# Interval in Complete Lattice is Complete Lattice

## Theorem

Let $\left({L, \preceq}\right)$ be a complete lattice.

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

Let $\left[{a \,.\,.\, b}\right]$ be the closed interval between $a$ and $b$.

Then $\left[{a \,.\,.\, b}\right]$ is also a complete lattice under $\preceq$.

## Proof

Let $I = \left[{a \,.\,.\, b}\right]$.

Let $S \subseteq I$.

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

Let $S \ne \varnothing$.

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

Although $\left({\left[{a \,.\,.\, b}\right], \preceq}\right)$ is a complete lattice, it is only a complete sublattice of $\left({L, \preceq}\right)$ if $a = \inf L$ and $b = \sup L$. That is, if it *equals* $\left({L, \preceq}\right)$.

## Sources

- 1955: Alfred Tarski:
*A lattice-theoretical fixpoint theorem and its applications*(*Pacific J. Math.***Vol. 5**,*no. 2*: pp. 285 – 309): $\S 1$