# 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?

• In a sense yes. Every metric space can be completed, basically by adding limits of Cauchy sequences to your space (you also have to quotient by an equivalence relation, but that's in the technical details). See here: proofwiki.org/wiki/Completion_Theorem_(Metric_Space) – Moya Oct 5 '15 at 4:13
• In a different sense no, because if you did, what good would it be to distinguish complete from non-complete metric spaces? – pjs36 Oct 5 '15 at 4:14
• @Vader That I disagree with. Pick a sequence in $\mathbb{Q}$ converging increasing upward to $\sqrt{2}$. This is definitely not convergent in $\mathbb{Q}$ since the limit must be in $\mathbb{Q}$. See Jason's answer: it's best not to talk about all Cauchy sequences converging even though there is this notion of a completion. – Moya Oct 5 '15 at 4:20
• @Moya I was just giving the OP something to think about, and liked the "competing viewpoints" feel :) – pjs36 Oct 5 '15 at 4:20
• How do you define convergent? However you define it should immediately give you an answer to your question. – Marcel Besixdouze Oct 5 '15 at 4:30

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".
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 $X$.