Nonempty Compact Sets are Tightly Bounded

Last updated Nov 6, 2022

Let $(Y, \leq)$ be a Total Ordering with the Order Topology. Let $K \subset Y$ be Nonempty and Compact. Then $K$ is a Tightly Bounded Set.

Consider $K$ as an Order-Preserving Net $(a_{a}){a \in K} \subset K$. Since $K$ is Compact and A Set is Compact iff all Nets have a Convergent Subnet, we know there exists a Cofinal Subset $S \subset K$ so that $(b{b}){b \in S}$ is a convergent Subnet. It is order-preserving because its parent is. Because Order-Preserving Nets Converge to their Supremum, we know $b{b} \to \sup\limits_{b \in S} b_{b} = \sup\limits S$. Because Supremum of a Cofinal Subset is the Supremum of the Set, it follows that $\sup\limits S = \sup\limits K$. Therefore $K$ has an Upper Bound (namely $\sup\limits K$). Since the Subnet converges in $K$, we know $\sup\limits K \in K$. Because the Reverse Ordering generates same Topology, we can apply the same argument on the Reverse Ordering to get $\inf\limits K \in K$. Therefore $K$ is a Tightly Bounded Set. $\blacksquare$

TODO - the corresponding statement for Metric Spaces

• Empty Set
• Supremum
• Infimum