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.

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