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Consider the metric $d(x,y)=|x-y|+|sgn(x)-sgn(y)|$.

To start, $d(x,y)$ is indeed a metric ($d(x,x)=0, d(x,y)>0, d(y,x)=d(x,y))$ are all fairly easy to show. For the triangle inequality, I used the following trick: $|x-z+z-y|+|sgn(x)-sgn(z)+sgn(z)-sign(y)|$ and used the regular triangle inequality to show that $d(x,z)\leq d(x,y)+d(y,z)$.

My space is the interval $X=[-a,0)\cup (0,a]$. Is the completion $X^*$ of this space just adding the point $\{0\}$ to $X$? I know compact euclidean spaces are complete metric spaces (assuming a metric has been defined). I know a space has to be complete iff it is compact and totally bounded, which applies here, but I am struggling to prove rigorously that ${0}$ is the point needed to complete the space.

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    $\begingroup$ "A space is complete iff it is compact and totally bounded" You got the words "complete" and "compact" reversed, which I assume is a typo. $\endgroup$ – DanielWainfleet Sep 11 '17 at 17:45
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Note that for your space $X$ with the given metric, given a subsequence $\{x_{n}\}_{n=1}^{\infty}$ of $[-a,0)$ or of $(0,a],$ the metric $d$ reduces to the ordinary metric on $\mathbb{R}.$ In particular, the sequences $\{a/n\}_{n=1}^{\infty}$ and $\{-a/n\}_{n=1}^{\infty}$ are Cauchy with respect to $d,$ so in the completion, they must have a limit. Let's call these limits $0^{+}$ and $0^{-}$ for the time being. We have to extend the definition of $d$ to $0^{\pm}$ in order to make sense of the metric in the completed space, which we can accomplish by defining $\mathrm{sgn}^{*}$ to be the usual $\mathrm{sgn}$ function on $\mathbb{R}$, and giving values of this function to $0^{\pm}$. We see that in order for $a/n\rightarrow 0^{+}$ with respect to $d,$ we need $\mathrm{sgn}^{*}(0^{+})=1$ (otherwise $d(a/n,0^{+})= a/n+|1-\mathrm{sgn}^{*}(0^{+})|\rightarrow |1-\mathrm{sgn}^{*}(0^{+})|>0$ as $n\rightarrow\infty$, so this convergence does not hold). Similarly, we need $\mathrm{sgn}^{*}(0^{-})=-1$.

Let's call $X^{*}=[-a,0)\cup(0,a]\cup\{0^{-},0^{+}\},$ with $d^{*}$ defined as $d$, but with $\mathrm{sgn}^{*}$ as defined above rather than $\mathrm{sgn}.$ We observe that $d^{*}=d$ when restricted to $X$, and that any Cauchy sequence in $d^{*}$ must stabilize in either $I^{-}=[-a,0)\cup\{0^{-}\}$ or $I^{+}=\{0^{+}\}\cup(0,a]$ (i.e., $\{x_{n}\}_{n=1}^{\infty}$ stabilizes in $I^{+}$ if there is some $N$ such that for all $n\geq N,$ $x_{n}\in I^{+}$), since otherwise for any $N,$ there is some pair $n,m\geq N$ such that $x_{n}\in I^{-},$ $x_{m}\in I^{+},$ in which case $d^{*}(x_{n},x_{m})\geq 2$. But if $\{x_{n}\}_{n=1}^{\infty}$ stabilizes in $I^{+},$ say, then there is some $N$ such that for all $n\geq N,$ $x_{n}\in I^{+},$ which means that $d^{*}(x_{n},x_{m})=|x_{n}-x_{m}|$ for $n,m\geq N.$ Clearly $I^{+}$ with $d^{*}$ restricted to this set is isomorphic to $[0,a],$ which is complete, and thus $\{x_{n}\}_{n=1}^{\infty}$ has a limit in $I^{+}.$ Repeating this argument for $I^{-},$ we see that all Cauchy sequences in $X^{*}$ have limits with respect to $d^{*}$, so $(X^{*},d^{*})$ is complete, as desired.

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    $\begingroup$ Note also that it is impossible to have just a single point completion: if we take $\{a/n\},\{-a/n\},$ both converging to 0, then for $\varepsilon>0$ and sufficiently large $n,$ we have $2\leq d^{*}(-a/n,a/n)\leq d^{*}(-a/n,0)+d^{*}(0,a/n)<2\varepsilon,$ which is a contradiction for $\varepsilon\leq 1.$ $\endgroup$ – RideTheWavelet Sep 11 '17 at 15:26
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(I). The "canonical" completion of $(X,d)$ is usually presented by as $(X\cup (C/ \sim),d')$ where $C$ is the set of $d$-Cauchy sequences in $X$ that do not converge to members of $X,$ and $\sim$ is the equivalence relation on $C$ where $(u_n)_n\sim (v_n)_n$ iff $\lim_{n\to \infty}d(u_n,v_n)=0,$ and $C/\sim$ is the set of $\sim$-equivalence classes. And (naturally) $d'(u,v)=d(u,v)$ for $u,v\in X.$

(II). In your Q, $\;C/\sim$ has $2$ members which I will call $0^-$ and $0^+.$ A sequence $(u_n)_n \in 0^-$ iff $u_n<0$ for all but finitely many $n,$ and in the reals $\lim_{n\to \infty}u_n=0.$ Also, $(u_n)_n\in 0^+$ iff $u_n>0$ for all but finitely many $n,$ and in the reals $\lim_{n\to \infty}u_n=0.$

If there are infinitely many $m$ with $u_m>0$ and infinitely many $n$ with $u_n<0$ then $(u_n)_n$ is not $d$-Cauchy because for every $r$ there are $m,n>r$ with $d(u_m,u_n)>2.$

We cannot append the single point $0$ and get a completion $(X',d')$ as then $$\lim_{n\to \infty}d'(0,-a/n)=0=\lim_{n\to \infty}d'(0,+a/n)$$ which, by the triangle inequality, implies $$0=\lim_{n\to \infty}d'(-a/n,+a/n)= \lim_{n\to \infty}d(-a/n,+a/n)$$ which is impossible because $d(-a/n,+a/n)>2.$

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