Bounded Set

Last updated Nov 1, 2022

Let $(M, d)$ be a Metric Space and let $A \subset M$. We say $A$ is a Bounded Set if $\sup\limits_{x,y \in A} d(x,y) < \infty$.

Let $(X, ||\cdot||)$ be a Normed Vector Space and let $A \subset X$. We say $A$ is a Bounded Set if $\sup\limits_{x \in A} ||x|| < \infty$.

1. Definition (1) $\Leftrightarrow$ Definition (2) in a Normed Vector Space: Suppose $\sup\limits_{x \in A} ||x|| < \infty$. Note $\sup\limits_{x,y \in A} ||x-y|| \leq \sup\limits_{x,y \in A} ||x|| + ||y|| \leq 2\sup\limits_{x \in A} ||x|| < \infty$. On the other hand, suppose $r := \sup\limits_{x,y \in A} ||x-y|| < \infty$. For any $x \in A$, $A \subset B_{r}(x)$. But then, $B_{r}(x) \subset (r + ||x||)B(X)$. Thus $\sup\limits_{y \in A} ||y|| < r + ||x|| < \infty$. $\blacksquare$

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

• Closed Interval