Finite Infima Set and Upper Closure is Filter
Theorem
Let $P = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $X$ be a non-empty subset of $S$.
Then
- $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
where
- $\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
- $X^\succeq$ denotes the upper closure of $X$.
Proof
By Finite Infima Set and Upper Closure is Smallest Filter:
- $X \subseteq \map {\operatorname {fininfs} } X^\succeq$
By definition of non-empty set:
- $\map {\operatorname {fininfs} } X^\succeq$ is a non-empty set.
We will prove that
- $\map {\operatorname {fininfs} } X$ is filtered.
Let $x, y \in \map {\operatorname {fininfs} } X$
By definition of finite infima set:
- $\exists A \in \map {\mathit {Fin} } X: x = \inf A \land A$ admits an infimum
and
- $\exists B \in \map {\mathit {Fin} } X: y = \inf B \land B$ admits an infimum
where $\map {\mathit {Fin} } X$ 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
- $C \in \map {\mathit {Fin} } X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
- $C \ne \O \implies C$ admits an infimum.
By Union is Empty iff Sets are Empty:
- $C = \O \implies A = \O$
So
- $C$ admits an infimum.
By definition of finite infima set:
- $\inf C \in \map {\operatorname {fininfs} } X$
- $A \subseteq C$ and $B \subseteq C$
Thus by Infimum of Subset:
- $\inf C \preceq x$ and $\inf C \preceq y$
Hence $\map {\operatorname {fininfs} } X$ is filtered.
$\Box$
By Filtered iff Upper Closure Filtered:
- $\map {\operatorname {fininfs} } X^\succeq$ is filtered.
By Upper Closure is Upper Section:
- $\map {\operatorname {fininfs} } X^\succeq$ is upper.
Hence $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
$\blacksquare$
Sources
- Mizar article WAYBEL_0:funcreg 11
- Mizar article WAYBEL_0:funcreg 16
- Mizar article WAYBEL_0:funcreg 22