Conditions to be an Algebra of subsets

Question

In my notes there are numerous equivalent conditions for a family of subsets $A$ of a set $X$ to be an algebra

In particular why is $X \in A$ and $ S \cup T , S \cap T \in A$ for every $S,T \in A$ not enough to guarantee $A$ being an algebra of subsets.

Answer 1

A simple example of non-algebra $(X,A)$ satisfying the required conditions is $X=\{0,1\}$ and $A=\{\{0\},\{0,1\}\}$.

Answer 2

An algebra on a set $\mathcal{S}$ needs to contain $\emptyset$ and $\mathcal{S}$, for the same reason a field $\Bbb{F}$ needs to have a $0$ and a $1$.

A set $\Bbb{F}$ might be closed under suitable operations of addition and multiplication, but it won't have all the properties we want fields to have, unless it also has an additive identity $0$ and a multiplicative identity $1$.

Similarly, it's not enough for our collection of subsets of $\mathcal{S}$ to be closed under union and intersection; to have all the properties we want "algebras of sets" to have, our algebra needs an identity for union and intersection. And $\emptyset$ is the identity for union, and $\mathcal{S}$ is the identity for intersection.