Interval of Totally Ordered Set is Convex
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Theorem
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $I \subseteq S$ be an interval in $S$.
Then $I$ is convex.
Proof
There are a number of cases to investigate.
Open Interval
Let $I = \openint a b$ be an open interval:
- $I = \set {x \in S: a \prec x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
- $a \prec s \prec x$
and:
- $x \prec t \prec b$
and so:
- $a \prec x \prec b$
and $x \in I$.
Thus we have:
- $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
$\Box$
Left Half-Open Interval
Let $I = \hointl a b$ be a left half-open interval:
- $I = \set {x \in S: a \prec x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
- $a \prec s \prec x$
and:
- $x \prec t \preceq b$
and so:
- $a \prec x \preceq b$
and $x \in I$.
Thus we have:
- $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
$\Box$
Right Half-Open Interval
Let $I = \hointr a b$ be a right half-open interval:
- $I = \set {x \in S: a \preceq x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
- $a \preceq s \prec x$
and:
- $x \prec t \prec b$
and so:
- $a \preceq x \prec b$
and $x \in I$.
Thus we have:
- $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
$\Box$
Closed Interval
Let $I = \closedint a b$ be a closed interval:
- $I = \set {x \in S: a \preceq x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
- $a \preceq s \prec x$
and:
- $x \prec t \preceq b$
and so:
- $a \preceq x \preceq b$
and $x \in I$.
Thus we have:
- $\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $1$