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.

  • $\begingroup$ What is $\phi$? Did you mean the empty set? This is denoted as $\varnothing$ or $\emptyset$. $\endgroup$ – Alexis Olson Nov 1 '16 at 16:56
  • $\begingroup$ Yeah edited. Sorry $\endgroup$ – pks Nov 1 '16 at 16:57
  • $\begingroup$ Did you mean $h(E) - E$ or $\Bbb R - E$? The difference set $E - E$ is empty. $\endgroup$ – Alexis Olson Nov 1 '16 at 16:59
  • $\begingroup$ See it again. Sorry for the confusion $\endgroup$ – pks Nov 1 '16 at 17:02
  • $\begingroup$ The key to the exercise is Baire's theorem. (No, there need not exist such a homeomorphism.) $\endgroup$ – Daniel Fischer Nov 5 '16 at 23:13

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.

  • $\begingroup$ Or more simply if $E$ is a dense $G_{\delta}$ set then $h(E)\cap E$ is again a dense $G_\delta$ set. $\endgroup$ – pks Nov 9 '16 at 6:21

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