# Prove $\bigcap (A_n \cup B_n) \supset (\bigcap A_n)\cup(\bigcap B_n)$

Prove that:

$$\bigcap (A_n \cup B_n) \supset \left(\bigcap A_n\right) \cup \left(\bigcap B_n\right)$$

Also find an example when there is no equality. Ok, so

$$\bigcap A_n := \{a \mid \forall Y \in A: a \in Y\}$$

$$\bigcap B_n := \{b \mid \forall X \in B: b \in X\}$$

$$\bigcup A_n := \{c \mid \exists Z \in A: c \in Z\}$$

$$\bigcup B_n := \{d \mid \exists T \in B: d \in T\}$$

Then how can I proceed?

• Normally one uses $\displaystyle \text{“}\cup\text{''}$ in things like $$A\cup B\quad\text{and}\quad A_1 \cup \cdots \cup A_n$$ and $\displaystyle \text{“}\bigcup\text{''}$ in things like $$\bigcup_{k=1}^n A_k.$$ I edited the question accordingly. $\qquad$ – Michael Hardy Nov 16 '19 at 0:50

Let $$S = \bigcap(A_n\cup B_{n})$$ and let $$T=(\bigcap A_n)\cup (\bigcap B_n)$$ denote the left and right hand sides. We want to prove that $$T\subset S$$. So let $$t\in T$$; we must show that $$t\in S$$.
Because $$t\in T$$, $$t\in \bigcap A_n$$ or $$t\in \bigcap B_n$$. That is, either $$t\in A_n$$ for all $$n$$, or $$t\in B_n$$ for all $$n$$. In either case, we have $$t\in A_n\cup B_n$$ for all $$n$$, which means $$t\in S$$.
• Normally one uses $\displaystyle \text{“}\cup\text{''}$ in things like $$A\cup B\quad\text{and}\quad A_1 \cup \cdots \cup A_n$$ and $\displaystyle \text{“}\bigcup\text{''}$ in things like $$\bigcup_{k=1}^n A_k.$$ I edited this answer accordingly. $\qquad$ – Michael Hardy Nov 16 '19 at 0:52
• I see. Could you please give me a hand on how to prove the equality instead of inclusion when $\forall n \in N: A_{n+1} \subset A_n$ and same for $B_n$? And about the inclusion, an example when it becomes an equality? – alladinsane Nov 16 '19 at 8:59
For each $$n$$, $$\; A_n\subset A_n\cup B_n$$, so $$\bigcap_n A_n \subset \bigcap_n( A_n \cup B_n),$$ and similarly for $$B_n$$.