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In metric space every convergent sequence is a Cauchy sequence, but what about the converse?

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    Take $X=(0,1)$ with the usual metrics and $x_n=\frac1n$. (This was asked several times on the site.) – 2017-01-12
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    http://math.stackexchange.com/q/471577/ – 2017-01-12

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Consider the metric space $\mathbb R\setminus\{0\}$ with the usual distance, and the sequence $$ 1, \frac12, \frac13, \frac14, \frac15, \ldots $$

In $\mathbb R$ this converges to $0$ (and so it is Cauchy), but $0$ is not an element of the metric space, so it is not convergent as a sequence in $\mathbb R\setminus\{0\}$.

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The sequence $(x_n)$ defined by

$$x_{n+1}=\left(x_n+\frac{2}{x_n}\right)\frac12$$

for $x_0>0$ converge to $\sqrt 2$. But if our metric space is $\Bbb Q$ then the sequence doesnt converge on this space.