# 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).

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

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

2. $$X$$ is dense in $$X'$$.

3. 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 $$\mathbb R$$.