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Let $(M, d)$ be a metric space. I have a conjecture that is something along the lines of: If $M$ is closed, then it is complete, and was wondering how true this statement is.

I know that it might not be sensible to say that a metric space $M$ is closed; the property of being closed is usually relative to a surrounding space, but I think I can reformulate this statement in two ways that might make sense.

1) If $M$ is closed in any metric space $M'$ that contains $M$, then $M$ is complete

2) If $M$ contains all of its limit points, then $M$ is complete.

The idea is that a Cauchy sequence should be convergent and the only situations where it is not are when the limit is outside of the metric space. Is this statement true? If not, I would really appreciate some counterexamples.

If it is true, why is the completion defined with equivalence classes when instead we can say the completion of a metric space $M$ is just it's closure $\overline{M}$? I am definitely missing something, so would love some clarification!

Note: I have not done too much Topology so would appreciate if we could keep the answer to metric spaces.

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  • $\begingroup$ You want to define the completion as the closure of $M$ in some ambient metric space, but the answer will depend a lot on which ambient metric space you choose. The usual definition involving equivalence classes is a universal solution to this problem; it constructs a closure which is as large as possible. $\endgroup$ – Qiaochu Yuan Nov 27 '17 at 3:23
  • $\begingroup$ @QiaochuYuan That makes a lot of sense. I am kind of struggling to see why the closure of $M$ needs to be relative to some ambient metric space. Is is not just the set of limit points of $M$, which shouldn't need the specification of a surrounding space to deal with? For ex. I can look at the seq $\{1/n\}$ in the metric $(0,2)$ and without knowing the surrounding space is $\mathbb{R}$, see that limit point is some object, call it 0, and add it to the new metric space. Is the issue that we need to extend the metric to include this object now? $\endgroup$ – gowrath Nov 27 '17 at 3:26
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    $\begingroup$ I think the point is that the closure of a metric space doesn't make sense intrinsically. For instance, the closure of any metric space $X$ is itself, since $X$ is closed, and a closed set equals its closure. So, if our metric space is $X=(0,1)$, we have that $\overline{X}=(0,1)$. If we instead view $(0,1)$ as embedded in $\mathbf{R}$, we know that $\overline{(0,1)}=[0,1]$, which is clearly not the same thing. Further, if we view $(0,1)$ as being embedded in the positive reals, $\mathbf{R}_{>0}$, we have that $\overline{(0,1)}=(0,1]$, a different result yet again. $\endgroup$ – Antonios-Alexandros Robotis Nov 27 '17 at 3:33
  • $\begingroup$ You are absolutely right, it is definitionally true that $M$ is closed with respect to $M$. So then the equivalence class approach to completion is the way to find the closure of $M$ with respect to the "largest" metric space containing $M$? $\endgroup$ – gowrath Nov 27 '17 at 3:36
  • $\begingroup$ Yes, that's basically the idea. $\endgroup$ – Qiaochu Yuan Nov 27 '17 at 3:50
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Statement 1 is correct. Completeness is equivalent to "closed in any larger metric space $M'$" (with the understanding that the two metrics agree on $M$).

Indeed, every metric space has a completion $\widehat{M}$, which is a complete metric space containing $M$ as a dense subset. By assumption, $M$ is closed as a subset of $\widehat{M}$. Being both closed and dense implies that $M=\widehat{M}$. Thus $M$ is complete.

Statement (2) does not really make sense. There is no concept of a limit point being just "outside", not contained in anything. In order to talk about limit points that are not in $M$, we need a larger space $M'$ in which those points are contained. And then we arrive back at statement (1).

For the same reason, we don't get anything new from "the closure of $M$" without specifying a larger space in which the closure is taken.

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