# Completion of metric-space, when “corresponding” Euclidean space is complete

I somehow couldn't find the answers to several probably simple questions. I am new to the topic, thus, please excuse any lack of knowledge.

Let $A_1\subseteq\mathbb{R}^n$, such that the metric-space $(A_1,d_1)$ is complete, with $d_1(x_1,x_2)=\|x_1-x_2\|_2$ the usual Euclidean metric.

Given that $(A_1,d_1)$ is complete, I try to understand properties of the completion $\bar{A}_2$ of $A_2$ with respect to the metric space $(A_2,d_2)$ for any metric $d_2$, for the specific case when $A_1=A_2$. Sorry for the terrible notation...

Short example: Let $A_1=\left[1,\infty\right)$; $(A_1,d_1)$, with $d_1(x,y)=|x-y|$ is complete; however, $(A_2,d_2)$, with $A_1=A_2$ and $d_2(x,y)=|\frac{1}{x}-\frac{1}{y}|$ is not complete, since Cauchy-series $(a_k)_{k\in\mathbb{N}}$, $a_k=k$ does not converge in $A_2$. Without respect to notation, I can probably write the completion of $A_2$ somewhat like $\bar{A}_2=A_2\cup\{\infty\}$.

My questions: 1) Is the "$\infty$" in $\bar{A}_2=A_2\cup\{\infty\}$ in my example only one equivalence class, or are there "more than one $\infty$'s"?

2) Assume additionally that $A_1$ in $(A_1,d_1)$ is bounded (i.e. compact), and $n=1$. Is it still possible that a $d_2$ exists, such that $(A_2,d_2)$, with $A_2=A_1$, is not complete? If yes, are there some moderate assumptions about $d_2$ preventing such cases?

3) For a not bounded $A_1$, $n=1$ and an arbitrary $d_2$, do all elements in $\bar{A}_2 - A_2$ confer to the interpretation "diverges to $\pm\infty$"?

4) Do the answers change for $n>1$?

1) There is only one $\infty$ in your example, i.e. $\overline{A}_2 = A_2 \cup \infty$. If $\{x_n\}\subset A_2$ is a Cauchy sequence in your metric and is bounded in the usual sense in $\mathbb{R}$, then it converges to something in $A_2$ (since on bounded subsets of $A_2$ your metric and the usual metric are equivalent up to a constant factor). If $\{x_n\}$ is unbounded and Cauchy (in the new metric), then you should be able to show that it is in the same equivalence class as the sequence $a_n=n$ that you already described.
2) This can't happen if the identity map from $(A_1, d_1)$ to $(A_1, d_2)$ is continuous. In that case, $(A_1, d_1)$ is compact, the image of a compact metric space under a continuous map is compact, so $(A_1, d_2)$ is compact and therefore complete.
Without any assumptions whatsoever on $d_2$, of course anything can happen, there is no reason to preserve any topological or metric properties at all. For example, let $A_1=[0,1]$ and $d_1$ be the usual Euclidean metric. Let $f(x) = x$ if $x\in (0,1]$ and let $f(0)=1$. Define $d_2(x,y)=|f(x)-f(y)|$ (check that it's a metric). Then $1/n$ is a Cauchy sequence in $([0,1], d_2)$ that doesn't converge.
3) This probably needs some more specifics. With respect to an arbitrary metric $d_2$, really anything can happen.