Finding an example of a bounded sequence in a complete metric space such that the sequence has no partial limit

Question

I'm working through an analysis text and I've come across this exercise:

Give an example of a complete metric space $X$ and a bounded sequence $\left(x_{n}\right)$ in $X$ such that the sequence $\left(x_{n}\right)$ has no partial limit.

The reason that I'm stuck with this one is because I'm shaky about what it means for a sequence to be bounded. Does this mean that for any positive integers $n$ and $m$, the set of all possible $|x_{n}-x_{m}|$ is a bounded set of real numbers?

If so, I'm not sure how to relate the boundedness of a sequence to the completeness of a metric space.

I know that a metric space $X$ is said to be complete if every Cauchy sequence in $X$ converges to a point in $X$... what does this have to do with bounded sequences?!? Thanks in advance for any help.

EDIT: I'm going to add a couple of definitions.

A sequence is said to be bounded if there exists a constant $C\gt 0$ such that for any positive integers $n$ and $m$ the inequality $|x_{n}-x_{m}|\lt C$ holds.

A point $x$ in a metric space $X$ is said to be a partial limit of a sequence $\left(x_{n}\right)$ if for every $\epsilon\gt 0$ there are infinitely many values of $n$ for which $x_{n}\in B\left(x,\epsilon\right)$.

Answer 1

Fix any point $y_0\in X$, and define a sequence $(x_n)$ to be bounded if there exists a constant $C>0$ such that $d(x_n,y_0)\leq C$ for all $n$. This definition does not depend on $y_0$, since if you choose any other $y_1\in X$, then by the triangle inequality, $$|d(x_n,y_0)-d(x_n,y_1)|\leq d(y_1,y_0),$$ where the right-hand-side is a constant, so the sequence $(x_n)$ is bounded "with respect to $y_0$" if and only if it is bounded "with respect to $y_1$".

Hint for the exercise: Consider a space with no limit points.

Answer 2

Let $X=L^{\infty}(\mathbb{R})$, let $f_n \in X$ and define

$$ f_n(x) = \begin{cases} 1 & x \in [n, n+1] \\ 0 & \text{else} \end{cases} $$ The sequence $(f_n)_{n=0}^{\infty}$ is bounded by 1, but has no convergent subsequence.