# Completeness of a Metric Space Given a Dense Subset with Converging Cauchy Sequences

I am stuck on an exercise which is: Let $$A$$ be a dense subset of a metric space $$X$$. Suppose that every Cauchy sequence in $$A$$ converges to a point in $$X$$. Prove that $$X$$ is complete.

I have seen that similar questions have been answered here: Proving that $X$ is complete if $A\subset X$ is dense and every cauchy sequence in $A$ converges to a point in $X$. and here: Every Cauchy sequence in $A$ converges in $X$, where $A$ is dense. Show that $X$ is complete.

I am having trouble understanding this proof. I understand that we have to show that any Cauchy sequence $$\{x_i\}$$ in $$X$$ converges in $$X$$. The part I don't understand is why we pick $$y_n \in A$$ such that $$d(x_n, y_n) < \frac{1}{n}$$. Is this because we know that every Cauchy sequence in $$A$$ converges to a point in $$X$$?

Any help or hints would be appreciated.

• Suppose you pick $y_{n}\in A$ in that way. Then what can you say about $d(y_{n},y_{m})$? Hint:- $d(y_{n},y_{m})\leq d(y_{n},x_{n})+d(x_{n},x_{m})+d(x_{m},y_{m})\leq \frac{1}{n}+d(x_{n},x_{m})+\frac{1}{m}$. Can you conclude now that $y_{n}$ is Cauchy? Can you find a limit for the sequence $y_{n}$ using the property that $A$ has? Suppose that you can and the limit is $y\in X$ . What can you say about $d(x_{n},y)$? Hint: $d(x_{n},y)\leq d(x_{n},y_{n})+d(y_{n},y)$. Commented Aug 1 at 10:52

Consider the completion $$\hat X$$ of $$X$$. Since $$A$$ is dense in $$X$$, it is also dense in $$\hat X$$. For each point $$p\in \hat X$$, choose a Cauchy sequence $$\{x_n\}$$ in $$A$$ converging to $$p$$. By hypothesis, every Cauchy sequence in $$A$$ converges to a point of $$X$$. In particular, $$\{x_n\}$$ converges to a point $$x\in X$$. But a convergent sequence cannot converge to more than one point. Hence $$p=x$$. It follows that $$\hat X=X$$, i.e., $$X$$ is complete.
• How can we be sure that for each $p \in \hat{X}$ there exists a Cauchy sequence $\{x_n\}$ converging to $p$? Is this because $A$ is a dense subset of $\hat{X}$, and that means every point in $\hat{X}$ can be approximated arbitrarily closely by points in $A$? This would mean that for any $\epsilon > 0$, there exists a positive integer $N$ such that $d(x_n, p) < \epsilon$ for all $n \geq N$. Would this interpretation be correct? The rest of your argument I understand. Commented Aug 1 at 12:03