# Bounded set

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Needs fleshing out
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
The message provided is:
I've found a definition!

## 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]:

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

## Notes

1. [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?
2. 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]"

# 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