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Let $\{A_n\}_{n\in\mathbb N}$ be an indexed family of sets. Then:

$(i) (\bigcup\limits_{n=1}^\infty A_n)' = \bigcap\limits_{n=1}^\infty (A'_n)$

$(ii) (\bigcap\limits_{n=1}^\infty A_n)' = \bigcup\limits_{n=1}^\infty (A'_n)$

I went from doing simple, straightforward indexed set proofs to this, and I don't even know where to start.

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    @Hagen von Eitzen: Except when the family is empty in which case a separate argument is needed.2017-08-09

4 Answers 4

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It’s not really any different from proving the finite versions: just show that each of the sets is a subset of the other. For (i), for instance, you want to show that

$\left(\bigcup_{n\ge 1}A_n\right)'\subseteq\bigcap_{n\ge 1}A_n'\tag{1}$ and that

$\bigcap_{n\ge 1}A_n'\subseteq\left(\bigcup_{n\ge 1}A_n\right)'\;.\tag{2}$

To show $(1)$, assume that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$; then $x\notin\bigcup_{n\ge 1}A_n$. This means that for every $n\ge 1$, $x\notin A_n$, which by the definition of complement means that $x\in A_n'$ for every $n\ge 1$. But that’s exactly what it means to say that $x\in\bigcap_{n\ge 1}A_n'$, so I’ve just proved $(1)$.

To prove $(2)$, assume that $x\in\bigcap_{n\ge 1}A_n'$. From the definition of intersection this means that $x\in A_n'$ for every $n\ge 1$, and hence that $x\notin A_n$ for every $n\ge 1$. This in turn means that $x\notin\bigcup_{n\ge 1}A_n$, i.e., that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$, so $(2)$ is also true. Finally the truth of $(1)$ and $(2)$ ensures that

$\left(\bigcup_{n\ge 1}A_n\right)'=\bigcap_{n\ge 1}A_n'\;.$

I’ll leave (ii) to you; you should be able to use much of what I did here as a model.

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If $a\in (\bigcup_{n=1}^{\infty}A_{n})'$ then $a\notin A_{n}$ for any $n\in \mathbb{N}$, therefore $a\in A_{n}'$ for all $n\in \mathbb{N}$. Thus $a\in \bigcap_{n=1}^{\infty}A_{n}'$. Since $a$ was arbitrary, this shows $(\bigcup_{n=1}^{\infty}A_{n})' \subset \bigcap_{n=1}^{\infty}A_{n}'$. The other containment and the other problem are similar.

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$(i)$ let $ x \in (\cup_{n=1}^\infty A_n)' \Rightarrow x \notin A_n $ for some $n$. Therefore, $x \in A_n$ for some $n$. and hence, $x \in \cap_{n=1}^\infty (A'_n)$

Now, using same reasoning, prove the other direction: $\cap_{n=1}^\infty (A'_n) \subseteq (\cup_{n=1}^\infty A_n)'$ This would give you the desired result. Second part is analogous.

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My good sir! We need to show: $\left(\bigcup_{n\ge 1}A_n\right)'\subseteq\bigcap_{n\ge 1}A_n'$ and $\bigcap_{n\ge 1}A_n'\subseteq\left(\bigcup_{n\ge 1}A_n\right)'.$

The following are all biconditional statements. See, if $x∈\left(\bigcup_{n\ge 1}A_n\right)'$, that is, if x is in the complement of all these sets collectively, then x certainly isn't in the union of all the sets, so we see $x∉\left(\bigcup_{n\ge 1}A_n\right).$ Now, if x is not in the union of all these indexed sets, then most certainly it is also not in the intersect of all these sets, or: $x∉\left(\bigcap_{n\ge 1}A_n\right).$ Then for a final step, another logical statement: if x is not in the intersection of all these indexed sets, then it is in the intersection of everything else (draw 3-part Venn-Diagrams to convince yourself this) thus, $x∈\left(\bigcap_{n\ge 1}A_n'\right).$

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    Perhaps. I'm providing a little more intuition. Is that going to be a problem?2014-12-07