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 $n>1$?


Here are some hints/sketches.

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.

4) The answer to (1) could certainly change. The answer I gave to (2) is the same, with the same proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.