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 $$ \operatorname{diam}\, {Y}_i\,\longrightarrow\,0\,\;. $$
QUESTION:
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?
