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Let $C_b(X,R)$ be the set of all bounded continuous functions on $X$ with supremum norm. Is $l(X)$ dense in $C_b(X,R)$ with respect to the map $l:X \to C_b(X,R)$ defined by $l(x)=f_x$ where $f_x(t)=d(x,t)-d(x_0,t)$ where $x_0$ is a fixed point in $X$.

This is a step proving every metric space has a completion from Volker Runde's "A taste of topology". I checked that the map $l$ is an isometry and $C_b(X,R)$ is complete. The book also have not shown this. It is not obvious to me. Thanks in advance for your help.

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  • $\begingroup$ What are your thoughts on the matter? In particular, where are running into trouble when you attempt to solve it yourself? (This is not a "do your hormework for you" site.) And to be clear on the notation, by $C_b(X, R)$ do you mean the set of all bounded functions from $X \to \Bbb R$, topologized by the supremum norm? $\endgroup$ – Paul Sinclair Apr 1 '17 at 14:47
  • $\begingroup$ I doubt it would be dense (consider e.g. $X=\Bbb R$ with Euclidean norm, then each $f_x$ is of the form $t\mapsto |x-t|\,-\,|x_0-t|$, which altogether seem a poor family of functions. Nevertheless, you can take the closure of $l(X)$ and that's it. $\endgroup$ – Berci Apr 1 '17 at 16:07
  • $\begingroup$ @PaulSinclair I have edited the post $\endgroup$ – Arindam Apr 1 '17 at 16:11
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    $\begingroup$ Well, if we have an isometric embedding of metric spaces $f:A\to B$, and $B$ is complete, then the completion of $A$ appears as the closure $\overline{f(A)}$ (closed subsets of a complete metric space are themselves complete). $\endgroup$ – Berci Apr 1 '17 at 16:36
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    $\begingroup$ What Berci is pointing out is that $l$ does not need to be dense. You were mis-interpreting Runde's proof. All that was needed is an isometry $l$ from $X$ into some complete metric space $C$. Because then $\overline {l(A)}$ is also complete and serves as a completion of $X$. It is not necessary that $\overline{l(A)} = C$. $\endgroup$ – Paul Sinclair Apr 1 '17 at 18:45

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