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Is this equation true?

$\mathcal C \bigcap_{M\in A}M=\bigcup_{M\in A}\mathcal CM$ where C is the complement, M is a set, A is a set of sets.

I don't know how to start proving or disproving it because it's not only two sets but all that belong to A. So I can't use Venn diagrams to show this ...

The intersection/union of sets is defined as:

$\bigcap_{M\in A}=\{x\in \Bbb R \mid M\in A\Rightarrow x\in A\}$ $\bigcup_{M\in A}=\{x\in \Bbb R \mid \exists M\;,M\in A\Rightarrow x\in A\}$

How can I start?

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    possible duplicate of [Infinite DeMorgan laws](http://math.stackexchange.com/questions/207570/infinite-demorgan-laws)2012-10-25

1 Answers 1

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To prove that $\mathcal C\bigcap_{M\in\mathcal{A}}M=\bigcup_{M\in\mathcal{A}}\mathcal CM\;,$

show that each side is a subset of the other:

$\mathcal C\bigcap_{M\in\mathcal{A}}M\subseteq\bigcup_{M\in\mathcal{A}}\mathcal CM\tag{1}$ and

$\bigcup_{M\in\mathcal{A}}\mathcal CM\subseteq\mathcal C\bigcap_{M\in\mathcal{A}}M\;.\tag{2}$

$(1)$ and $(2)$ can be proved by ‘element-chasing’: assume that some object $x$ is an element of the lefthand side, and prove that it is necessarily an element of the righthand side.

To prove $(1)$, for instance, suppose that $\displaystyle{x\in\mathcal C\bigcap_{M\in\mathcal{A}}M}$. Then $x\notin\bigcap\limits_{M\in\mathcal A}M$. By the definition of intersection this means that there is at least one $M_0\in\mathcal A$ such that $x\notin M_0$. But then

$x\in\mathcal CM_0\subseteq\bigcup_{m\in\mathcal A}\mathcal C M\;,$

and since $x$ was an arbitrary element of $\mathcal C\bigcap\limits_{M\in\mathcal A}M$, it follows that

$\mathcal C\bigcap_{M\in\mathcal{A}}M\subseteq\bigcup_{M\in\mathcal{A}}\mathcal CM\;.$

I’ll leave $(2)$ to you.