de Morgan’s laws for sets (proof)

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

${\left({\cup }_{i\in I}{A}_i\right)}^{\prime }$$=$${\cap }_{i\in I}{A}_i^{\prime },$

where $A^{\prime }$ denotes the complement of $A$.

Let us define $S={\left({\cup }_{i\in I}{A}_i\right)}^{\prime }$and $T={\cap }_{i\in I}{A}_i^{\prime }$. To establish the equality $S=T$, we shalluse 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 anelement in $S$.Then $x\notin {\cup }_{i\in I}{A}_i$, so $x\notin A_i$ for any $i\in I$.Hence $x\in A_i^{\prime }$ for all $i\in I$, and $x\in {\cap }_{i\in I}{A}_i^{\prime }=T$.Conversely, suppose $x$ is anelement in $T={\cap }_{i\in I}{A}_i^{\prime }$. Then $x\in A_i^{\prime }$ 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,

${\left({\cap }_{i\in I}{A}_i\right)}^{\prime }$$=$${\cup }_{i\in I}{A}_i^{\prime },$

follows by applying the first claim to the sets $A_i^{\prime }$.