I need a detailed proof for the following: Suppose that xn is in Z (integers) fo

Question

I need a detailed proof for the following:

Suppose that xn is in Z (integers) for n in N (natural numbers). If xn is Caugchy, prove that xn is eventually constant; that is, that there exist numbers a in Z (integers) and N in N (second N represents natural numbers) such that xn=a for all n >= N.

PLEASE NOT THAT I DID NOT HAVE THE SIGN TO SHOW THAT SOMETHING LIVES IN INTEGERS, NATURAL NUMBERS ETC SO I JUST SAID IS IN.

ALSO PLEASE NOTE THAT SINCE I CAN NOT BOLD ITEMS THAT THE CAPITAL Z REPRESENTS INTEGERS, AS SHOWN IN PARENTHESIS, AND ALL OF THE N'S EXCEPT FOR THE FIRST ONE IN "N in N" REPRESENT NATURAL NUMBERS.

Explanation / Answer

Suppose that {xn} is Cauchy and each xn is an integer. Then for every eps > 0 there exists a natural number M such that whenever, i and j are both greater than M, we have

|xi - xj| < eps

(read "eps" as epsilon)

In particular, this holds when eps = (1/2). Let be the natural number such that for any i, j > N we have that

|xi - xj| < (1/2)

But since xi and xj are integers, the only way this can happen is if xi = xj. (Distinct integers have differences of at least 1)

Thus we have that for any i, j > N,

xi = xj = a for some integer a.

Thus whenever, i > N, we have xi = a. Thus the sequence {xn} is eventually constant.

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