Meet in Inclusion Ordered Set

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Theorem

Let $P = \left({X, \subseteq}\right)$ 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 lower bound for $\left\{ {A, B}\right\}$

We will prove that

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

Let $C \in X$ such that

$C$ is lower bound for $\left\{ {A, B}\right\}$.

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:

$\inf \left\{ {A, B}\right\} = A \cap B$

Thus by definition of meet:

$A \wedge B = A \cap B$

$\blacksquare$


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