Bounded set
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
