Closed Interval in Complete Lattice is Complete Lattice

Theorem
Let $\struct {L, \preceq}$ be a complete lattice.

Let $a, b \in L$ with $a \preceq b$.

Let $I = \closedint a b$ be the closed interval between $a$ and $b$.

Then $\struct {I, \preceq}$ is also a complete lattice.

Proof
Let $S \subseteq I$.

If $S = \O$, then it has a supremum in $I$ of $a$ and an infimum in $I$ of $b$.

Let $S \ne \O$.

Since $S \subseteq I$, $a$ is a lower bound of $S$ and $b$ is an upper bound of $S$.

Since $L$ is a complete lattice, $S$ has an infimum, $p$, and a supremum, $q$, in $L$.

Thus by the definitions of infimum and supremum:
 * $a \preceq p$ and $q \preceq b$

Let $x \in S$.

Since an infimum is a lower bound:
 * $p \preceq x$

Since a supremum is an upper bound:
 * $x \preceq q$

Thus $a \preceq p \preceq x \preceq q \preceq b$.

Since $\preceq$ is an ordering, it is transitive, so by Transitive Chaining:


 * $a \preceq p \preceq b$ and $a \preceq q \preceq b$.

That is, $p, q \in I$.

Thus $p$ and $q$ are the infimum and supremum of $S$ in $I$.

As every subset of $I$ has a supremum and infimum in $I$, $I$ is a complete lattice.

Remark
Although $\struct {I, \preceq}$ is a complete lattice, it is only a complete sublattice of $\struct {L, \preceq}$ if $a = \inf L$ and $b = \sup L$. That is, if it equals $\struct {L, \preceq}$.