$[0,1]$ cannot be partititioned into two sets with given properties Show that the interval $[0,1]$ cannot be partitioned into two disjoint sets $A$ and $B$ such that $B=A+a$ for some real number $a$.
My proof: Suppose by contradiction that $[0,1]=A\sqcup B$ where $B=A+a$ and WLOG let $a>0$. It's easy to verify that $0\in A$ and $1-a\in A$. Also it's easy to prove that $A\subset [0,1-a]$ and $B=A+a\subset [a,1]$.
First case: If $1-a<a$ then $a>\frac{1}{2}$. It is easy to show that in this case $\frac{1}{2}\notin [0,1-a]$ and $\frac{1}{2}\notin [a,1]$ $\Rightarrow$ $1/2\notin A$ and $1/2\notin B$ but $1/2\in [0,1]$. This is a contradiction.
Second case: If $1-a=a$ then $a=\frac{1}{2}$. In this case $A\subset [0,\frac{1}{2}]$ and $B\subset [\frac{1}{2},1]$. Since $0\in A$ then $\frac{1}{2}\in B$. Since $1\in B$ then $1-\frac{1}{2}\in A$. Thus $\frac{1}{2}\in A\cap B$. Contradiction.
Third case: If $1-a>a$ then $a<\frac{1}{2}$. Let $I=[a,1-a]$ and it can be shown that $I\cap A\neq \varnothing$ and $I\cap B\neq \varnothing$. I have stuck here and not able to derive contradiction. 
Can anyone please give a hint how to overcome this obstacle?
 A: (Just working around the points $0$ and $1$ is not sufficient. We have to cross the abyss in between in a controlled way.)
We go ahead and  find out how two such sets would have to look like. We therefore assume that $A$ and $B$ form such a partition of $[0,1]$, whereby $0<a<1$. Then necessarily $[0,a[\>\subset A$, hence $[a,2a[\>\subset B$. Therefore the points in $[2a,3a[\>$ have to lie in $A$, and so forth. It turns out that the half open intervals
$$[0,a[\>,\quad [2a,3a[\>,\quad [4a,5a[\>,\quad\ldots$$ (up to the upper limit $1$) all belong to $A$ and the intervals
$$[a,2a[\>,\quad [3a,4a[\>,\quad [5a,6a[\>,\quad\ldots$$ (up to the upper limit $1$) all belong to $B$. The number $1$ necessarily belongs to one of the $B$-intervals. Therefore we have
$$(2m-1)a\leq 1<2m a$$
for some $m\geq1$. There is a $\beta$ with $1<\beta<2ma$. As $\beta$ is in the cutoff of a $B$-interval it follows that $\alpha:=\beta-a\in A$. But $\alpha+a>1$, hence $\alpha+a\notin B$. This violates the condition $B=A+a$.
A: Fix $a \in (0,1)$ - those are the only options we need to consider - and define
$$x \sim_a y \iff \frac{x-y}{a} \in \mathbb{Z}.$$
Then we can partition $[0,1]$ in the specified way if and only we can partition each equivalence class of $\sim_a$ in the specified way. That means, if $\xi = \min \{ x \in [0,1] : x \sim_a \xi\}$, we must have $\xi + ka \in A$ for even $k$ with $\xi + ka \in [0,1]$ and $\xi + ka \in B$ for odd $k$. In particular, $k_{\max}(\xi) = \bigl\lfloor \frac{1-\xi}{a}\bigr\rfloor$ must be odd for every $\xi \in [0,a)$. Choosing $\xi = 0$ we see that there is an $n$ such that
$$\frac{1}{2n} < a \leqslant \frac{1}{2n-1}.$$
But $\bigl\lfloor \frac{1-\xi}{a}\bigr\rfloor$ is even (namely, it's $2n-2$) when
$$1 - (2n-1)a < \xi < a.$$
This is possible since $1 < 2na \implies 1-(2n-1)a < a$, and $(2n-1)a \leqslant 1 \implies a \leqslant 1 - (2n-2)a$ so we then have $\bigl\lfloor \frac{1-\xi}{a}\bigr\rfloor = 2n-2$ indeed.
So there always is an equivalence class that cannot be partitioned in the required way.
