Image by a translate/homeomorphism of a set E of Lebesgue measure 0 is disjoint with E

Question

This is a question on Walter Rudin's RCA (exercise 5.21):

Suppose $E\subset\mathbb{R}$ is a measurable set with $m(E)=0$. Must there be a translate $E+x$ of $E$ that does not intersect $E$? Must there be a homeomorphism $h:\mathbb{R}\rightarrow\mathbb{R}$ such that $h(E)$ does not intersect $E$?

I am guessing the answer to the first one is no, because it being true would immediately answer the second one ($t\mapsto t+x$ would be a valid homeomorphism). However I can't even start to prove neither of the questions.

PS: this was asked before I know, but an example of $E$ for which a translate doesn't exist wasn't given, and also the existence of $h$ was proven using a little bit of category theory which I know nothing about.

Answer 1

The answer to both questions is no.

1. For an explicit counterexample consider the Cantor set $C \subseteq [0,1]$ and set $$E := \bigcup_{k\in\mathbb{Z}}(E+k).$$ Clearly $E$ is a measurable set with $m(E) = 0$. We'll show that every translate of $E$ intersects $E$.

   Indeed, for $x \in \mathbb{R}$ we have $y \in E \cap (E+x)$ if and only if $y = e = e'+x$ for some $e,e' \in E$. Equivalently, $x =e' - e$ for some $e,e' \in E$. Recall that $C-C =[-1,1]$ so $$E-E \supseteq (C+k)-C = [k-1,k+1], \quad\text{for all } k \in \mathbb{Z}$$ and therefore $E-E = \mathbb{R}$. We conclude that every $x \in \mathbb{R}$ can be written as a difference of two elements of $E$ and hence $E \cap (E+x)$ is nonempty.

2. For the second question, let $E$ be a dense set of measure zero which can be written as a countable intersection $E = \bigcap_{n=1}^\infty U_n$ of open sets $U_n$. For an explicit example we can take $$U_n = \bigcup_{k=1}^\infty\left\langle q_k-\frac1{n2^k}, q_k+\frac1{n2^k}\right\rangle$$ where $(q_k)_k$ are the rational numbers. We'll show that for every homeomorphism $h : \mathbb{R} \to \mathbb{R}$ holds that $h(E) \cap E$ is nonempty. We have $$h(E) \cap E = h\left(\bigcap_{m=1}^\infty U_m\right) \cap \left(\bigcap_{n=1}^\infty U_n\right) = \left(\bigcap_{m=1}^\infty h(U_m)\right) \cap \left(\bigcap_{n=1}^\infty U_n\right) = \bigcap_{m,n \in \mathbb{N}} h(U_m) \cap U_n.$$ Since $h$ is a homeomorphism, the sets $h(U_m) \cap U_n$ are also open dense sets in $\mathbb{R}$ so by the Baire Category Therorem, the countable intersection $\bigcap_{m,n \in \mathbb{N}} h(U_m) \cap U_n$ is nonepmty. Therefore $h(E) \cap E \ne \emptyset$. Obviously every translate of $E$ intersects $E$ as well.