De Morgan's Laws
Lemma 1: \((\bigcup_n S_n)^c=\bigcap_n S_n^c\)
Proof for Lemma 1:
\[\because \forall x \in (\bigcup_n S_n)^c, x \notin \bigcup_n S_n, \therefore x \notin S_i, \forall i \in \{1,2,\dots,n\}. \therefore x \in S_i^c, \forall i \in \{1,2,\dots,n\}, x \in \bigcap_n S_n^c.\\ \therefore (\bigcup_n S_n)^c \subset \bigcap_n S_n^c .
\]
\[\because \forall x \in \bigcap_n S_n^c, x \in S_i^c, \forall i \in \{1,2,\dots n\}, x \notin S_i, \forall \{1,2,\dots,n\}, \therefore x \notin \bigcup_n S_n, \therefore x \in (\bigcup_n S_n)^c. \\ \therefore \bigcap_n S_n^c \subset (\bigcup_n S_n)^c.
\]
\[\therefore (\bigcup_n S_n)^c=\bigcap_n S_n^c.
\]
Lemma 2: \((\bigcap_n S_n)^c=\bigcup_n S_n^c\)
Proof for Lemma 2 :
\[\because \forall x \in (\bigcap_n S_n)^c, x \notin \bigcap_n S_n, \therefore \exists i \in \{1,2,\dots,n\}, x \in S_i^c, x \in \bigcup_n S_n^c. \therefore (\bigcap_n S_n)^c \subset \bigcup_n S_n^c .
\]
\[\because \forall x \in \bigcup_n S_n^c, \therefore \exists i \in\{1,2,\dots,n\}, x \notin S_i, x \notin \bigcap_n S_n, \therefore x \in (\bigcap_n S_n)^c.\therefore \bigcup_n S_n^c \subset (\bigcap_n S_n)^c.
\]
\[\therefore (\bigcap_n S_n)^c=\bigcup_n S_n^c.
\]
Summary:
The complement of the union of the sets equals the intersection of the sets' complement.
The complement of the intersection of the sets equals the union of the sets' complement.
