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let $(A,d)$ be a metric space which is complete and let $B$ be closed in A. Then prove $(B,p)$ is complete, with $p = d|_{B\times B}$ (i.e. restriction of d onto $B\times B$)

thoughts:

since $B$ is closed in $A$ and $(A,d)$ is complete i figured that $(B,y)$ is complete - from here I know I have to show for $x_n,x_m \in B$,$ p(x_n,x_m) < \epsilon $ for $x,m\geq N$ but I am not sur ehow to deal with the restriction of $d$ on $B\times B$ - I mean, how would I go about evaluating $p(x_n,x_m)$?

edit: progress:

$d|_{YXY} ((x_n,x_m),(y_n,y_m)) = p((x_n,x_m),(y_n,y_m)) $, with $(x_n,x_m), (y_n,y_m) \in Y$ now if I let $x_n,y_n$ be cauchy I can see that $p<\epsilon$ but no idea how ot prove it

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    $\begingroup$ In fact any closed subset of a complete metric space is itself a complete metric space. $\endgroup$ – Math1000 Jan 25 '15 at 18:25
  • $\begingroup$ @Math1000 any ideas how to solve the problem? $\endgroup$ – forgotten89 Jan 25 '15 at 18:40
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Let $\{x_n\}_n\subseteq B\subseteq A$ be a Cauchy sequence (in $B$). As it is a Cauch sequence also in $A$, and as $A$ is complete (i.e. Cauchy $\implies$ convergence), $\{x_n\}_n$ converges to some point $x$. But $B$ were closed (under convergence), so $x$ must be in $B$. Hence any arbitrary Cauchy sequence in $B$ is convergent, that is, $B$ is complete.

Note: $p$ is simply the restriction of $d$ to $B$. That roughly means that the notion of distance between points of $B$ is defined exactly as the notion of distance between points of $A$, only you disregard those points which are not elements of $B$. So if $x,y\in B$, then $p(x,y)=d(x,y)$, but if either one of $x,y$ are not in $B$, then $p$ is meaningless in measuring the distance between them.

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  • $\begingroup$ in that case is the question then not moot? I mean, if $p = d$ for elements in $B$ then why would the question ask to show that $(B,p)$ is complete? because I have shown that $(B,d)$ is complete $\endgroup$ – forgotten89 Jan 25 '15 at 18:20
  • $\begingroup$ Well, you could say that the question is relatively easy. The trick is simply in keeping in mind that $B$ is closed. $\endgroup$ – Alp Uzman Jan 25 '15 at 18:32
  • $\begingroup$ There are two additional exercises that you can do: 1) you could try to come up with non-complete subspaces; 2) you could consider the set $B$ with a different metric $p$ (which is maybe equivalent to $d$, i.e. a sequence is convergent in $d$ iff it is so in $p$). $\endgroup$ – Alp Uzman Jan 25 '15 at 18:33
  • $\begingroup$ im sorry I really don't understand what you've done - you've shown that $(B,d)$ is complete right, I have also shown this as it is a theorem we had. But I'm aksing how do I show $(B,p)$ is complete? I mean, if you take a look at my edited OP how can I use that definition to show $(B,p)$ is complete? I ask because I do not understand how you have shown $(B,p)$ is complete $\endgroup$ – forgotten89 Jan 25 '15 at 18:40
  • $\begingroup$ Oh, now I see your problem. $(B,d)$ is simply a "notational abuse", denoting that we are supposed to consider $B$ as a subspace of $A$. $(B,p)$, on the other hand, is more rigorous in differentiating a function (in this case $d$) and its restriction (in this case $p=d|_{B\times B}$. So your theorem and the question differs in notational preferences only. $\endgroup$ – Alp Uzman Jan 25 '15 at 18:46
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Let $X$ be a complete metric space and $F$ a closed set in $X$. If $x_n$ is a Cauchy sequence in $F$, then it has a limit $x\in X$. Since $F$ is closed, $x\in F$. Hence $F$ is complete.

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