Let $X$ be a set with subsets $A_i \subset X$ for $i\in I$ where $I$ is an arbitrary index-set. In other words, $I$ can be finite, countable, or uncountable. We first show that \begin{eqnarray*} \displaystyle \big( \cup_{i\in I} A_i \big)' &=& \cap_{i\in I} A_i', \end{eqnarray*}where $A'$ denotes the complement of $A$

Let us define $S=\big( \cup_{i\in I} A_i \big)'$ and $T=\cap_{i\in I} A_i'$ To establish the equality $S=T$ we shall use a standard argument for proving equalities in set theory. Namely, we show that $S\subset T$ and $T\subset S$ For the first claim, suppose $x$ is an element in $S$ Then $x\notin \cup_{i\in I} A_i$ so $x\notin A_i$ for any $i\in I$ Hence $x\in A_i'$ for all $i\in I$ and $x\in \cap_{i\in I} A_i'=T$ Conversely, suppose $x$ is an element in $T=\cap_{i\in I} A_i'$ Then $x\in A_i'$ for all $i\in I$ Hence $x\notin A_i$ for any $i\in I$ so $x\notin \cup_{i\in I} A_i$ and $x\in S$

The second claim, \begin{eqnarray*} \big( \cap_{i\in I} A_i \big)' &=& \cup_{i\in I} A_i', \end{eqnarray*}follows by applying the first claim to the sets $A_i'$