Linearly Ordered Space is Connected iff Linear Continuum

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Theorem

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.


Then $S$ is a connected space if and only if it is a linear continuum.


Proof


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Necessary Condition

Suppose that $T$ is disconnected and a linear continuum.

Then there are non-empty open sets $U, V$ that separate $T$.

Let $a \in U$ and $b \in V$.

Without loss of generality, suppose that $a \prec b$.

Let $B = \set {p \in S: \closedint a p \subseteq U}$

$B$ contains $a$, so it is non-empty, and it is bounded above by $b$.

In fact, if $y$ is any point not in $U$, it must be an upper bound for $B$:

Because, if it were not bounded above by $y$, there would be some $p \in B$, such that $y < p$.

But $p \in B$ means by definition that $\closedint a p \subseteq U$.

Now, $a < y < p$, so $y \in \closedint a p$ and so $y \in U$, which was not the case by assumption.

So any $y \notin U$ is an upper bound for $B$, and so in particular $b$ is.

So $B$ has a supremum $m$, as a bounded above, non-empty set has a supremum.

This is part of the definition of linear continuum, namely being Dedekind complete.


First suppose that $m \in B$.

Then $m \in U$.

Since $U$ is open, there exists a $p \in S$ such that $\hointr m p \subseteq U$.

Since $S$ is densely ordered, it has an element $q$ strictly between $m$ and $p$.

Then $m \in B$ implies that:

$\closedint a m \subseteq U$

We also have that:

$\closedint m q \subseteq \hointr m p \subseteq U$

so:

$\closedint a q \subseteq U$

Then $q \in B$, contradicting the fact that $m$ is an upper bound for $B$.

Thus $m \notin B$.


So $\closedint a m \nsubseteq U$ (or else $m \in B$ by definition).

So there is some element $w \in S$ that is in $\closedint a m$ and that is not in $U$.

Thus $w$ lies in $V$, as these two sets cover $S$.

As already proved above, $w \notin U$ implies it is an upper bound for $B$.

But then $m$ is the minimal one upper bound by the definition of supremum, so $w < m$ is impossible.

We can only have $w = m$.

So $m \notin U$, so by necessity, $m \in V$.

Then as $V$ is open, we have some $w' < m$ such that $\hointl {w'} m \subseteq V$.

Again, by $S$ being densely ordered, we have some $w$ with $w' < w < m$.

This $w$ is in $V$, so not in $U$

So $w$ is an upper bound for $B$ again, and strictly smaller than $m$.

This is a contradiction.

This shows that the separation cannot exist.

$\Box$


Sufficient Condition

Suppose that $T$ is not a linear continuum.

Then either $T$ is not densely ordered or $T$ is not Dedekind complete.

Suppose first that $T$ is not densely ordered.

Then there are points $a, b \in S$ such that $a \prec b$ and no point lies strictly between $a$ and $b$.

Thus:

$X = a^\preceq \cup b^\succeq$

and the components of this union are disjoint.


By Lower Closure is Strict Lower Closure of Immediate Successor:

$a^\preceq = b^\prec$

where:

$a^\preceq$ denotes the lower closure of $a$
$b^\prec$ denotes the strict lower closure of $b$

By Upper Closure is Strict Upper Closure of Immediate Predecessor:

$b^\succeq = a^\succ$

where:

$b^\succeq$ denotes the upper closure of $b$
$a^\succ$ denotes the strict upper closure of $a$.


Thus these two sets are open sets that separate $T$.

Therefore $T$ is disconnected.

Suppose next that $T$ is not Dedekind complete.

Then there is a non-empty set $B \subset S$ which is bounded above in $S$ but has no supremum in $S$.

Let $U$ be the set of upper bounds of $B$ (non-empty by assumption).


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Since $B$ has no supremum, $U$ is open.

Let $A = S \setminus U$ be the set of points that are not upper bounds for $B$.

Let $p \in S \setminus U$.

Then there is an element of $s \in B$ such that $p \prec s$.

Then:

$p \in s^\prec \subset S \setminus U$

Thus $S \setminus U$ is also open.

It is non-empty because it contains all elements of $B$.

So $U$ and $S \setminus U$ are open sets separating $S$.

So $T$ is disconnected.

$\blacksquare$


Sources

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