Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant

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Theorem

Let $\Z_{>0}$ be the set of (strictly) positive integers.

Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.


Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.

Then:

$\exists m, n \in \Z_{>0}: \forall r > n: x_r = m$

That is, $\sequence {x_n}$ is eventually constant.


Proof

Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$.

By definition:

$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$

Let $\epsilon < 1$, say: $\epsilon = \dfrac 1 2$.

By the definition of $d$:

$\forall m, n \in \N: x_m \ne x_n \implies \map d {x_m, x_n} \ge 1$

So the only possible way for:

$\forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$

is for $x_m = x_n$.

Hence the result.

$\blacksquare$


Sources

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