Difference between revisions of "Bounded set"
(Created page with " ==Notes== {{Todo|Surely this can be generalised to an arbitrary Metric space}} ==Of {{M|\mathbb{R}^n}}== Given a set {{M|A\subseteq\mathbb{R}^n}} we say {{M|A}} is bound...") |
(Being refactoring, actually have a reference to metric space version instead! Going to create equivalent conditions) |
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| − | + | {{Stub page|grade=A|msg=Needs fleshing out}} | |
| + | {{Refactor notice|grade=A|msg=I've found a definition!}} | ||
| + | __TOC__ | ||
| + | ==Definition== | ||
| + | Let {{M|(X,d)}} be a [[metric space]]. Let {{M|A\in\mathcal{P}(X)}} be an arbitrary [[subset of]] {{M|X}}. Then we say "{{M|A}} is bounded in {{M|(X,d)}}" if{{rFAVIDMH}}: | ||
| + | * {{M|1=\exists C<\infty\ \forall a,b\in A[d(a,b)<C]}} - where {{M|C}} is real<ref group="Note">{{M|C\in\mathbb{R}_{\ge 0} }} should do as {{M|0}} could be a bound, I suppose on a one point set?</ref> | ||
| + | {{anchor|unbounded}}If a set is not bounded it is "''[[unbounded]]''" (that link redirects to this line) | ||
| + | ==[[Equivalent conditions to a set being bounded|Equivalent conditions]]== | ||
| + | {{:Equivalent conditions to a set being bounded/Statement}} | ||
| + | ==See also== | ||
| + | * [[Diameter (set)]] | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
| + | ==References== | ||
| + | <references/> | ||
| + | {{Definition|Metric Space|Functional Analysis|Analysis}} | ||
| + | =OLD PAGE= | ||
==Notes== | ==Notes== | ||
{{Todo|Surely this can be generalised to an arbitrary [[Metric space]]}} | {{Todo|Surely this can be generalised to an arbitrary [[Metric space]]}} | ||
Latest revision as of 23:31, 29 October 2016
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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]"
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
