Let $(a_n)_{n \in \Bbb N}$ be a Cauchy sequence .Show there exists a $N\in \Bbb N$ such that $a_n \in B(a_N; 1)$ for all $n\geq N$.

Question

Let $(a_n)_{n \in \Bbb N}$ be a Cauchy sequence in metric space V.Show there exists a $N\in \Bbb N$ such that $a_n \in B(a_N; 1)$ for all $n\geq N$.

It seems to me that this sort of follows from the definition, since you have that for all $n,m \geq N$ $d(a_n,a_m) < \epsilon$, so for every $n,m \geq N$ $a_n$ and $a_m$ get closer and closer to eachother as $N$ gets bigger.

Some hints on where to start are appreciated!

Answer 1

Yes it absolutely follows from the definition, take $\epsilon=1$, and apply the definition!