Alternative characterization of complete metric space

Question

Let $(X,d)$ be a metric space. It is complete if every Cauchy sequence for $d$ on $X$ is convergent.

I've heard an alternative definition of completeness for $(X,d)$: it is complete iff the following condition holds:

Let $\{x_n\}\subset X$ be a sequence of points which leaves every compact set of $X$ (we say $\{x_n\}$ goes to infinity), then $\displaystyle{\lim_{n\rightarrow \infty}d(x,x_n)=\infty}$ for every point $x\in X$.

Is this alternative characterization right? Where can I find a proof of the equivalence of the two characterizations?

Thank you!

Answer 1

Consider $X = \mathbb R$ in the discrete metric. Then a set is compact if and only if it is finite. Now, take for example $x_n$ an enumeration of $\mathbb Q$. Then $x_n$ leaves eventually every finite subset of $\mathbb R$.

Also note that $d(x,x_n) \to 1$ for all $x \in \mathbb R$.

Answer 2

They are not alternative (First condition on metric space does not imply second)

Consider $(X,d)$, cone over $(S^1,d_0)$ where for $x\neq y\in S^1$, $d_0(x,y)=\infty$ That is as a set $X$ is a unit ball in ${\bf R}^2$. Here $$ d(re^{i\theta },r_2 e^{i\theta_2})=r+r_2 $$ for $\theta\neq \theta_2$ and $$ d(re^{i\theta },r_2 e^{i\theta_2})=|r-r_2| $$ for $\theta =\theta_2$

Note that $X$ is closed so that it is complete. But $\{x_n=e^{i\theta_n} \}$ where $\theta_n$ is distinct does not convergent subsequence since $d(x_n,x_m)=2$ for $n\neq m$ So it is not in any compact set. But $d(x,x_n)\leq 2$ for all $x\in X$