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 $b$.
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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.
$\blacksquare$
Remark
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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 $\struct {L, \preceq}$.
Sources
- 1955: Alfred Tarski: A lattice-theoretical fixpoint theorem and its applications (Pacific J. Math. Vol. 5, no. 2: pp. 285 – 309): $\S 1$