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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?

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  • $\begingroup$ 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. $\qquad$ $\endgroup$ – Michael Hardy Nov 16 '19 at 0:50
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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 $t\in S$.

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  • $\begingroup$ Perfectly explained. Thanks! $\endgroup$ – alladinsane Nov 15 '19 at 22:39
  • $\begingroup$ 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. $\qquad$ $\endgroup$ – Michael Hardy Nov 16 '19 at 0:52
  • $\begingroup$ 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? $\endgroup$ – alladinsane Nov 16 '19 at 8:59
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Hint:

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 $B_n$.

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