# Bounded set

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## Contents

## Definition

Let [ilmath](X,d)[/ilmath] be a metric space. Let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. Then we say "[ilmath]A[/ilmath] is bounded in [ilmath](X,d)[/ilmath]" if^{[1]}:

- [ilmath]\exists C<\infty\ \forall a,b\in A[d(a,b)<C][/ilmath] - where [ilmath]C[/ilmath] is real
^{[Note 1]}

If a set is not bounded it is "*unbounded*" (that link redirects to this line)

## Equivalent conditions

Let [ilmath](X,d)[/ilmath] be a metric space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. Then the following are all logical equivalent to each other^{[Note 2]}:

- [ilmath]\exists C<\infty\ \forall a,b\in A[d(a,b)<C][/ilmath] - [ilmath]A[/ilmath] is bounded (the definition)
- [ilmath]\forall x\in X\exists C<\infty\forall a\in A[d(a,x)<C][/ilmath]
^{[1]}

## See also

## Notes

- ↑ [ilmath]C\in\mathbb{R}_{\ge 0} [/ilmath] should do as [ilmath]0[/ilmath] could be a bound, I suppose on a one point set?
- ↑ Just in case the reader isn't sure what this means, if [ilmath]A[/ilmath] and [ilmath]B[/ilmath] are logically equivalent then:
- [ilmath]A\iff B[/ilmath]. In words "[ilmath]A[/ilmath]
*if and only if*[ilmath]B[/ilmath]"

- [ilmath]A\iff B[/ilmath]. In words "[ilmath]A[/ilmath]

## References

# OLD PAGE

## Notes

TODO: Surely this can be generalised to an arbitrary Metric space

## Of [ilmath]\mathbb{R}^n[/ilmath]

Given a set [ilmath]A\subseteq\mathbb{R}^n[/ilmath] we say [ilmath]A[/ilmath] is bounded^{[1]} if:

- [ilmath]\exists K\in\mathbb{R} [/ilmath] such that [ilmath]\forall x\in A[/ilmath] (where [ilmath]x=(x_1,\cdots,x_n)[/ilmath]) we have [ilmath]\vert x_i\vert\le K[/ilmath] for [ilmath]i\in\{1,\cdots,n\} [/ilmath]

## Immediate results

(Real line) [ilmath]A\subseteq[-K,K]\subset\mathbb{R} [/ilmath] (where [ilmath]K> 0[/ilmath] and [ilmath][-K,K][/ilmath] denotes a closed interval) *if and only if* [ilmath]A[/ilmath] is bounded.

This follows right from the definitions, the [ilmath]K[/ilmath] is the bound.

**Proof:** [ilmath]\implies[/ilmath]

- If such a [ilmath]K[/ilmath] exists then it is the bound. We are done.

**Proof:** [ilmath]\impliedby[/ilmath]

- If it is bounded then it is easy to see that given a bound [ilmath]K[/ilmath], [ilmath]A\subseteq[-K,K][/ilmath] (use the implies-subset relation)

*Note:*^{[1]} stipuates the [ilmath]A\subseteq[-K,K]\implies[/ilmath] bounded direction only.

(Real line) Every closed interval ([ilmath][a,b][/ilmath] for [ilmath]a,b\in\mathbb{R} [/ilmath] and [ilmath]a\le b[/ilmath]) is bounded.^{[1]}

Choose [ilmath]K=\text{Max}(\vert a\vert,\vert b\vert)[/ilmath] - result follows

## See also

## References

- ↑
^{1.0}^{1.1}^{1.2}Introduction to topology - Bert Mendelson - Third Edition