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Is the extended $\overline{\mathbb R}^2$ written as $\mathbb R^2 \cup \{-\infty,+\infty\}$ or $(\mathbb R\cup \{-\infty,+\infty\})^2$ ? What is the name for the latter set?

A.k.a Are there four infinite points or just two or infinite number of infinite points?

Thanks for the information. So the most common compactification used is the Riemann sphere $\overline{\mathbb C}$ which adds only one point to $\mathbb C$.

The question I am more interested in is the natural generalization of the two points compactification of $\mathbb R$ into multiple dimensions. Intuitively, it should add more than four points to $\mathbb R^2$. It is not the Real projective space $\mathbb R\mathbb P^n$. Is it related to any named compactification? It has a poor algebraic structure, i.e. not a group or semi-group, so it might attract less interests.

Related questions:

Is there a name for complex numbers over affinely extended reals?

Do the two types of extended real number systems originate in algebraic geometry?

High GPA
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    Viewing $\Bbb R^2=\Bbb C$ we do a one-point-compactification, see here. – Dietrich Burde May 16 '19 at 18:35
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    One can add just one point Riemann sphere, a whole line of points real projective plane where each line point towards only one of the new points. A line of points: Poincare plane, but here each line points to two of the new added points. And other options each useful for different purposes. – logarithm May 16 '19 at 18:45
  • @logarithm This is exactly what I want! How do I check for the known properties for the compactifications similar to this one? – High GPA May 16 '19 at 19:11
  • @logarithm Why are you deleting your answer? The extended plane $\overline{\mathbb R}\times\overline{\mathbb R}$ is widely used. See https://math.stackexchange.com/questions/1406652/is-there-a-name-for-complex-numbers-over-affinely-extended-reals – High GPA May 16 '19 at 19:20
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    I was trying to replace it with a link to a proper reference, but haven't found one. – logarithm May 16 '19 at 19:27
  • @logarithm Thanks for your rigorousness -- a rare attribute. – High GPA May 16 '19 at 19:31
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    You might have misread this link https://math.stackexchange.com/questions/1406652/is-there-a-name-for-complex-numbers-over-affinely-extended-reals. There is no assertion there that $\overline{\mathbb R} \times \overline{\mathbb R}$ is widely used. – Lee Mosher May 16 '19 at 20:42

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There are many compactifications of $\mathbb R ^n$. Perhaps the most common is the one-point compactification $\mathbb R ^n \cup \{\infty\}$ which is homeomorphic to the sphere $S^n$. You may also take $\overline{\mathbb R}^n$ whose remainder is homeomorphic to $S^{n-1}$. Another variant is $\overline{\mathbb R}^{n-m} \times (\mathbb R \cup \{\infty\})^m$.

If you are interested in compactifications $C$ of $\mathbb R ^n$ such that the remainder $C \setminus \mathbb R ^n$ is finite, then the answer is simple (although the proof is non-trivial): For $n \ge 2$ the one-point compactification is the only such compactification. See my answer to Is the two-point compactification the second-smallest compactification?.

Paul Frost
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  • 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? – High GPA May 16 '19 at 21:54
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    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 $\prod_{i=1}^n X _i$. – Paul Frost May 16 '19 at 21:57
  • Then how do I find more properties on this compactification? Are there any people worked on this structure? – High GPA May 16 '19 at 21:59
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    Which properties are you interested in? Why do think this compactification is interesting? – Paul Frost May 16 '19 at 22:02
  • 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. – High GPA May 17 '19 at 09:23
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    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. – Paul Frost May 17 '19 at 10:28