# About weakly Cauchy sequences in complete metric spaces

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?

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.