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May 17, 2019 at 10:28 comment added Paul Frost Ok, you can select an adaquate compactification for various purposes. As I explained, there are no compactifications such that the remainder is finite with more than one point. Following the link in my answer, you will moreover see that if the remainder is infinite, it must even be uncountable. Your "favourite" compactification $\overline{\mathbb R} ^2$ has the property that each ray $tv$, $v \in \mathbb R^2, t \in [0,\infty)$ approaches an individual limit point in the remainder. This is not true for many other compactifications.
May 17, 2019 at 9:28 vote accept High GPA
May 17, 2019 at 9:23 comment added High GPA For example, someone might prefer to have both $+\infty$ apples and one orange $(+\infty,1)$, but she won't like to lose $+\infty$ apples just to get one orange $(-\infty,1)$. This could be captured by the $\overline{\mathbb R}^2$ compactification, but in the one point compactification, she must be indifferent between $(-\infty,1)$ and $(+\infty,1)$ which does not really make sense.
May 16, 2019 at 22:02 comment added Paul Frost Which properties are you interested in? Why do think this compactification is interesting?
May 16, 2019 at 21:59 comment added High GPA Then how do I find more properties on this compactification? Are there any people worked on this structure?
May 16, 2019 at 21:57 comment added Paul Frost Concerning the remainder: Yes. As far as I know, this compactification does not have a special name. The general idea is this. If $Y_i$ ís a compactification of $X_i$, then $\prod_{i=1}^n Y_i$ is a compactification of $\prod_{i=1}^n X _i$.
May 16, 2019 at 21:54 comment added High GPA Thanks for your answer! By "remainder is homomorphic to $S^{n-1}$ do you mean that $\overline{\mathbb R}^n\setminus \mathbb R^n$ is the remainder? This is exactly what I am looking for. What is the name for this type of compactification?
May 16, 2019 at 21:44 history answered Paul Frost CC BY-SA 4.0