The definition of a complete metric space Is the following not equivalent to the traditional definition of a complete metric space (i.e. every Cauchy sequence converges)?
$X$ is complete if for every $\{x_n\} \subset X$ such that $x_n \rightarrow x$, then $x \in X$.
If this is not the same, what is a counterexample? If it is the same, why don't we use this definition instead of involving Cauchy sequences?
 A: This is not the same; in fact, your version is the definition of a closed set in a metric space. For example, take the rational numbers $\mathbb{Q}$. Then if $\{x_n\}\subset \mathbb{Q}$ converges to some $x \in \mathbb{Q}$, it is trivially true that $x \in \mathbb{Q}$ and thus $\mathbb{Q}$ is closed in $\mathbb{Q}$ (which is also trivial). However, $\mathbb{Q}$ is not $complete$ since one can define a Cauchy sequence that does not converge to $any$ rational number; for example, take $a_0 = 1$ and $a_{n+1} = \frac{1}{2}\left(\frac{2}{a_n} + a_n\right)$. This sequence $should$ converge to $\sqrt{2}$, but it does not because $\sqrt{2} \not\in \mathbb{Q}$; this space has holes and is therefore not complete.
Edit: If the whole space $X$ is complete, i.e. has no holes, then a subset $A$ is complete iff $x_n\to x \in X \implies x \in A$, which may be where the confusion comes in. This is why closed subsets are equivalent to a complete subset when the whole space $X$ is complete. In general, however, a subset being complete implies that it's closed but not the other way around (completeness is stronger).
