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 $0$ to speak of. However I've not found a reference for either.

Definition

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

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

Equivalent statements

1. $$\exists B\in\mathbb{R}_{\ge0}\forall n\in N[\Vert a_n\Vert<B]$$

   Proof: (above $\implies$ this)

   • Need to show that $\Vert a_n-a_m\Vert < B\implies \Vert a_n\Vert < B'$ for some $B'$
     • Note $B>\Vert a_n-a_m\Vert \ge \Vert a_n \Vert - \Vert a_m \Vert$ thus $B + \Vert a_m\Vert >\Vert a_n\Vert$ always
       • That is WE ALWAYS HAVE $B + \Vert a_m\Vert >\Vert a_n\Vert$, define $B':=B+\Vert a_1\Vert$, then it follows that for all $n$, $B'>\Vert a_n\Vert$

   Proof: (this $\implies$ above)

   • Need to show $\Vert a_n\Vert < B \implies d_{\Vert\cdot\Vert}(a_n,a_m)<B'$ for some $B'$.
     • Choose $B'=2B$ then $d_{\Vert\cdot\Vert}(a_n,a_m):=\Vert a_n-a_m\Vert \le \Vert a_n\Vert + \Vert a_m\Vert < B + B = 2B$