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Not all sequences that are Cauchy are convergent. Here is what I think the example should be. Somehow the metric space is open but does not contain its limit points. Is this the right direction of thought?

d13
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  • What is the definition of a "convergent space" and a "Cauchy space"? Do you mean "sequence" instead of "space"? In that case, try constructing a space in which nothing is Cauchy? – akkkk Oct 28 '12 at 15:57
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    The terms Cauchy and convergent refer to sequences, not spaces. Are you asking for a metric space in which there are Cauchy sequences that do not converge? One familiar example is $\Bbb Q$. – Brian M. Scott Oct 28 '12 at 15:57
  • @akkkk: A constant sequence is Cauchy in any metric space. – Brian M. Scott Oct 28 '12 at 15:58
  • yes yes, sorry! i am asking for a space that has Cauchy sequences that does not converge, so to show that there should be a subset of that space which is open and therefore doesnt contain limit points, hence does not converge. – d13 Oct 28 '12 at 16:00
  • am i right in stating the above? – d13 Oct 28 '12 at 16:01
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    @d13: You should be able to edit your question and make this correction. – Nate Eldredge Oct 28 '12 at 16:04

3 Answers3

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Just take any sequence of rational numbers that converges to an irrational number. Then the sequence is Cauchy in $ \mathbb{Q} $, but does not converge in $ \mathbb{Q} $.

Haskell Curry
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6

There can be sequences that are Cauchy but do not converge; for example, the sequence $(1,\frac{1}{2},\frac{1}{3},\ldots)$ does not converge in the metric space $(0,2)$. A metric space in which every Cauchy sequence converges is said to be complete.

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In $\mathbb{K}[X]$.

$\forall P \in \mathbb{K}[X], P = \sum\limits_{i \in\mathbb{N}} p_i X^i, \|P\|=\max\limits_{i \in\mathbb{N}}( | p_i | )$

Let $P_n = \sum\limits_{i = 1}^n \frac{X^i}{i!}$

$P_n$ is Cauchy but doesn't converge in $\mathbb{K}[X]$.

xavierm02
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