Finite Suprema Set and Lower Closure is Ideal

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Theorem

Let $P = \struct {S, \vee, \preceq}$ be a join semilattice.

Let $X$ be a non-empty subset of $S$.


Then

$\map {\operatorname{finsups} } X^\preceq$ is ideal in $P$.

where

$\map {\operatorname{finsups} } X$ denotes the finite suprema set of $X$,
$X^\preceq$ denotes the lower closure of $X$.


Proof

By Finite Suprema Set and Lower Closure is Smallest Ideal:

$X \subseteq \map {\operatorname{finsups} } X^\preceq$

By definition of non-empty set:

$\map {\operatorname{finsups} } X^\preceq$ is a non-empty set.

We will prove that

$\map {\operatorname{finsups} } X$ is directed.

Let $x, y \in \map {\operatorname{fininfs} } X$

By definition of finite suprema set:

$\exists A \in \map {\operatorname {Fin} } X: x = \sup A \land A$ admits a supremum

and

$\exists B \in \map {\operatorname {Fin} } X: y = \sup B \land B$ admits an supremum

where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$.

Define $C = A \cup B$.

By Union of Subsets is Subset:

$C \subseteq X$

By Union of Finite Sets is Finite:

$C$ is finite.

Then

$C \in \map {\operatorname {Fin} } X$

By Existence of Non-Empty Finite Suprema in Join Semilattice:

$C \ne \O \implies C$ admits a supremum.

By Union is Empty iff Sets are Empty:

$C = \O \implies A = \O$

So

$C$ admits a supremum.

By definition of finite suprema set:

$\sup C \in \map {\operatorname{finsups} } X$

By Set is Subset of Union:

$A \subseteq C$ and $B \subseteq C$

Thus by Supremum of Subset:

$x \preceq \sup C$ and $y \preceq \sup C$

Hence $\map {\operatorname{finsups} } X$ is directed.

$\Box$


By Directed iff Lower Closure Directed:

$\map {\operatorname{finsups} } X^\preceq$ is directed.

By Lower Closure is Lower Section:

$\map {\operatorname{finsups} } X^\preceq$ is lower.

Hence $\map {\operatorname{finsups}} X^\preceq$ is ideal in $P$.

$\blacksquare$


Sources