About weakly Cauchy sequences in complete metric spaces

Question

Let $(X,d)$ be a complete metric space. Call a sequence $(x_n)\subseteq X$ a weakly Cauchy sequence in $X$ if there is some $y\in X$ such that $(d(y,x_n))_n$ is a Cauchy sequence in $\mathbb{R}$. It is clear from the estimate $$|d(y,x_m)-d(y,x_n)|\leqslant d(x_m,x_n)$$ that a Cauchy sequence in $X$ is a weakly Cauchy sequence. It is also true that a Cauchy sequence $(x_n)$ is bounded i.e. there is $z\in X$ such that $\sup_nd(z,x_n)<+\infty$.

Is it true that a bounded and weakly Cauchy sequence in $X$ is a Cauchy sequence in $X$?

If not, under what additional conditions is a weakly Cauchy sequence in $X$ a Cauchy sequence?

Answer 1

The first statement is not correct. You can simply take $(X, d)=(\mathbb{R}, |\cdot|)$ (standard real line) and consider sequence $x_n=(-1)^n$. Then, for $y=0$ we have $d(y, x_n)=1$ for all $n\in\mathbb{N}$. However, the sequence $\{x_n\}_{n=1}^{\infty}$ is not the Cauchy sequence in metric space $(\mathbb{R}, |\cdot|)$ (because this metric space is complete and $\{x_n\}_{n=1}^{\infty}$ isn't convergent).

Note that $d(y, x_n)$ is a Cauchy sequence iff sequence $\{d(y, x_n)\}_{n=1}^{\infty}$ is convergent. That's why you need some unusual conditions for $(X, d)$ to make this statement true.