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Find the sets $$ \bigcup_{N=1}^\infty\left(\bigcap_{n=N}^\infty A_n\right) \text{ and } \bigcap_{N=1}^\infty\left(\bigcup_{n=N}^\infty A_n\right) $$ if

(1) $A_1,A_2,\dots$ are pairwise disjoint

(2) if $A_n=\left\{\begin{array}{ll} B \text{ if } n \text{ is odd}\\ C \text{ if } n \text{ is even} \end{array} \right.$

For (1) my argument is that $\bigcup_{N=1}^\infty\left(\bigcap_{n=N}^\infty A_n\right)=\emptyset$ (as I first take the intersection of disjoint sets which is empty) and $\bigcap_{N=1}^\infty\left(\bigcup_{n=N}^\infty A_n\right)=\infty$ (as I first take the union of a infinitely large set, and that the intersection of infinitely large set is infinite) Am I correct?

For (2) $\bigcup_{N=1}^\infty\left(\bigcap_{N=1}^\infty A_n\right)=\bigcup_{N=1}^\infty\left(B_1\cap C_2 \cap B_3 \cap C_4\cap \ldots \right)$ and $\bigcap_{N=1}^\infty\left(\bigcup_{n=N}^\infty A_n\right)=\bigcap_{N=1}^\infty\left(B_1\cup C_2 \cup B_3 \cap C_4\cup \ldots \right)$, but I havve really no idea how to proceed from here.

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  • $\begingroup$ In your own answer $(1)$ you mixed up some indices I think (in the second part). $\endgroup$ – GenericNickname Nov 17 '14 at 9:18
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First, $\cap_{N=1}^{\infty} ( \cup_{n=N}^{\infty} A_n) = \infty$ makes no sense as a set can be of infinite cardinality but not be equal to infinity.

Now regarding your task: I assume that there is some set $A$ given such that $A_n \subseteq A$ for all $n$ (if not, let $A = \cup_{n=1}^{\infty} A_n$). I think it would help you to understand that

$$\bigcup_{N=1}^{\infty} \left( \bigcap_{n=N}^{\infty} A_n \right) = \left\lbrace a \in A ~ | ~ \text{There exists } n_0 \text{ such that } a \in A_n \text{ for all } n \geq n_0 \right\rbrace$$ and $$\bigcap_{N=1}^{\infty} \left( \bigcup_{n=N}^{\infty} A_n \right) = \left\lbrace a \in A ~| ~ a \in A_n \text{ for infinitely many } n \right\rbrace $$ no matter what $A_n$ actually is.

This way it is easier to see that for (1) we have $$\bigcup_{N=1}^{\infty} \left( \bigcap_{n=N}^{\infty} A_n \right) = \emptyset = \bigcap_{N=1}^{\infty} \left( \bigcup_{n=N}^{\infty} A_n \right) $$ and for (2) we have $$\bigcup_{N=1}^{\infty} \left( \bigcap_{n=N}^{\infty} A_n \right) = B \cap C ~, ~ \bigcap_{N=1}^{\infty} \left( \bigcup_{n=N}^{\infty} A_n \right) = B \cup C.$$

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The following characterization can be helpful:

$$a\in\bigcup_{N=1}^{\infty}\bigcap_{n=N}^{\infty}A_{n}\iff\left\{ n\in\mathbb{N}\mid a\notin A_{n}\right\} \text{ is a finite set}$$

$$a\in\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}A_{n}\iff\left\{ n\in\mathbb{N}\mid a\in A_{n}\right\} \text{ is not a finite set}$$

(1) Set $\left\{ n\in\mathbb{N}\mid a\notin A_{n}\right\} $ is not finite and $a\in A_{n}$ cannot be true for more than one index $n$. So both sets are empty.

(2) Set $\left\{ n\in\mathbb{N}\mid a\notin A_{n}\right\} $ is finite if and only if $a\in B\cap C$ and set $\left\{ n\in\mathbb{N}\mid a\in A_{n}\right\} $ is infinite if and only if $a\in B\cup C$.

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