Is every Cauchy sequence in a non-complete metric space convergent? A metric space $X$ is called complete if every Cauchy sequence in $X$ has a limit in $X$.
For a non-complete metric space $X$, can we say that every Cauchy sequence is convergent? (even though the limit is not in $X$)
In other word, is every Cauchy sequence convergent?
 A: No. By definition, the limit of a sequence must be an element of the metric space. 
Recall the definition: We say that a sequence $\{x_n\}$ in a metric space $(X, d)$ converges to $L$ if for every $\epsilon > 0$, we can find an $N$ such that $d(x_n, L) < \epsilon$ when $n \geq N$. 
If the limit $L$ were not in $X$, the expression $d(x_n, L)$ wouldn't make any sense because our metric is defined only for elements of $X$. 
In other words, if the limit were not in the space, we wouldn't know how to measure the distance between the limit and the elements of the sequence. So we wouldn't be able to say that the elements are getting closer to the limit, because we would never know how far away from it they are!
A: A sequence is convergent if and only if it has a limit, so no, Cauchy sequences are not necessarily convergent in non-complete spaces. However, there is the notion of a completion. Given a metric space $X$ a completion of $X$ is a complete metric space $\hat X$ in which $X$ is densely and isometrically imbedded. It turns out every metric space has a unique (up to isometric bijection) completion. So in this sense, a non-convergent Cauchy sequence in $X$ will converge in $\hat X$. This is non-trivial however and you are best not to use phrases such as "converges but in another space".
