TITLE: How to show a metric space is not complete
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.