Completely Irreducible Element equals Infimum of Subset implies Element Belongs to Subset

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Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.

Let $X \subseteq S$, $p \in S$ such that

$p$ is completely irreducible and $p = \inf X$

Then $p \in X$


Aiming for a contradiction, suppose

$p \notin X$

By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element:

$\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$

where $p^\succeq$ denotes the upper closure of $p$.

We will prove that

$q$ is lower bound for $X$

Let $x \in X$.

By definitions of infimum and lower bound

$p \preceq x$

By assumption:

$x \ne p$

By definition of strictly precede:

$p \prec x$

Thus by step condition:

$q \preceq x$


By definition of strictly precede:

$p \preceq q$

We will prove that

$\forall b \in S: b$ is lower bound for $X \implies b \preceq q$

Let $b \in S$ such that

$b$ is lower bound for $X$.

By definition of infimum:

$b \preceq p$

Thus by definition of transitivity:

$b \preceq q$


By definition of infimum:

$q = \inf X = p$

Thus this contradicts $p \ne q$ by definition of strictly precede.