In a metric space $(X,d)$, for every Cauchy Sequence in $X$, and $z \in X$, the numeric succession $\{d(x_n;z)\}$ converges

Question

Good night,

I'm studying for a test and out teacher gives us a guide for it, but after trying i can't solve the last exercise, so any help would be appreciated.

Prove that, for every Cauchy sequence $\{x_n\}$ in a metric space $(X,d)$ and for every $z \in X$, the numeric succession of real numbers $\{d(x_n;z)\}$ converges.

Answer 1

Prove that $d(x_n,z)$ is a Cauchy sequence in $\mathbb R$. What do we know about Cauchy sequences in $\mathbb R$?

Full proof:

Let $\epsilon>0$. Since $x_n$ is a cauchy sequence there exists $N\in \mathbb N$ such that if $n,m>N$ we have $d(x_n,x_m)<\epsilon$. Now notice $d(x_m,x_n)+d(x_n,z) \geq d(x_m,z)$ from where $d(x_m,x_n)\geq d(x_n,z)-d(x_m,z)$. Analogously $d(x_n,x_m)+d(x_m,z)\geq d(x_n,z)$ and so $d(x_n,x_m)\geq d(x_n,z)-d(x_m,z)$. From here $\epsilon>d(x_n,x_m)\geq|(x_n,z)-(x_m,z)|$. Proving $\{d(x_n,z)\}$ is a Cauchy sequence, and every Cauchy Sequence in $\mathbb R$ converges.