Every Cauchy sequence in $A$ converges in $X$, where $A$ is dense. Show that $X$ is complete.

Question

My question is: Let $( X, d)$ be a metric space and $A$ a dense subset of $X$ such that every Cauchy sequence in $A$ converges in $X$. Prove that $( X, d)$ is complete.

Solution:

Case 1: If $X = A$ then it's trivial.

Case 2: If $X = A'$ for some X belonging to $X$ then there are sequences $a_{in}$ converging to those $x_{n}$ and so how should I deal with sequence of sequences?

Answer 1

Let $\{x_n\}$ be any Cauchy sequence. Choose $y_n \in A$ such that $d(x_n,y_n) <\frac 1 n$. Then $d(y_n,y_m) \leq \frac 1 n +\frac 1 m +d(x_n,x_m)$ so $(y_n)$ is a Cauchy sequence in $A$. I will let you verify that if $y_n \to y$ the $x_n \to y$. Hence $X$ is complete.