> [!NOTE] Lemma > There doesn't exist a [[Finite Partition of Closed Real Interval|partition]] $A,B$ of $[0,1]$ so that $B=A+a$ for some $a\in\mathbb{R}.$ > **Proof** Suppose there exists such a partition. WLOG suppose $a$ is positive otherwise rename the sets. Then $(1-a,1] \subset B$; hence $(1-2a,1-a] \subset A.$ Now $1-2a\not \in A$ since $1-2a+a=1-a\in B$ but $1-a\in A$ hence $1-2a\in B.$ Thus $(1-3a,1-2a]\subset B$ otherwise $A \cap B$ is non-empty. In general, for all positive integers $n,$ $(1-(2n+1)a,1-2na]\subset B$ and $(1-(2n+2)a,1-(2n+1)a]\subset A$ which we can prove by induction. Thus, the interval of the form $(1-(n+1)a,1-na]$ which contains $0$ must be a subset of either $A$ or $B$ which is impossible since this interval exits $[0,1].$