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Let $(X, d)$ be a metric space and $(\overline{X}, \overline{d})$ its completion. If I have a finite covering of $X = \bigcup_i B(x_i, \delta)$, would this also be a covering of the completion of the metric space?

I would say yes. I tried something like this:

Let $\overline{x}:=(x_n)_n \in \overline{X}$. Then for $\epsilon > 0$ we have $N \in \mathbb{N}$, so that for $m \geq N$ $d(x_N, x_m) < \epsilon$ holds because it's a Cauchy sequence. Then this $x_N$ is within $\delta$-distance of some $x_i$ in our covering, so we have $d(x_i, x_N) < \delta$. Then

$d(x_i, \overline{x}) = \lim_{n \rightarrow \infty}d (x_i, x_n) < \lim_{n \rightarrow \infty} d(x_i, x_N) + d(x_N, x_i) < \delta + \epsilon$

For $\epsilon \rightarrow 0$ our covering holds?

oac
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  • This is obviously false. – Kavi Rama Murthy Nov 10 '21 at 10:19
  • Then help me understand where I am wrong. – oac Nov 10 '21 at 10:21
  • First of all distinguish between balls in $X$ and balls in $\overline X$. As it stands you are asking something like this: If a set is equal to $X$ is it also equal to $\overline X$. Besides, your last step (letting $\epsilon \to 0$ and concluding that you still have strict inequality is blatantly false. – Kavi Rama Murthy Nov 11 '21 at 05:25

1 Answers1

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Of course, $X$ could be already complete, in which case the statement would be obviously true.

Otherwise consider $X=]0,1[$ with absolute value of difference as distance.

Alp Uzman
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