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Exercise: Suppose that we have an arbitrary metric space $(M,d)$ where $d$ is the Euclidian metric. If $(A,d)$ is complete, show that $A$ is closed in $M$.

My approach:

I know that:

  • $(A,d)$ is complete if every Cauchy sequence in $A$ converges to a point in $A$.
  • If a sequence $(x_n)\subset A$ converges to some point $x\in M$, implies $x\in A$, then $A$ is closed.

So what I want to do is show that if every Cauchy sequence in $A$ converges to a point in $A$, then every sequence in $A$ converges to a point in $A$. I'm not sure how though. A sequence is Cauchy if given $\epsilon >0, \exists N>0$ such that $d(x_n, x_m) < \epsilon$ when $m,n\geq N$, and I'm not sure how to proceed from here.

Question: Is my approach going somewhere? How should I proceed to solve this exercise?

Thanks!

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  • $\begingroup$ I'm sorry, but what is the Euclidean metric in an arbitrary metric space? $\endgroup$ – Xander Henderson Feb 23 '18 at 14:59
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To show that $A$ is closed it is sufficient to show that the limit of every convergent sequence in $A$ is an element of $A$. Now if $x_n\rightarrow x$ with $x_n\in A$, then you first only know that $x\in M$. But since $x_n$ is convergent, it is also Cauchy. So it converges in $A$, if $A$ is complete. Conseuquently, $x\in A$ so $A$ is closed.

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Some hints:

  1. use the sequential characterization of closed subsets
  2. recall that a convergent sequence is always a Cauchy sequence.

So, start with $x_n \to x$, $x_n \in A$, $x \in M$. Your task is to prove that $x \in A$ as well. Use 2 and 1 to conclude.

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