Bounded Sets and Bounded Functions in a Metric Space

Definition: Let $(M,d)$ be a metric space. A subset $S\subseteq M$ is said to be Bounded if there exists a positive real number $r>0$ such that $S\subseteq B(x,r)$ for some $x\in M$. The set $S$ is said to be Unbounded if it is not bounded.

By the definition above, we see that $S$ is bounded if there exists some open ball with a finite radius that contains $S$.

For example, consider the metric space $(M,d)$ where $d$ is the discrete metric defined for all $x,y\in M$ by:

${\begin{aligned}\quad d(x,y)=\left\{{\begin{matrix}0&\mathrm {if} \;x=y\\1&\mathrm {if} \;x\neq y\end{matrix}}\right.\end{aligned}}$

Let $S\subseteq M$. Then by the definition of the discrete metric, for all $x,y\in S$ we have that $d(x,y)\leq 1$. Therefore, if we consider any point $x\in S$ and take $r=2$ then:

${\begin{aligned}\quad S\subseteq B(x,2)\end{aligned}}$

Therefore, $S$ is bounded. This shows that every subset of $M$ is bounded with respect to the discrete metric. In fact, the wholeset $M$ is also bounded and $M\subseteq B(x,2)$ for any $x\in M$.

For another example, consider the metric space $(\mathbb {R} ^{3},d)$ where $d$ is the Euclidean metric. Consider the following set:

${\begin{aligned}\quad S=\{(x,y,z)\in \mathbb {R} ^{3}:x,y,z\geq 0\}\subseteq \mathbb {R} ^{3}\end{aligned}}$

The set $S$ above is the first octant of $\mathbb {R} ^{3}$, and is actually unbounded. To prove this, suppose that instead $S$ is bounded. Then there exists a maximal distance between some pair of points $\mathbf {x} ,\mathbf {y} \in S$, say:

${\begin{aligned}\quad d_{\mathrm {max} }=d(\mathbf {x} ,\mathbf {y} )\end{aligned}}$

Then $S\subseteq B(x,d_{\mathrm {max} })$. For $k>1$ consider the point $k\mathbf {y} =(kx_{1},kx_{2},kx_{3})\in \mathbb {R} ^{3}$. Then:

${\begin{aligned}\quad d(\mathbf {x} ,k\mathbf {y} )={\sqrt {(x_{1}-ky_{1})^{2}+(x_{2}-ky_{2})^{2}+(x_{3}-ky_{3})^{2}}}\end{aligned}}$

Since $(x_{i}-ky_{1})^{2}\geq (x_{i}-y_{i})^{2}$ for each $i\in \{1,2,3\}$ we see that:

${\begin{aligned}\quad d(\mathbf {x} ,k\mathbf {y} )\geq d(\mathbf {x} ,\mathbf {y} )=d_{\mathrm {max} }\end{aligned}}$

But then $\mathbf {y} \not \in B(\mathbf {x} ,d_{\mathrm {max} })$ implies $\mathbf {y} \not \in S$ which is a contradiction. Therefore our assumption that $S$ was bounded is false.

Licensing

Content obtained and/or adapted from:

• Bounded Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license