A continuous automorphism of the additive group of $\mathbb{R}$ is of the form $f(x)=cx$ for some real scalar $c$.
A measurable automorphism of $\mathbb{R}$ is necessarily continuous.

so if our function is measurable then it is of the form $f(x)=cx$ for some scalar $x$. Now let $a,b$ be the two elements of the basis which the function permutes. Then $f(a)=ca$, $f(b)=cb$, $f(a)=b$ and $f(b)=a$. Thus $ca=b$ and $cb=a$. So $a^2=cba=cab=b^2$, hence $a = \pm b$, which gives a $\mathbb{Q}$-linear relation between the basis elements.

As for your last question, the only field automorphism of $\mathbb{R}$ is the identity. To see this, suppose $f$ is a field automorphism. Then $f(x^2)=f(x)^2$, so $f$ takes positive numbers to positive numbers. Since $f$ is linear it follows that it is monotone. Thus it has countably many discontinuities. In particular it is measurable. Use the above two exercises to finish the argument.

Dense subset of the plane that intersects every rational line at precisely one point?

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