# Finite Infima Set and Upper Closure is Filter

## Theorem

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

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

Then

${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filter in $P$.

where

$\operatorname{fininfs}\left({X}\right)$ denotes the finite infima set of $X$,
$X^\succeq$ denotes the upper closure of $X$.

## Proof

$X \subseteq {\operatorname{fininfs}\left({X}\right)}^\succeq$

By definition of non-empty set:

${\operatorname{fininfs}\left({X}\right)}^\succeq$ is a non-empty set.

We will prove that

$\operatorname{fininfs}\left({X}\right)$ is filtered.

Let $x, y \in \operatorname{fininfs}\left({X}\right)$

By definition of finite infima set:

$\exists A \in \mathit{Fin}\left({X}\right): x = \inf A \land A$ admits an infimum

and

$\exists B \in \mathit{Fin}\left({X}\right): y = \inf B \land B$ admits an infimum

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

Define $C = A \cup B$.

$C \subseteq X$
$C$ is finite.

Then

$C \in \mathit{Fin}\left({X}\right)$
$C \ne \varnothing \implies C$ admits an infimum.
$C = \varnothing \implies A = \varnothing$

So

$C$ admits an infimum.

By definition of finite infima set:

$\inf C \in \operatorname{fininfs}\left({X}\right)$
$A \subseteq C$ and $B \subseteq C$

Thus by Infimum of Subset:

$\inf C \preceq x$ and $\inf C \preceq y$

Hence $\operatorname{fininfs}\left({X}\right)$ is filtered.

$\Box$

${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filtered.
${\operatorname{fininfs}\left({X}\right)}^\succeq$ is upper.

Hence ${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filter in $P$.

$\blacksquare$