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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.

Mathstudent123
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    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: $d(x_{n},y)\leq d(x_{n},y_{n})+d(y_{n},y)$. – Mr. Gandalf Sauron Aug 01 '24 at 10:52

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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.

Mikhail Katz
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  • 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.

     – Mathstudent123 Aug 01 '24 at 12:03
  • Yes. A convergent sequence is always a Cauchy sequence. – Mikhail Katz Aug 01 '24 at 12:05
  • I should have realized that, it was the first theorem my teacher taught us on the subject of completeness. I really appreciate your answer. Thank you, Mikhail! – Mathstudent123 Aug 01 '24 at 12:12
  • OK but if you need this for a class, make sure you are allowed to use the notion of the completion of a metric space. – Mikhail Katz Aug 04 '24 at 07:59