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How can I prove that $A_1 \cup A_2 \cup \dots \cup A_n = U$ if and only if $A_1^c \cap A_2^c \cap\dots \cap A_n^c = \emptyset$ ?

I know I have to prove $A \implies B$ and $\lnot B\implies \lnot A$, but I don't understand how.

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    De Mogan's Law? Are you familiar with it? – 2017-01-30
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    Yes, but I don't know how to use it in this problem. – 2017-01-30
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    Then apply it to both sides and note that $U^c=\emptyset$ and $\emptyset^c=U$. You can dot it. Just take your time. – 2017-01-30
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    By the way, it should be De Morgan's Law and not De Mogan's in my first comment.--:) – 2017-01-30
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    You're wrong in the second line: the $(\lnot B\implies \lnot A)$ is equivalent to $(A \implies B)$ so what you've wrote is twice the rightwards implication with a lack of a leftwards implication. An 'iff' expression: '_A_ if and only if _B_' is an equivalence, or both-way implication, so $(A\implies B) \land (B\implies A)$ is what you actually need to prove. – 2017-02-02
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    To proof $A \Longleftrightarrow B$ which is the same as "$A$ iff $B$" and you have to proof $A \implies B$ and $B \implies A$. Note that $\lnot B\implies \lnot A$ is equivalent to $A \implies B$. – 2017-02-02

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A sum (a union) of $A_i$ sets is by definition such a set, that each of its elements belongs to at least one of $A_i$: $$\forall(x\in U)\ \exists i: x\in A_i$$ An intersection (a common part) of ${A_i}^c$ sets is a set, whose every element belongs to all ${A_i}^c$ sets. However, for any $x$ if $x$ belongs to some set $S$, then, by definition of a complement set, is does not belong to $S^c$. $$(x\in S)\iff(x\notin S^c)$$

And as every element $x\in U$ belongs to some $A_i$, it also does not belong to the corresponding ${A_i}^c$ $$\forall(x\in U)\ \exists i: x\notin {A_i}^c$$ hence it does not belong to all ${A_i}^c$. Consequently, none $x$ belongs to an intersection of all ${A_i}^c$ $$\forall(x\in U): x\notin \bigcap_i {A_i}^c$$ which means the intersection is empty $$\bigcap_i {A_i}^c = \emptyset$$

Which proves your $\Longrightarrow$ implication.

Same steps worked backwards prove the $\Longleftarrow$ implication.