Simple Proof Methods

1. Theorem. For any set $A$ and a universe $U$ in which $A \subseteq U$, the following holds:$$A \cap U = A.$$ Proof. Let $A$ be any set and $U$ be a universe that contains $A$. To show that $A \cap U = A$, we need to show that $A \cap U \subseteq A$ and that $A \subseteq A \cap U$:
   • [Showing that $A \cap U \subseteq A$] Let $x$ be any element in the set $A \cap U$. This means that $x$ is in the intersection of $A$ and $U$. This means that $x \in A$ and that $x \in U$. Since, as we just said, $x \in A$, it means that $A \cap U \subseteq A$.
   • [Showing that $A \subseteq A \cap U$] Let $x$ be any element in the set $A$. Because $A \subseteq U$, this means that $x \in U$, too. Since $x \in A$ and $x \in U$, it means that $x$ is in the intersection of $A$ and $U$, so $x \in A \cap U$, which shows that $A \subseteq A \cap U$.
   Because we showed that $A \cap U \subseteq A$ and $A \subseteq A \cap U$, we conclude that $A \cap U = A$ by the definition of set equality.◼