Let {xn} be a Cauchy sequence such that every term xn is an integer. Show that {xn} is ""eventually constant"" – i. e. there exist and N > 0 such that xn = xm for all n > m ≥ N

Given epsilon equal to 1/2, then should be n0 such that, for any values n and m greater than or equal to n0, we have that |an-am| < 1/2

Since the difference between 2 integers is an integer, and the module may not be negative, we have no other choice than |an-am| = 0. Thus, an = am. This means that for any k bigger than n0, we have that ak = an0, as a result, the sequence is eventually constant.

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step-by-step explanation:

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