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The sequence $\frac{1}{n}$ is convergent under euclidean metric. But not convergent with discrete metric.

Is there a non-constant convergent sequence with discrete metric ?

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  • $\begingroup$ It need not be constant right from the start. $\endgroup$ Oct 22 '15 at 15:53
  • $\begingroup$ you mean that there is $N$ such that for each $n>N$ , sequence coverges ? That is it has a constant tail after some $N$ . $\endgroup$ Oct 22 '15 at 15:56
  • $\begingroup$ Yes. In a discrete space, a convergent sequence is eventually constant. But of course it can behave arbitrarily at the beginning. $\endgroup$ Oct 22 '15 at 15:57
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No, in discrete metric, every point is isolated – every singleton $\{x\}$ is open, and hence every sequence converging to $x$ must have a final segment contained in $\{x\}$. So it is eventually constant.

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Same as other answer, except note that "eventually constant" is not the same as "constant." So it's easy to find non-constant examples.

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  • $\begingroup$ An example if possible ? $\endgroup$ Oct 22 '15 at 15:59
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    $\begingroup$ @AngeloMark Look at the sequence $0,1,1,1,\cdots$ meaning after the initial $0$ all terms are $1$. It converges to $1$ in discrete metric yet is not a constant sequence. $\endgroup$
    – coffeemath
    Oct 22 '15 at 16:12

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