2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.