Tightly Bounded Set

Last updated Nov 6, 2022

Let $(M, d)$ be a Metric Space and let $A \subset M$. We say $A$ is a Tightly Bounded Set if$\sup\limits_{x,y \in A} d(x,y) < \infty$ and there exist $x’,y’ \in A$ s.t. $d(x’,y’) = \sup\limits_{x,y \in A} d(x,y)$.

Let $(X, ||\cdot||)$ be a Normed Vector Space and let $A \subset X$. We say $A$ is a Tightly Bounded Set if $\sup\limits_{x,y \in A} ||x-y|| < \infty$ and there exists $x’,y’ \in A$ so that $||x’ - y’|| = \sup\limits_{x,y \in A}||x - y||$.

1. Definition (1) $\Leftrightarrow$ Definition (2) in a Normed Vector Space.

Let $(X, \leq)$ be a Total Ordering endowed with the Order Topology and let $A \subset X$. We say $A$ is a Tightly Bounded Set if there exists $s,t \in A$ so that $A \subset [s,t]$. That is, for some $s,t \in A$, $s \geq a \geq t$ for all $a \in A$.

1. Definitions (1) and (2) are carrying definition (3) over to sets that are not Total Orderings by a Continuous Function into $\mathbb{R}$ (namely $d$). The Lower Bound is always achieved in a Metric Space because $d(x,x) = 0$ for any $x \in A$.
2. If $A$ is a Tightly Bounded Set, this is equivalent $\sup\limits A \in A$ and $\inf\limits A \in A$. This is because $A$ contains is Upper Bound and its Lower Bound if it is Tightly Bounded Set.

• Closed Interval
• Supremum
• Infimum