Lattice is Complete iff it Admits All Suprema
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Then $\struct {S, \preceq}$ is a complete lattice if and only if
- $\forall X \subseteq S: X$ admits a supremum.
Proof
Sufficient Condition
Let $\struct {S, \preceq}$ be a complete lattice.
Thus by definition of complete lattice:
- $\forall X \subseteq S: X$ admits a supremum.
$\Box$
Necessary Condition
Let $\struct {S, \preceq}$ be such that
- $\forall X \subseteq S: X$ admits a supremum.
We will prove that
Let $X \subseteq S$.
Thus by assumption:
- $X$ admits a supremum.
Define:
- $Y := \leftset {s \in S: s}$ is a lower bound for $\rightset X$
By assumption:
- $Y$ admits a supremum
We will prove that
- $\sup Y$ is a lower bound for $X$
Let $x \in X$.
By definition of lower bound:
- $x$ is an upper bound for $Y$
By definition of supremum:
- $\sup Y \preceq x$
Thus by definition:
- $\sup Y$ is a lower bound for $X$
We will prove that
- $\forall s \in S: s$ is a lower bound for $X \implies s \preceq \sup Y$
Let $s \in S$ such that:
- $s$ is a lower bound for $X$.
By definition of $Y$:
- $s \in Y$
By definition of supremum:
- $\sup Y$ is an upper bound for $Y$.
Thus by definition of upper bound:
- $s \preceq \sup Y$
Thus by definition
- $X$ admits an infimum
By definition
- $\struct {S, \preceq}$ is a lattice.
Thus the result follows by definition of complete lattice.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article YELLOW_2:24