# Partitioning intervals in the real line

Did you know that the closed interval $[0,1]$ cannot be partitioned into two sets $A$ and $B$ such that $B = A + t$ for some real number $t$? Of course, the half-open interval $[0,1)$ can so be partitioned: $A = [0,1/2)$ and $t = 1/2$. Why is this? I will leave the full details to the reader but I am sure they can be reconstructed without much difficulty using the following sketch:

Assume such a partition can be so made, and assume without loss of generality that $t \gt 0$. Then we must have $[0,t)\subseteq A$ and $(1-t,1]\subseteq B$. This shows that $t \lt 1/2$. Now, by assumption, $B = A + t$. Therefore, $[t,2t)\subseteq B$ since $[0,t)\subseteq A$ and similarly $(1-2t,1-t]\subseteq A$. Because $A$ and $B$ are disjoint, this implies that $t \lt 1/4$. We can continue to play this game, which shows that $t$ is strictly less than $1/(2n)$ for any $n$ and hence $t = 0$. This shows that such a partition cannot be made.

Pretty good. But did you know that the closed interval $[0,1]$ cannot be partitioned into two nonempty, disjoint open sets? Neither can any interval, whether open, closed, or half-open. In the language of topology, intervals of real numbers are connected. Proof?