How to show a metric space is not complete In order to show that a metric space $(X, d)$ is not complete one may apply the definition and look for a Cauchy sequence $\{x_n\}\subset X$ which does not converge with respect to the metric $d$. Now I have often seen (on books, e.g.) another approach: one may show that a sequence $\{x_n\}\subset X$ converges with respect to the metric $d$ to a limit $x$ which is not contained in $X$.  
A common example may be the following: since $x_n:= (1+1/n)^n\in \mathbb{Q}$ for every $n \in \mathbb{N}$ and $x_n \to e$, but $e \notin \mathbb{Q}$, one can conclude that $\mathbb{Q}$ is not complete. 
I've always considered this to be obvious but I now realize I can't explain why this works. The quantity $d(x_n, x)$ itself need not be well-defined, in general, if $x \notin X$. So my question is: why (and under which conditions) this criterion for not-completeness of a metric space ("limit is not in the same space as the sequence") can be used? 
 A: It can be used when you are aware of the existence of a metric space $(Y,d^\ast)$ such that:


*

*$X\subset Y$;

*$(\forall x,x'\in X):d^\ast(x,x')=d(x,x')$.


In the example that you have mentioned, that space is $\mathbb R$, endowed with its usual metric.
Actually, such a metric space always exist (take the completion of $X$, for instance), but if you don't know how to work with it, that's a useless piece of information.
A: The point here is that there are two metric spaces involved.
Basically, the following situation happens. Let $X$ be a metric space and $Y$ be a metric subspace of $X$ (thus $Y \subseteq X$ and the distance in $Y$ is the distance in $X$ restricted to $Y$).
In our situation we have a sequence $(y_n)_n$ in $Y$ and this converges to a point $x \notin Y$. But thus $(y_n)_n$ is a convergent sequence in $X$ and thus Cauchy in $X$ and thus Cauchy in $Y$. However, if $(y_n)_n$ would converge in $Y$, there is a limit $y \in Y$ and by uniqueness of limits we get $y =x \notin Y$, which is impossible.
Thus, such a sequence cannot converge in $Y$ and we have found a Cauchy sequence that does not converge. Hence, $Y$ can't be complete.
A: For the second method you mentioned note that $(1+1/n)^n$ is a Cauchy sequence on the rationals with respect to the usual  metric  restricted on $\Bbb{Q}$
Even  if the sequence $x_n$ converged to a point $x \notin Y \subseteq X$, it is still Cauchy with the metric restricted on $Y$(name it $d_Y$)
If $(Y,d_Y)$ was complete then would exist $z \in Y$ such that $x_n \to z$.
But every metric space is Hausdorff and thus the limit o a sequence is unique,so $z=x \notin Y$ which is a contradiction. 
