## The classes of left, right, and two-sided ideals of a semigroup are closed under finite intersection and arbitrary union

Let $S$ be a semigroup, and let $L(S)$, $R(S)$, and $I(S)$ denote the sets of left- right- and two-sided ideals of $S$. Let $X \in \{L(S), R(S), I(S)\}$ and let $Y \subseteq X$ be nonempty. Show that $\bigcup Y \in X$ and that if $\bigcap Y \neq \emptyset$, then $\bigcap Y \in X$. Further, show that if $Y \subseteq I(S)$ is finite, then $\bigcap Y$ is not empty (and so is an ideal).

Suppose $Y \subseteq L(S)$ is a nonempty collection of left ideals of $S$. Certainly $\bigcup Y \neq \emptyset$. If $a \in \bigcup Y$, then we have $a \in T$ for some $T \in Y$. Now if $s \in S$, then $sa \in T \in \bigcup Y$. So $S(\bigcup Y) \subseteq \bigcup Y$, and thus $\bigcup Y$ is a left ideal of $S$. Now suppose $\bigcup Y \neq \emptyset$, and say $a \in \bigcap Y$. So $a \in T$ for all $T \in Y$. If $s \in S$, then $sa \in T$ for all $T \in Y$, and so $sa \in \bigcap Y$. Thus $\bigcap Y$ is a left ideal.

Likewise, the results hold for $R(S)$. Since every two-sided ideal is also a left and a right ideal, the results also follow for $I(S)$.

Now suppose $Y = \{T_i\}_{i=1}^n$ is a finite collection of ideals of $S$. Now $\prod T_i \subseteq T_k$ for each $k$, so that $\prod T_i \subseteq \bigcap Y$. Since each $T_i$ is nonempty, $\prod T_i$ is nonempty, and so $\bigcap Y$ is nonempty.