# Finite Suprema Set and Lower Closure is Smallest Ideal

## Theorem

Let $L = \left({S, \vee, \preceq}\right)$ be a join semilattice.

Let $X$ be a subset of $S$.

Then $X \subseteq \operatorname{finsups}\left({X}\right)^\preceq$ and

for every ideal $I$ in $L$: $X \subseteq I \implies \operatorname{finsups}\left({X}\right)^\preceq \subseteq I$

where

$\operatorname{finsups}\left({X}\right)$ denotes the finite suprema set of $X$,
$X^\preceq$ denotes the lower closure of $X$.

## Proof

$X \subseteq \operatorname{finsups}\left({X}\right)$
$X^\preceq \subseteq \operatorname{finsups}\left({X}\right)^\preceq$
$X \subseteq X^\preceq$

Thus by Subset Relation is Transitive:

$X \subseteq \operatorname{finsups}\left({X}\right)^\preceq$

Let $I$ be a ideal in $L$ such that

$X \subseteq I$

Let $x \in {\operatorname{finsups}\left({X}\right)}^\preceq$

By definition of lower closure of subset:

$\exists y \in \operatorname{finsups}\left({X}\right): x \preceq y$

By definition of finite suprema set:

$\exists A \in \mathit{Fin}\left({X}\right): y = \sup A \land A$ admits a supremum

where $\mathit{Fin}\left({X}\right)$ denotes the set of all finite subsets of $S$.

$A \subseteq I$
$A \ne \varnothing \implies \sup A \in I$
$A = \varnothing \implies \sup A = \bot \in I$

where $\bot$ denotes the smallest element of $S$.

So

$y \in I$

Thus by definition of lower set:

$x \in I$

$\blacksquare$