How do you prove that $\mathbb{Z}$ (the set of integers) equipped with the Euclidean metric (induced from real numbers) is complete? I am having trouble with this question, I don't really know where to start!
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3Apply the definition? – Mark Bennet Oct 24 '14 at 15:56
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1Note I wrote "apply the definition" because I don't know what definition is being used. The answers so far assume "every Cauchy sequence converges", but also possible, for example is "every set which is bounded above has a least upper bound" – Mark Bennet Oct 24 '14 at 16:03
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Directly: try and show a Cauchy sequence in $\mathbb{Z}$ is eventually constant.
A little more theoretical: If you know $\mathbb{R}$ is complete, and $\mathbb{Z}$ is closed in $\mathbb{R}$, try and show that closed subsets of complete spaces are complete.
jxnh
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Complete means if we start with a cauchy sequence, it actually converges. So, let $x_n$ be a cauchy sequence in $\mathbb Z$. Then, for for any $\epsilon >0$, $\exists N\in \mathbb N$,$\forall n,m\ge N,|x_n -x_m |<\epsilon$. So, lets pick an $\epsilon$. Say. $\frac 1 {10}$. Then after a certain point, $|x_n-x_m|<\frac 1 {10}$. So we have $-\frac 1 {10}<x_n - x_m <\frac 1 {10}$, hence
$x_m - \frac 1 {10}<x_n<x_m+\frac 1 {10}$. But integers are at least 1 apart, so the only way for this to be true is for $x_n=x_m$. Thus, $\forall n\ge N$,$x_n =x_N$
So your sequence is eventually constant, and thus converges to that constant.
Alan
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Let $(x_n)$ be a Cauchy sequence. Then there exists $k$ such that, for all $m,n\ge k$, $$ |x_m-x_n|<\frac{1}{2} $$ which means $x_m=x_n$. Then…
egreg
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