Subsets partitioning $\mathbb R^d$ into $2^k$ disjoint sets

Question

I have been going through Terrance Tao's Introduction to Measure Theory book and I have been having trouble developing intuition behind a certain statement he makes after defining a (complex-valued) Simple function as a linear combination of indicator functions of Lebesgue measurable sets and complex numbers.

Within the definition, he makes the comment that it is not required that the Lebesgue measurable sets be disjoint but that it can be easy enough to arrange this. He says that any $k$ subsets $E_1, \ldots, E_k$ of $\mathbb R^d$ partition $\mathbb R^d$ into $2^k$ disjoint sets, each of which is an intersection of $E_i$ or the complement $\mathbb R^d - E_i$. I do not understand how the $k$ subsets partition $\mathbb R^d$ into $2^k$ disjoint sets and I am not able to create a simple case which can help me understand the general one. Any help would be much appreciated. Thank you in advance.

Answer 1

For each $I\subseteq[k]=\{1,2,\ldots,k\}$ let

$$E_S=\bigcap_{i\in S}E_i\setminus\bigcup_{i\in[k]\setminus S}E_i\,;$$

$E_S$ consists of those point of $\Bbb R^d$ that are in $E_i$ if and only if $i\in S$. For instance, $S_\varnothing=\Bbb R^d\setminus\bigcup_{i=1}^kE_i$, $S_{[k]}=\bigcap_{k=1}^kE_i$, and $E_{\{1\}}$ is the set of points of $\Bbb R^d$ that are in $E_1$ but not in any of $E_2,\ldots,E_k$. Since $[k]$ has $2^k$ subsets, there are $2^k$ of these sets $E_S$.

• Show that if $S,T\subseteq[k]$, and $S\ne T$, then $E_S\cap E_T=\varnothing$.

(It might be helpful to think about $3$ sets in $\Bbb R^2$ and draw a typical Venn diagram: it divides the plane into $2^3=8$ regions.)