Q.Write the following in roster form.

A set is a well-defined, distinguishable collection of objects -- given any object, we must be able to decide for certain whether it belongs. "Beautiful flowers" is not well-defined (beauty is subjective); "red flowers in a named garden" is.

Subsets. $A\subseteq B$ means every element of $A$ lies in $B$. Mutual inclusion ($A\subseteq B$ and $B\subseteq A$) forces $A=B$. For any $A$: $\varnothing\subseteq A$ and $A\subseteq A$ are its trivial subsets (the second makes $A$ its own improper subset); $A\subsetneq B$ ("proper subset") additionally requires $A\ne B$, i.e. $B$ has at least one extra element. The number-system chain is $N\subset W\subset Z\subset Q\subset R$.

A set can even be an element of another set: if $A=\{1,2\}$ and $B=\{1,\{1,2\},3,4\}$, then $A\in B$, since the single object $\{1,2\}$ is literally listed as one of $B$'s four members. (This does not automatically make $A\subseteq B$ -- here it does not, since $2\notin B$ as an individual element.)

Union, intersection, complement, difference. For a fixed universal set $U$:

$A,B$ are disjoint if $A\cap B=\varnothing$. Indexed forms $\bigcup_{i=1}^nA_i,\ \bigcap_{i=1}^nA_i$ extend union/intersection to many sets at once.

Power set. $P(A)=\{B:B\subseteq A\}$, the set of all subsets of $A$; if $n(A)=n$, then $n(P(A))=2^n$ (each of the $n$ elements is independently in or out of a candidate subset).

Algebraic laws. Commutative, associative and distributive laws hold for $\cup,\cap$ exactly as $+,\times$ do for numbers; identity laws $A\cup\varnothing=A,\ A\cap U=A$; idempotent $A\cup A=A\cap A=A$; absorption $A\cup(A\cap B)=A\cap(A\cup B)=A$. De Morgan's laws $(A\cup B)'=A'\cap B'$ and $(A\cap B)'=A'\cup B'$ (and their set-difference forms) let a complement/difference of a compound set be rewritten in terms of the complements/differences of its pieces -- indispensable for simplifying set expressions.

Cardinality (inclusion-exclusion). $n(A\cup B)=n(A)+n(B)-n(A\cap B)$; for disjoint sets, $n(A\cup B)=n(A)+n(B)$; for three sets, $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$. This formula is the workhorse behind every "how many people study at least one of two/three subjects" word problem: total counts minus double-counted overlaps, plus back the triple-counted centre.

Finite vs. infinite, singleton. $n(X)=k$ (finite) if $X$ has exactly $k$ elements for some whole number $k$; otherwise $X$ is infinite. $n(A)=1$ makes $A$ a singleton; note $n(\varnothing)=0$ but $n(\{\varnothing\})=1$ -- a set containing the empty set as its one member is not itself empty.