If I'm trying to prove that a Metric space $(M,d)$ is not complete I have to find a Cauchy a sequence that doesn't converge in $M$.
Using the following Metric Space as an example:
$M = \{ x \in \mathbb{R}^2 : ||x||<1 \}$
$d(x,y) = 2-||x||-||y||$ if $x \neq y$, and $0$ otherwise
I know that the point $(1,1)$ lies outside this space, as $||(1,1)|| = \sqrt{2} >1 $
So now I want to see if a sequence exists with this limit point.
I figured, in the usual metric the sequence $(\frac{\sqrt{2}}{2}-\frac{1}{n},\frac{\sqrt{2}}{2}-\frac{1}{n}$ would converge to $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$
How would I check to see if the limit point is the same given the above metric $d$?
Or even better, how would I determine exactly what the limit of this sequence is given the above metric $d$.
Take for example the sequence $x_n = \left (1-\frac{1}{n},0\right )$ this sequence clearly belongs to $M$
Now in order to prove that the sequence is a Cauchy sequence we regard the distance between two terms of the sequence and show that the distance tends to $0$ as $n,m \to \infty$
$$d(x_n,x_m) = 2 - ||x_n|| - ||x_m|| = 2 - \left(1-\frac{1}{n}\right) - \left(1-\frac{1}{m}\right) = \frac{1}{n} + \frac{1}{m} \longrightarrow 0 $$
After that, we should prove that the sequence not converges to any point, or more general, prove that there are not convergent sequences (showing that the space is not complete) , this can be easily done just taking an arbitrary point $x$ and an arbitrary sequence $(x_n)$ and showing that $(x_n)$ can not converge to $x$ (take a look at the distance between $x$ and a general point of the sequence)