Meet in Set of Ideals

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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 Meet is Intersection in Set of Ideals:

$I_1 \wedge I_2 = I_1 \cap I_2$

($\subseteq$)

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

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

$\Box$

($\supseteq$)

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:

$x \in I_1 \cap I_2$

$\blacksquare$


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