Dense subset in which Cauchy sequences are convergent

Question

Let $S$ be a dense set of a metric space $X$, such that all Cauchy sequences in $S$ are convergent (not necessarily in $S$). Then $X$ is complete space.

How can I show that $X$ is complete space given this condition?

Answer 1

Let us denote the metric of $X$ by $d$. Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $X$. For each positive integer $n$, there is some $y_n \in S$ with $d(x_n,y_n) < 1/n$. Hence $$d(y_m,y_n) \leqq d(y_m,x_m) + d(x_m,x_n) + d(x_n,y_n) < 1/m + d(x_m,x_n) + 1/n,$$ for every pair of positive integers $m$ and $n$, whence $(y_n)_{n=1}^\infty$ is also a Cauchy sequence. By assumption, there exists some $x \in X$ for which $(y_n)_{n=1}^\infty$ converges to. Since $$ d(x_n,x) \leqq d(x_n,y_n) + d(y_n,x) <1/n + d(y_n,x),$$ it follows that $x_n \rightarrow x$. This establishes the result.