# Connected Subspace of Linearly Ordered Space

## Theorem

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

Let $Y \subseteq S$.

Then $Y$ is connected in $\struct {S, \tau}$ if and only if both of the following hold:

- $(1): \quad Y$ is convex in $S$
- $(2): \quad \struct {Y, \preceq \restriction_Y}$ is a linear continuum, where $\restriction$ denotes restriction.

## Proof

### Necessary Conditions

Let $Y$ be connected in $\struct {S, \tau}$.

Aiming for a contradiction, suppose $Y$ is not convex in $S$.

Then there exist $a, b, c \in S$ such that:

- $a \prec b \prec c$
- $a, c \in Y$ but $b \notin Y$

Recall that:

- $b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \set {u \in S: u \prec b}$
- $b^\succ$ denotes the (strict) upper closure of $b$: $b^\succ = \set {u \in S: b \prec u}$

We have that $Y \cap b^\prec$ and $Y \cap b^\succ$ are separated in $Y$.

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Hence is disconnected.

Hence by Proof by Contradiction it follows that $Y$ is convex in $S$.

$\Box$

Aiming for a contradiction, suppose $Y$ is convex in $S$, but not a linear continuum.

Then by Order Topology on Convex Subset is Subspace Topology, the subspace topology on $Y$ is the same as the order topology on $Y$.

Thus by Linearly Ordered Space is Connected iff Linear Continuum, $Y$ is disconnected.

Hence by Proof by Contradiction it follows that $Y$ is a linear continuum.

$\Box$

### Sufficient Condition

Suppose that $Y$ is convex in $S$ and a linear continuum.

Then the result follows from:

- Order Topology on Convex Subset is Subspace Topology
- Linearly Ordered Space is Connected iff Linear Continuum.

$\blacksquare$