Every integer valued Cauchy sequence is ultimately constant

Question

Question: Let $x(_n)_n$ be a Cauchy sequence and assume that $x_n \in ℕ$ for all $n \in ℕ$. Show that there is an integer $n_0$ such that $x_{n_0} = x_n$ for all $ n \ge n_0 $

proof. (I'm not sure how to start or where to go from here)

Since $(x_n)_n$ is Cauchy, it is bounded so the Bolzano-Weierstrass Theorem implies that there is a converging subsequence $x(_{m_k})_k$ and $(x_n)_n$ itself converges.

Answer 1

I'll let you complete the proof, but we're given two facts about the sequence:

• It's a Cauchy sequence, implying that we can find a certain point after which all elements are within an arbitrary distance of each other (that could be as small as you want, say of the order $10^{-10}$). Think what the sequence will be like after this point.

Every element is a natural number, and natural numbers occur in discrete gaps of $1$. Then the only way a set of natural numbers can be arbitrarily close to each other is if they are equal.