This is taken from Universal Algebra Text book by Stan Burris.
I have a question regarding the last conclusion as to how does the author conclude that Sg is an algebraic closure operator. How do we conclude that there exists a finite $Y$. ?
Definition which I know of algebraic closure operator. A closure operator C on the set A is an algebraic closure operator if for every X $\subseteq$ A $\\$ C(X) = $\cup${$C(Y) : Y \subseteq X $ and $Y$ is finite}
We'll use induction by $n$. At first, notice that $E(\cdot)$ is isotone: $X \subseteq Y$ implies $E(X) \subseteq E(Y)$. Hence by induction $E^k(X) \subseteq E^k(Y)$ for all $k \in \omega$.
Since $X \subseteq X \cup Y$ and $Y \subseteq X \cup Y$ we have $E^k(X) \cup E^k(Y) \subseteq E^k(X \cup Y)$, for all $k \in \omega$.
Basis: $a \in E^{0}(X) = X \Rightarrow a \in E^0(\{a\})$ and $Y = \{a\}$.
Inductive step: assume that the statement is true for all $k < n$.
$a \in E^n(X) \Leftrightarrow a \in E(E^{n-1}(X)) = E^{n-1}(X) \cup \{f(a_1, \dots, a_t) : \text{$f$ is fundamental $t$-ary operation on $A$, $t \in \omega$, and $a_1, \dots, a_t \in E^{n-1}(X)$}\} = E^{n-1}(X) \cup B_n.$
So, either $a \in E^{n-1}(X)$ or $a \in B_n$. If $a \in E^{n-1}(X)$ we are done by induction hypothesis. If $a \in B_n$ then $a = f(a_1, \dots, a_t)$ for some fundamental $t$-ary operation on $A$ and $a_1, \dots, a_t \in E^{n-1}(X)$. It means that $a \in E(\{a_1, \dots, a_t\})$.
By induction hypothesis $a_i \in E^{n-1}(Y_i)$ for some finite $Y_i \subseteq X,\ i = \overline{1, t}$. Hence $a \in E(\{a_1, \dots, a_t\}) \subseteq E(\bigcup_{i = 1}^tE^{n-1}(Y_i)) \subseteq E(E^{n-1}(\bigcup_{i = 1}^tY_i)) = E^{n}(Y)$, where $Y = \bigcup_{i = 1}^tY_i$ is a finite subset of $X$.