Existence of homeomorphism $h: \mathbb{R}\to \mathbb{R}$ such that $h(E)\cap E=\emptyset$ Suppose $ E\subset \mathbb{R}$ and $m(E)=0$ ($m$ is the usual lebesgue measure). 
Must there exist a homeomorphism $h: \mathbb{R}\to \mathbb{R}$ such that $h(E)\cap E=\emptyset$ $?$
What i got till now is if the difference set $E-E=\{x-y : x\in E,y\in E\}$ is not the whole $\mathbb{R}$, then for any $ l\in (E-E)^c$,the translate $h(x)=x+l$ is the required homeomorphism. 
Any hints would be appreciated. This is an exercise from the Banach spaces chapter from Big rudin's and i have no idea how to use the theory of the chapter here.
 A: This exercise illustrates that, although the Lebesgue measure is closely connected to the geometry and topology of $\mathbb{R}$, the measure-theoretic notion of smallness is quite different from a topological notion of smallness.
From the measure-theoretic angle, a set $E\subset \mathbb{R}$ with $m(E) = 0$ is small. Very very small. And its complement is big, really big.
Topologically, we can call a set $M \subset \mathbb{R}$ small if it is meagre (of the first category). A non-meagre subset of $\mathbb{R}$ (a set of the second category) is big, and a co-meagre subset - that is, a subset whose complement is meagre - is very big, topologically.
Now you can take your null set $E$ to be a dense $G_\delta$-set in $\mathbb{R}$. Then $E$ is non-meagre, and $\mathbb{R}\setminus E$ is meagre, and thus too small to contain a subset that is of the second category in $\mathbb{R}$. So for such an $E$ we must have $h(E) \cap E \neq \varnothing$ for every homeomorphism $h \colon \mathbb{R}\to \mathbb{R}$, since homeomorphisms preserve category.
