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

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?

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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 9 '11 at 19:56
@AsafKaragila: Plus all those other weird ones that depend on the Axiom of Choice. –  Arturo Magidin Oct 9 '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 9 '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 9 '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

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.

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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:

1. A continuous automorphism of the additive group of $\mathbb{R}$ is of the form $f(x)=cx$ for some real scalar $c$.
2. 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.

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