Maximal Element need not be Greatest Element
Theorem
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $M \in S$ be a maximal element of $S$.
Then $M$ is not necessarily the greatest element of $S$.
Proof
Let $S = \set {a, b, c}$.
Let $\preccurlyeq$ be defined as:
- $x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$
A straightforward but laborious process determines that $\preccurlyeq$ is a partial ordering on $S$.
We have that:
- $c \preccurlyeq x \implies c = x$
and:
- $b \preccurlyeq x \implies b = x$
and so by definition, both $b$ and $c$ are maximal elements of $S$.
Suppose $b$ is the greatest element of $S$.
Then from Greatest Element is Unique it follows that $c$ can not be the greatest element of $S$.
Hence the result.
In fact, from the definition of the greatest element of $S$:
- $x$ is the greatest element of $S$ if and only if $\forall y \in S: y \preccurlyeq x$
it can be seen directly that neither $b$ nor $c$ is the greatest element of $S$.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
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- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles