# Meet is Intersection in Set of Ideals

## Theorem

Let $\mathscr S = \struct {S, \wedge, \preceq}$ be a meet semilattice.

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

Let $P = \struct {\mathit{Ids}\left({\mathscr S}\right), \precsim}$ be an ordered set where $\mathord\precsim = \mathord\subseteq\restriction_{\mathit{Ids}\left({\mathscr S}\right) \times \mathit{Ids}\left({\mathscr S}\right)}$

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

Then

$I_1 \wedge_P I_2 = I_1 \cap I_2$

## Proof

$I_1 \cap I_2 \in \mathit{Ids}\left({\mathscr S}\right)$

We will prove that

$I_1 \cap I_2$ is lower bound for $\set {I_1, I_2}$

Let $x \in \set {I_1, I_2}$.

Then by definition of unordered tuple:

$x = I_1$ or $x = I_2$
$I_1 \cap I_2 \subseteq x$

Thus by definition of $\precsim$:

$I_1 \cap I_2 \precsim x$

$\Box$

We will prove that

$\forall I \in \mathit{Ids}\left({\mathscr S}\right): I$ is lower bound for $\set {I_1, I_2} \implies I \precsim I_1 \cap I_2$

Let $I \in \mathit{Ids}\left({\mathscr S}\right)$ such that

$I$ is lower bound for $\set {I_1, I_2}$

By definition of lower bound:

$I \precsim I_1$ and $I \precsim I_2$

By definition of $\precsim$:

$I \subseteq I_1$ and $I \subseteq I_2$
$I \subseteq I_1 \cap I_2$

Thus by definition of $\precsim$

$I \precsim I_1 \cap I_2$

$\Box$

By definition of infimum:

$\inf \set {I_1, I_2} = I_1 \cap I_2$

Hence by definition of meet:

$I_1 \wedge I_2 = I_1 \cap I_2$

$\blacksquare$