I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(\bar A,d)$ is complete. But are there spaces s.t. indeed for all $\varepsilon>0$, there is $N$ s.t. $d(x_n,x_{m})<\varepsilon$ for all $n\geq N$, and all $r\geq 0$, but the ball also move to $+\infty $ ?
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@JoséCarlosSantos: It's not duplicate because I'm not in a normed space but in a metric space. – user349449 Jul 15 '18 at 14:05
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It is a duplicate, since it is the same argument. – José Carlos Santos Jul 15 '18 at 14:07
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The definition of Cauchy sequence essentially states that the tail of the sequence is eventually contained in some ball, regardless of whether you define the ball thorugh metric or through norm. – Bihu Duo Dec 29 '20 at 19:33
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No, that cannot happen. Suppose that $(x_n)_{n \geq 0}$ is a Cauchy sequence in some metric space. By definition, there exists an integer $N \geq 0$ such that for all $n,m \geq N$ we have $d(x_n,x_m) \leq 1$. Thus, all the points $\{x_n ~|~ n \geq N\}$ are contained in the ball of radius 1 around $x_N$. By adding the finitely many points $x_1,\dots,x_{N-1}$, this sequence cannot become unbounded.
Florian R
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Note that $d(x_n,x_N)<\epsilon$ for all $n\ge N$ and that only finitely many $n<N$ are to be considered beyond that.
Hagen von Eitzen
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