Prove that the limit of a Cauchy sequence of integers is an integer

Question

I am trying to prove that the limit of a Cauchy sequence of integers must be an integer. I have attempted a proof which I shall write out below:

Let the convergent (as a Cauchy sequence must converge to a limit) sequence be defined as $(a_n)_{n\in N_+}, a_n\in Z$ $\forall n\in N_+$, now let us suppose that it does not converge to some $\ell \in Z$ for a contradiction.

By the definition of a number's ceiling and floor, we can say that: $$|x-\ell|=min{\{\lceil\ell\rceil-\ell, \ell-\lfloor\ell\rfloor\}} \forall x\in Z$$

If we set $\varepsilon :=\frac{1}{2}min{\{\lceil\ell\rceil-\ell, \ell-\lfloor\ell\rfloor\}}>0$ then by the definition of a limit $\exists N \in N_+$ s.t $n>N\implies |a_n-\ell|<\varepsilon$

But we are given that $a_n \in Z$ and so from the above we can say that $|a_n-\ell|=min{\{\lceil\ell\rceil-\ell, \ell-\lfloor\ell\rfloor\}}>\varepsilon$, so we have a contradicition and $\ell$ must be an integer.

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Could anyone tell me if this is correct/infallible? Any advice on making it more concise etc is welcome, thank you!

Answer 1

This is correct, and is a somewhat convoluted formalization of a one-sentence answer:

"Since the difference of two integers is never less than one, any Cauchy sequence of integers after a certain point (specifically after $\varepsilon$ gets smaller than $1$) must be constant."

This should make it clear that this fact holds in any discrete space, but holds in a somewhat trivial way.

Answer 2

Suppose $(a_n)$ is Cauchy and $a_n\in \Bbb Z$.

take $\epsilon=\frac{1}{4}$.

then $\exists N\in \Bbb N \;:\; $

$$\forall p>q>N \;\; |a_p-a_q|<\frac{1}{4}$$

but $|a_p-a_q| \in \Bbb N$, thus

$\forall p>q>N \; a_p=a_q=\lim_{n\to +\infty}a_n \in \Bbb Z$.