De Morgan’s laws

Textbook

All sets $A_i$ lie within a fixed universe $U$, with indices in a set $I$. Every complement is relative to $U$. De Morgan’s laws state:

$$U\setminus\bigcup_{i\in I}A_i=\bigcap_{i\in I}(U\setminus A_i),\qquad U\setminus\bigcap_{i\in I}A_i=\bigcup_{i\in I}(U\setminus A_i).$$

For the first identity, an arbitrary element is $x\in U$. Being outside the union means there is no index $i$ for which $x\in A_i$. Thus for every $i$, it follows that $x\notin A_i$: the element belongs to every complement. This is precisely membership in the intersection on the right. Every step is reversible, proving both inclusions.

For the second identity, $x\notin\bigcap_{i\in I}A_i$ means membership in $A_i$ fails for at least one index. Then $x\in U\setminus A_i$ for at least one index, exactly the condition for the union on the right. Again, the conditions are equivalent.

For two sets, the identities read $(A\cup B)^c=A^c\cap B^c$ and $(A\cap B)^c=A^c\cup B^c$. For an empty family, the convention is an empty union of $\varnothing$ and a relative empty intersection of $U$. Both laws still hold.

Related concepts: Set complement, Set union, Set intersection.

Every complement in an identity must use the same universe. Changing $U$ changes which elements count as external and invalidates a comparison with the previous expressions. For two sets, four membership cases exhaust the possibilities: in both, only the first, only the second, or neither. Within each case, the two sides of either identity give the same decision about whether to include the element.