I stumbled upon this problem on Putnam and Beyond by Gelca et. al. It’s problem 9 in chapter 1.1.
I have tried to do proof by contradiction by supposing that in fact $[0, 1]$ can be partitioned into two disjoint sets $A$ and $B$ such that $B = A + a$ for some real number $a$.
Assume $a$ is positive W.L.O.G. The first thing I did was to try to split $[0, 1]$ into two disjoint sets. There exists $k > 0$ such that $[0, 1] = [0, k] \cup (k, 1]$. Let $A = [0, k]$. I tried to reach a contradiction by showing that there is no real positive number $a$ such that $[0 + a, k + a] = (k, 1]$. So, $a > k$ and $k + a = 1$. Combining these two conditions, $a > \frac{1}{2}$ and $k < \frac{1}{2}$. But there seems to be no problem with $a$.
And I can’t think of any other way of contradicting the statement. A hint instead of a solution would be much appreciated.
Best Answer
It is trivial to show the statement for $|a|\ge1$. W.L.O.G we can say that $0<a<1$.
Now, let's suppose that sets $A$ and $B$ s.t. they satisfy the given conditions exist.
1. case
$a< \frac{1}{2}$
Considering that $A$ and $B$ are disjoint, the fact that $a>0$ and that element $1$ has to be in one of them it must be $1\in B$, this means that $\max A=1-a$. Similarly $\min B=a$. This implies, not only $[0,a)\subset A$, but also $2a\in A$. Continuing like this we can see that all subsets of $[0,1]$ of the form $[2ka,(2k+1)a)$ are subsets of $A$ and all subsets of the form $[(2k+1)a,(2k+2)a)$ are subsets of $B$. Keep in mind that $k\in \mathbb{N}$, until $2ka=1$. This means that there exists some $n\in \mathbb{N}$, s.t. $2na=1$, which would mean $1\in A$, but of course that is impossible, therefore a contradiction.
2. case
$a>\frac{1}{2}$
You can simply show that both $A$ and $B$ in this case would have to be intervals, since $\min B=a>\frac{1}{2}$ and $2a\notin A$, because if it were that $A$ wouldn't be a subset of $[0,1]$. Because $max A=1-a$, that leaves $(1-a,a)\not\subset A\cup B$. So obviously $A\cup B \neq [0,1]$. Contradiction!
3. case
$a=\frac{1}{2}$
In this case, $\max A=\min B=\frac{1}{2}$. Implying $A \cap B=\{\frac{1}{2}\}$. And again, a contradiction.
From all of this we can conclude that such sets $A$ and $B$ that satisfy the given conditions don't exist.