Every convergent sequence $(x_n)$ is bounded in $(M,d)$

Question

I wanted to know if my approach is feasible, and if it is, if I am on the right path.

Suppose $(x_n)$ is unbounded and converges to $x$. From the definition of a limit in a metric space, I gather that $x_n \rightarrow x \implies d(x_n, x) \rightarrow 0$, and that $d(x_n, x) < \epsilon$, for $n \ge N$, where $\epsilon > 0$ and $N \in \mathbb{N}$. The definition of an unbounded sequence, if I remember correctly, is that $\forall k > 0 \ \ \exists \ \ x \in (x_n)$ s.t. $|x| > k$. So if we let $\epsilon = k$ then $|x_n - x| < k \implies x_n < k + x$, a contradiction to $(x_n)$ being unbounded. I would appreciate hints as to how to tackle the proof, rather than answers. Thank you.

Answer 1

Your intuition is correct to an extent, but recall that metric spaces are a very general concept. In particular, there is no such thing as an absolute value, or even a distinguished point (as is $0$ for the reals). So,

Definition. A subset $S$ of a metric space $(M,d)$ is bounded if there exists a real constant $K > 0$ such that $d(s,s') < K$ for any $s,s' \in S$.

In other terms, $S$ is bounded if its diameter, $\operatorname{diam}(M) := \sup\{d(s,s') : s,s' \in S\}$ is finite.

Definition. A sequence $(x_n)_{n \geq 1} \subset (M,d)$ is bounded if $\{x_n\}_{n \geq 1}$ is bounded in $M$.

That is, a sequence is bounded if its terms, as a subset of $M$, are bounded in the 'metric' sense given in the previous definition.

Now, extend your intuition to a general case, but adjusting your tools to this context. A possible (direct) solution follows.

Proposition. Let $(x_n)_{n \geq 1}$ be a convergent sequence on a metric space $(M,d)$. Then, the sequence $(x_n)_{n \geq 1}$ is bounded.

Proof. Since the sequences converges, writing $l := \lim_{n \to \infty}x_n$, there exists $n_0 \in \mathbb{N}$ such that $n \geq n_0$ implies $d(x_n,l) < 1$. Thus, if $K := \max\{d(x_i,l): 1 \leq i\leq n_0\}$,

$$ d(x_n,x_m) \leq d(x_n,l) + d(l,x_m) \leq 2(K+1) $$

because $d(x_k,l) \leq K+1$ for any $k$, by construction of $K$ and $n_0$. Thus, the sequence has finite finite diameter. $\square$