Proof for \(\displaystyle (S \cup T)^c = S^c \cap T^c \)

Let \(\displaystyle x\in (S\cup T)^c \implies x \notin S \land x\notin T \implies x\notin S \cup T\)

Since

\(\displaystyle x\notin S\implies x\in S^c \)

\(\displaystyle x\notin T \implies x\in T^c\)

So \(\displaystyle \{ x: (x\in S^c) \land (x\in T^c)\} = S^c \cap T^c\)

\(\displaystyle x\in S^c \cap T^c \implies (S \cup T)^c \subseteq S^c \cap T^c \)

Let \(\displaystyle y \in S^c \cap T^c \implies y \in S^c \land y\in T^c \)

\(\displaystyle \implies y \notin S \cup T \implies y \in (S \cup T)^c \)

\(\displaystyle \implies S^c \cap T^c \subseteq (S \cup T)^c\)

Therefore \(\displaystyle (S \cup T)^c = S^c \cap T^c \quad \square\)

\(\displaystyle x\in S^c \cap T^c \implies (S \cup T)^c \subseteq S^c \cap T^c \)

Definition of subset: For two sets \(\displaystyle S \) and \(\displaystyle T \) we say that \(\displaystyle S \) is a subset of \(\displaystyle T \) if each element of \(\displaystyle S \) is also an element of \(\displaystyle T \).

In formal notation \(\displaystyle S \subseteq T \) \(\displaystyle \implies \forall x (x\in S \implies x \in T )\).

I know that every set is a subset of itself. So this holds

\(\displaystyle (S \cup T)^c \subseteq (S \cup T)^c\) and \(\displaystyle S^c \cap T^c \subseteq S^c \cap T^c\).

I also believe that I understand these rules:

\(\displaystyle S \cap T = \{ x: (x \in S) \land (x\in T)\}\)

\(\displaystyle S \cup T = \{ x: (x \in S) \vee (x\in T)\}\)

\(\displaystyle S^c = \{ x: (x \in u) \land (x\notin S) \} \)

Let \(\displaystyle x\in (S\cup T)^c \implies x \notin S \land x\notin T \implies x\notin S \cup T\)

Since

\(\displaystyle x\notin S\implies x\in S^c \)

\(\displaystyle x\notin T \implies x\in T^c\)

So \(\displaystyle \{ x: (x\in S^c) \land (x\in T^c)\} = S^c \cap T^c\)

\(\displaystyle x\in S^c \cap T^c \implies (S \cup T)^c \subseteq S^c \cap T^c \)

Let \(\displaystyle y \in S^c \cap T^c \implies y \in S^c \land y\in T^c \)

\(\displaystyle \implies y \notin S \cup T \implies y \in (S \cup T)^c \)

\(\displaystyle \implies S^c \cap T^c \subseteq (S \cup T)^c\)

Therefore \(\displaystyle (S \cup T)^c = S^c \cap T^c \quad \square\)

**My question is this part of the proof:**\(\displaystyle x\in S^c \cap T^c \implies (S \cup T)^c \subseteq S^c \cap T^c \)

Definition of subset: For two sets \(\displaystyle S \) and \(\displaystyle T \) we say that \(\displaystyle S \) is a subset of \(\displaystyle T \) if each element of \(\displaystyle S \) is also an element of \(\displaystyle T \).

In formal notation \(\displaystyle S \subseteq T \) \(\displaystyle \implies \forall x (x\in S \implies x \in T )\).

I know that every set is a subset of itself. So this holds

\(\displaystyle (S \cup T)^c \subseteq (S \cup T)^c\) and \(\displaystyle S^c \cap T^c \subseteq S^c \cap T^c\).

I also believe that I understand these rules:

\(\displaystyle S \cap T = \{ x: (x \in S) \land (x\in T)\}\)

\(\displaystyle S \cup T = \{ x: (x \in S) \vee (x\in T)\}\)

\(\displaystyle S^c = \{ x: (x \in u) \land (x\notin S) \} \)

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