Question of Cauchy sequence convergence I am having some problems understanding when Cauchy sequence don't converge, I know that if the metric space ($X$) is not complete the sequence don't converge IN $X$, every example I can think is embedded in $\mathbb{R}^n$, so it converge in the "big" space. For example:
$$
a_n = \left(1+\frac{1}{n}\right)^n
$$
does not converge in $\mathbb{Q}$, but it converges.
But I don't know if it is true or trivial that this happens always. The intuition tells me that a Cauchy sequence allways converge (even if it doesnt do it in the subspace that I am considering).
 A: This is true. Call a metric space complete if all Cauchy sequnces converge (with limit in the given space).
For each metric space $(X,d)$ there exists a metric space $(X',d')$ such that

*

*$X \subset X'$ and $d'(x,y) = d(x,y)$ for $x,y \in X$, i.e. the inclusion map $i : X \hookrightarrow X'$ is an isometry.


*$X$ is dense in $X'$.


*Each Cauchy sequence in $X'$ converges in $X'$, i.e. $(X',d')$ is complete.
This is called the completion of $(X,d)$ and it is essentially unique which means that if $(X'',d'')$ has the same properties 1. - 3., then there exists a unique isometry $\phi : (X',d') \to (X'',d'')$ such that $\phi(x) = x$ for all $x \in X$. Note that this universal property implies that $\phi$ is a bijection (take the analogous isometry $\psi :  (X'',d'') \to (X',d')$ and observe that $\psi \circ \phi$ and $\phi \circ \psi$ must be the identity maps).
I am not going into details of the construction, you should consult a textbook. Also see here.
Note that if $(X,d)$ is already complete, then it is its own completion. The completion of $\mathbb Q$ is $\mathbb R$.
