Set Theory
A set is a collection of objects, considered as a whole.
The objects that make up a set are called its elements or its members.
If $X$ and $Y$ are sets such that every element of $X$ is also an element of $Y,$ then we say $X$ is a subset of $Y,$ denoted as $X \subset Y.$ Formally,
$$X \subset Y \iff (\forall x)(x \in X \implies x \in Y)$$
If $X$ and $Y$ are sets such that every element of $Y$ is also an element of $X,$ then we say $X$ is a superset of $Y,$ denoted as $X \supset Y.$ This is the same as $Y \subset X.$
Two sets $X$ and $Y$ are equal if and only if $X$ is a subset of $Y$ and $Y$ is a subset of $X.$
Suppose $X$ and $Y$ are sets with $X \subset Y$ and $Y \subset X.$ Now, suppose $x \in X.$ Then, $x \in Y.$ Conversely, suppose $y \in Y.$ Then $y \in X.$ Thus, $(\forall x)(x \in X \iff x \in Y),$ and $X = Y$.
$\square$If $f(n) = x_n,$ for $n \in J,$ it is customary to denote the sequence $f$ by the symbol $\{x_n\},$ or sometimes by $x_1, x_2, x_3, \dots.$
De Morgan's Laws
The complement of a union is equal to the intersection of complements.
Let $A$ and $B$ be sets. We want to show that
$$ (A \cup B)^c = A^c \cap B^c. $$
Suppose $x \in (A \cup B)^c.$ Then, if $x \in A$ or $x \in B,$ then $x \in A \cup B$ and $x \notin (A \cup B)^c,$ a contradiction. Therefore, $x \notin A$ and $x \notin B.$ That is, $x \in A^c$ and $x \in B^c,$ therefore $x \in A^c \cap B^c.$
$\square$The complement of an intersection is equal to the union of complements.
Let $A$ and $B$ be sets. We want to show that
$$ (A \cap B)^c = A^c \cup B^c. $$
Suppose $x \in (A \cap B)^c.$ Then, $x$ is not in $A \cap B,$ that is, $x$ is either not in $A$ or it is not in $B$ or it is in neither. If $x$ is not in $A,$ then it is in $A^c,$ and therefore it is in $A^c \cup B^c.$ The same approach works with $B,$ and therefore $x \in A^c \cup B^c,$ and we have shown $(A \cap B)^c = A^c \cup B^c.$
$\square$