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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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Do you have a formal definition of what the notation $\bigcap _{M\in \mathcal A} M$ means? Your proof would need to start with that, and the style and level of formalism in the proof should match that of the definition you're working with. –  Henning Makholm Oct 25 '12 at 21:24
    
I've added these definitions to the question, thank you! –  Marco W. Oct 25 '12 at 23:22
    
possible duplicate of Infinite DeMorgan laws –  Asaf Karagila Oct 25 '12 at 23:26
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1 Answer 1

up vote 1 down vote accepted

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.

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