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Here is what I came up with.

Since $M$ is closed it follows that $U=X \backslash M$ is open. Then $(X \backslash M, d)$ can not be complete, because if $(X\backslash M, d)$ is complete this would implie that $X \backslash M$ is closed, which is not true. Since $(X \backslash M, d)$ is not complete there exists atleast one cauchy seq. $(x_n)_{n \in \mathbb{N}}$ s.t if $x_1, x_2,\dots , x_k \in X \backslash M$ then $x_{k+1}, \dots x_{n} \in M$. Since $X \backslash M$ not complete. Hence the seq. $(x_j)_{j=k+1}^{n}$ is cauchy and converge in $(M,d)$. On the other hand if $(x_n)_{n \in \mathbb{N}}$ is a cauchy seq. in $X$ with $x_1, x_2, \dots x_k \in M$ then..?

Olba12
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  • But why do you need such a roundabout logic? You can just use the definition of closure, right? If $x_n$ is a Cauchy sequence in $M$, then by definition it is so in $X$, and by completion in $X$, there is a limit $x$ of this sequence, which by closure of $M$, lies in $M$, making $x_n$ convergent in $M$,, therefore $M$ is complete. – Sarvesh Ravichandran Iyer Nov 03 '16 at 01:00
  • Yes, I could, but we have not defined it in class. @астонвіллаолофмэллбэрг – Olba12 Nov 03 '16 at 01:06
  • Oh I see.But then this question is impossible without knowing what closure is, right? You can take it down: every Cauchy sequence in $M$ is convergent in $M$,that is the limit lies in $M$. In addition, $M$ is the complement of an open set. Both these statements are equivalent. – Sarvesh Ravichandran Iyer Nov 03 '16 at 01:09
  • Maybe my professor have just assumed that I know what it is, I will do the proof with closure instead. Those two equivalent statements seems helpfull! @астонвіллаолофмэллбэрг – Olba12 Nov 03 '16 at 01:11
  • I am very happy to help, my friend. – Sarvesh Ravichandran Iyer Nov 03 '16 at 01:11

1 Answers1

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Just take a Cauchy sequence $(x_n)_{n\in\Bbb{N}}$ in $(M,d)$. That sequence is also a Cauchy sequence in $(X,d)$. Since that last space is complete, there exist $\bar{x} \in X$ such that $x_n \to \bar{x}$. On the other hand, since $(x_n)_n$ is convergent and $M$ closed, we obtain $\bar{x} \in M$.

Sorombo
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  • I understand that you have used that $M = \text{closure of } M$. I was trying to avoid that because we have not define the closure in class. But otherwise I understand your proof. – Olba12 Nov 03 '16 at 01:05