Consider a metric space $(S,d)$, its completion $\tilde S$, and its closure $\overline{S}$. (By closure I imply: closure in completion.)

In the standard topology (one induced by the metric, i.e., based on open balls), we have $\tilde{S} = \overline{S}$. (Please correct me if I am wrong.)

Also consider a Cauchy sequence $\left\{ y_n \right\}$ and its tails $$ Y_i\,=\,\left\{ y_i,\, y_{i+1},\;.\,.\,.\, \right\}\,\;. $$ The tails are nested, $$ Y_{i+1} \subset Y_i\,\;, $$ but are not necessarily closed. Their diameters are defined as $$ \operatorname{diam}\, Y_i\,\equiv\,\operatorname{sup}_{m,n\leq i}\,d(y_m,\,y_n)\,\;. $$ Likewise, the diameters of their completions (= closures) are $$ \operatorname{diam}\, \overline{Y}_i\,\equiv\,\operatorname{sup}_{m,n\leq i}\,\left(\,d(y_m,\,y_n), \, d(y_m,\,y)\,\right)\,\;, $$ $y$ being the limit of $\left\{y_i\right\}$. This limit is not necessarily in the space $S$, but is in $\tilde{S}$.

The sequence is Cauchy, so the tails satisfy $$ \operatorname{diam}\, {Y}_i\,\longrightarrow\,0\,\;. $$


How to prove the same for the completions (= closures, on this occasion) -- i.e., to show that $$ \operatorname{diam}\, {\overline{Y}}_i\,\longrightarrow\,0\,\;. $$

My understanding is that $$ \operatorname{diam}\, \overline{Y}_i\,\leq \operatorname{diam}\, Y_i\,+\,\sup_{m\geq i}\,d(y_m,\,y)\,\;. $$ If this is correct, my question may be put in a simpler form: how to prove that $$ \sup_{m\geq i}\,d(y_m,\,y)\,\longrightarrow\,0~~~\mbox{for}~~~i\longrightarrow\infty $$ Is this so simply by definition of the limit?

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    $\begingroup$ I am unable to correct the sentence “In the standard topology, we have $\tilde{S} = \overline{S}$” since I don't know what it means. $\endgroup$ Feb 3 at 16:16
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    $\begingroup$ What do you mean by the "closure" of $S$? $S$ is of course closed, i.e., $\operatorname{cl}_S S = S$. So you need to be more precise -- closure with respect to what? $\endgroup$
    – MPW
    Feb 3 at 16:40
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    $\begingroup$ Presumably, you mean the closure of $S$ in the completion? $\endgroup$ Feb 3 at 16:55
  • $\begingroup$ @MPW Yes, good point, I shall explain that this a closure w.r.t. completion. $\endgroup$ Feb 3 at 17:26
  • $\begingroup$ @ThomasAndrews Yes, in completion. I shall fix this in the text. $\endgroup$ Feb 3 at 17:27

1 Answer 1


HINT: What does it mean for $y$ to be the limit of the sequence $(y_i)$?


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