Completeness proof of metric spaces

Question

Let (X, d) be a metric space and let Y be a dense subset of X. If every Cauchy sequence from Y conerges in (X, d), then (X, d) is a complete metric space.

Thank you in advance!

Answer 1

Suppose $x_n$ is a Cauchy sequence in $X$. Then there is a sequence $y_n$ in $Y$ s.t. $d(x_n,y_n)<1/n$. This sequence will be Cauchy (why?). Hence $y_n$ converges and so does $x_n$.