Meet in Inclusion Ordered Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $P = \struct {X, \subseteq}$ be an inclusion ordered set.

Let $A, B \in X$ such that

$A \cap B \in X$


Then $A \wedge B = A \cap B$


Proof

By Intersection is Subset:

$A \cap B \subseteq A$ and $A \cap B \subseteq B$

By definition:

$A \cap B$ is a lower bound for $\set {A, B}$

We will prove that

$\forall C \in X: C$ is a lower bound for $\set {A, B} \implies C \subseteq A \cap B$

Let $C \in X$ such that:

$C$ is a lower bound for $\set {A, B}$.

By definition of lower bound:

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

Thus by Intersection is Largest Subset:

$C \subseteq A \cap B$

$\Box$


By definition of infimum:

$\inf \set {A, B} = A \cap B$

Thus by definition of meet:

$A \wedge B = A \cap B$

$\blacksquare$


Sources