Meet in Set of Ideals

Theorem
Let $\mathscr S = \left({S, \preceq}\right)$ be a meet semilattice.

Let ${\it Ids}\left({\mathscr S}\right)$ be the set of all ideals in $\mathscr S$.

Let $I_1, I_2$ be ideals in $\mathscr S$.

Then
 * $I_1 \wedge I_2 = \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$

where
 * $I_1 \wedge I_2$ denotes the meet in $\left({ {\it Ids}\left({\mathscr S}\right), \subseteq}\right)$

Proof
By Intersection of Ideals is Ideal:
 * $I_1 \wedge I_2 = I_1 \cap I_2$

($\subseteq$)
Let $x \in I_1 \cap I_2$.

By definition of intersection:
 * $x \in I_1 \land x \in I_2$

By Meet is Idempotent:
 * $x = x \wedge x$

Thus
 * $x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$

($\supseteq$)
Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$.

Then
 * $\exists i_1 \in I_1, i_2 \in I_2: x = i_1 \wedge i_2$

By Meet Precedes Operands:
 * $x \preceq i_1 \land x \preceq i_2$

By definition of lower set:
 * $x \in I_1 \land x \in I_2$

Thus by definition of intersection:
 * $x \in I_1 \cap I_2$