A Nonempty Set is Compact in the Order Topology iff it is Tightly Bounded and Complete
# Statement
Let $(X, \leq)$ be a Total Ordering endowed with the Order Topology. Then Nonempty $K \subset X$ is Compact If and Only If $K$ is a Tightly Bounded Set and $(K, \leq {\big|}_{K^{2}})$ is a Complete Ordering.
# Proof
$(\Rightarrow)$: Suppose Nonempty $K \subset X$ is Compact. Recall that Nonempty Compact Sets are Tightly Bounded, so $K$ is a Tightly Bounded Set. Since The Order Topology on a Subset is the Subspace Topology, we know $K$ equipped with the Subspace Topology is $(K, \leq {\big|}{K^{2}})$ equipped the Order Topology. Therefore, the latter is Compact. Because Compact Spaces are Complete, we have that $(K, \leq {\big|}{K^{2}})$ is a Complete Ordering $\checkmark$.
$(\Leftarrow)$: Let $K \subset X$ be a Tightly Bounded Set so that $(K, \leq {\big|}{K^{2}})$ is a Complete Ordering and let ${U_i}{i \in I}$ be an Open Cover for $K$ with Index Set $I$. We know $\sup\limits K, \inf\limits K \in K$. Denote $a:= \inf\limits K$ and $b := \sup\limits K$. Let $$S := {x \in K : [a,x] \cap K \text{ has a finite open subcover}}.$$ $S$ is Nonempty since there exists $i_{0} \in I$ for which $a \in U_{i_{0}}$. Therefore, ${a} = [a,a] \cap K$ has a finite Open Subcover and $a \in S$. $S$ has $b \in K$ as an Upper Bound because $S \subset K = K \cap [a,b]$. Therefore, $\sup\limits S \in K$ exists. We claim two things
- $\sup\limits S \in S$: Since $\sup\limits S \in K$, there is some $j_{0}$ so that $\sup\limits S \in U_{j_{0}}$. By construction of the Order Topology, there is some Open Interval (or Open Ray or $X$) that is contained in $U_{j_{0}}$ and contains $\sup\limits S$. If it was $(\leftarrow, y)$ for some $y \in X$ (or $y = \rightarrow$), then $a \in (\leftarrow, y)$ (since $\sup\limits S \in (\leftarrow, y)$ and $\sup\limits S \geq a$). But this means ${U_{j_{0}}}$ is an Open Cover of $[a, \sup\limits S] \cap K$, so $\sup\limits S \in S$. Otherwise if the Open Interval in $U_{j_{0}}$ is $(x,y) \ni \sup\limits S$ for some $x \in X$ and $y \in X$ or $y = \rightarrow$, then because $x$ is not an Upper Bound of $S$ ($x < \sup\limits S$), there exists some $z \in S$ so that $x < z \leq \sup\limits S$. Combining $U_{j_{0}}$ with the finite open subcover for $[a,z] \cap K$, we get a finite open subcover for $[a, \sup\limits S]$. Therefore $\sup\limits S \in S$ $\checkmark$.
- $\sup\limits S = b$: Suppose $\sup\limits S < b$. Let ${U_{i_{1}}, \dots, U_{i_{n}}}$ be a finite Open Subcover of $[a, \sup\limits S]$, with $n \in \mathbb{N}$. Then Without Loss of Generality, let $\sup\limits S \in U_{i_{1}}$. Then there exists some $(x,y) \subset U_{i_{1}}$ with $x \in X \cup {\leftarrow}$ and $y \in X \cup {\rightarrow}$ so that $\sup\limits S \in (x,y)$. We must have $y \leq b$, otherwise $b \in (x,y)$ and ${U_{i_{1}}, \dots, U_{i_{n}}}$ is a finite Open Subcover of $[a,b] \cap K$. Thus $y \in [a,b] \cap K$. Therefore there exists $U_{i_{n+1}}$ so that $y \in U_{i_{n+1}}$. But then ${U_{i_{1}}, \dots, U_{i_{n+1}}}$ forms a finite Open Subcover of $[a, y] \cap K$ and $y \in S$. But $y > \sup\limits S$ $\unicode{x21af}$.
These two points tell us that $b = \sup\limits S \in S$ and $[a,b] \cap K = K$ admits a finite Open Subcover. Since our Open Cover was arbitrary, $K$ is Compact $\checkmark$. $\blacksquare$
# Proof 2
($\Rightarrow$) See Proof (1).
$(\Leftarrow)$: TODO I want to use Every Net on a Total Ordering has a Monotone Subnet, but I’m not sure if it is true until I complete it.
- Every Net on a Total Ordering has a Monotone Subnet.
- Construct an order-preserving Subnet, and show it converges.