# Bounded sequence

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I have shown that a more conventional definition (that I encountered in my first year, using absolute value) is equivalent to the definition I've given here when on a normed space and there is a [ilmath]0[/ilmath] to speak of. However I've not found a reference for either.

## Definition

A sequence, [ilmath](a_n)_{n=1}^\infty[/ilmath] in a metric space, [ilmath](X,d)[/ilmath] is *bounded* if:

- [math]\exists B\in\mathbb{R}_{\ge 0}\ \forall n,m\in\mathbb{N}[d(a_n,a_m)<B][/math]

## Equivalent statements

- [math]\exists B\in\mathbb{R}_{\ge0}\forall n\in N[\Vert a_n\Vert<B][/math]
- Proof: (above [ilmath]\implies[/ilmath] this)
- Need to show that [ilmath]\Vert a_n-a_m\Vert < B\implies \Vert a_n\Vert < B' [/ilmath] for some [ilmath]B'[/ilmath]
- Note [ilmath]B>\Vert a_n-a_m\Vert \ge \Vert a_n \Vert - \Vert a_m \Vert[/ilmath] thus [ilmath]B + \Vert a_m\Vert >\Vert a_n\Vert[/ilmath] always
- That is WE ALWAYS HAVE [ilmath]B + \Vert a_m\Vert >\Vert a_n\Vert[/ilmath], define [ilmath]B':=B+\Vert a_1\Vert[/ilmath], then it follows that for all [ilmath]n[/ilmath], [ilmath]B'>\Vert a_n\Vert[/ilmath]

- Note [ilmath]B>\Vert a_n-a_m\Vert \ge \Vert a_n \Vert - \Vert a_m \Vert[/ilmath] thus [ilmath]B + \Vert a_m\Vert >\Vert a_n\Vert[/ilmath] always

- Need to show that [ilmath]\Vert a_n-a_m\Vert < B\implies \Vert a_n\Vert < B' [/ilmath] for some [ilmath]B'[/ilmath]
- Proof: (this [ilmath]\implies[/ilmath] above)
- Need to show [ilmath]\Vert a_n\Vert < B \implies d_{\Vert\cdot\Vert}(a_n,a_m)<B'[/ilmath] for some [ilmath]B'[/ilmath].
- Choose [ilmath]B'=2B[/ilmath] then [ilmath]d_{\Vert\cdot\Vert}(a_n,a_m):=\Vert a_n-a_m\Vert \le \Vert a_n\Vert + \Vert a_m\Vert < B + B = 2B[/ilmath]

- Need to show [ilmath]\Vert a_n\Vert < B \implies d_{\Vert\cdot\Vert}(a_n,a_m)<B'[/ilmath] for some [ilmath]B'[/ilmath].

- Proof: (above [ilmath]\implies[/ilmath] this)