Showing that the intersection of simple sets is a simple set

Question

The set of canonical intervals: $\mathcal{I}=\{[a,b) \quad | \quad a,b \in \mathbb{R} \quad \text{and} \quad a < b \}$

Simple sets: A subset $S$ of $\mathbb{R}$ is said to be simple if it’s the union of a finite number of canonical intervals.

-Attempt-

Let $s_1 =\cup_{i=1}^n I_i$ and $s_2 = \cup_{i=1}^m J_i$ I need to show that $s_1 \cap s_2$ and the difference between $s_1$ and $s_2$ is a simple set.

1. $s_1 \cap s_2$

Let $D= s_1 \cup s_2$ $\implies |D|=m+n$ intervals

Let $X=\{ \inf(K_i) | \; i \in \{1,2,3,...m+n\}, \; K_i \in D\}$

Let $Y=\{ \sup(K_i) | \; i \in \{1,2,3,...m+n\}, \; K_i \in D\}$

then $s_1 \cap s_2 = [\max{(X)},\min{(Y)}) \in \mathcal{I}$

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2. $s_1 - s_2$

case 1: $s_1 \cap s_2 = \emptyset \implies s_1 - s_2 =s_1$

case2: $s_1 \cap s_2 \neq \emptyset \implies s_1 - s_2 = s_1 - (s_1 \cap s_2)$

But, $s_1 \cap s_2 =I \in \mathcal{I}$

Now let $j$ be the minimal among all the indices of intervals in $s_1$ st $a_j \leq \inf(I) \implies I \supseteq [a_j,bj)$. Now $[a_j,b_j) = \underbrace{[\min(a_j , \inf(I)), \sup(I))}_\text{I}\cup [\sup(I),b_1)$

Now delete $I$ from $[a_j,b_j) \implies s_1 - s_2 = s_1 - (s_1 \cap s_2) = [\sup(I),b_1) \cup_{i=1}^{j-1} I_i \cup_{i=j+1}^{n} I_i$

Answer 1

It’s easier just to use the fact that intersection distributes over union:

$$\begin{align*} \bigcup_{i=1}^nI_i\cap\bigcup_{i=1}^mJ_i&=\bigcup_{i=1}^nI_i\cap\bigcup_{j=1}^mJ_j\\ &=\bigcup_{i=1}^n\left(I_i\cap\bigcup_{j=1}^mJ_j\right)\\ &=\bigcup_{i=1}^n\bigcup_{j=1}^m(I_i\cap J_j)\;. \end{align*}$$

Each $I_i\cap J_j$ is an intersection of canonical intervals, which you can easily check is either empty or a canonical interval, so $\bigcup_{i=1}^n\bigcup_{j=1}^m(I_i\cap J_j)$ is a finite union of canonical intervals and therefore a simple set.