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?
Best Answer
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.