# 3 directions of infinity ?

$N$ the positive natural numbers has one infinity. $Z$ the integers has 2 infinities.

What object would as "naturally" as possible have 3 infinities?

This probably can be answered in many ways. Yet for me the algebraic side would be more important than the topological one, though this does not exclude both.

What troubles me is that $Z$ is natural as being final in the category of rings) and moreover it is the completion ( in fractional sense) of $N$.

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I would like to point out Stallings theorem that 3 is impossible for groups. He showed there are only 4 possibilities for the "number of infinities" in your terminology, or the space of ends, in standardd definition. (The "ends" of a space is the inverse limit of the collection of components of complements of compact sets). Stallings proved that every group has either 0 (for finite groups), 1, 2 (implies virtually Z) or a Cantor set of ends. Cf. en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups – Bill Thurston Sep 22 '10 at 1:10
Precisely if we require only order I guess any K infinity can be achieved. But where does the limit stands between order and group ? – Jérôme JEAN-CHARLES Sep 22 '10 at 10:50

## 3 Answers

The obvious first answer: take three copies of $\mathbb{N}$ as total orders, then join them at the bottom element to get an unbounded poset with bottom. This of course isn't satisfactory as it doesn't give $\mathbb{Z}$ for two copies of $\mathbb{N}$. This strikes me as a sort of 'what about a 3-dimensional version of the complex numbers?' question, and could benefit from considering 'four infinities'...

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You may take $k$ copies of$N$ with an added bottom element. I think this still works with more structure than that of an order, for example that of a semi-lattice. The question is not so much for $k=3$ but more about how much structure you may add (more than semi lattice) while retaining the existence of $k$ infinity say for each $k$ ? When you impose a group structure this does not work as mentioned above by Bill Thurston. So what could be found in between that still works?. – Jérôme JEAN-CHARLES Sep 23 '10 at 14:54
Well take $\mathbb{Z} \cup_{\{0\}} i\mathbb{Z}$: the axes in the complex numbers. This is a monoid under multiplication and has a partially defined addition that distributes over arbitrary multiplication. Not sure what you'd call this though. – David Roberts Sep 24 '10 at 0:50
What does the union sign with 0 in index means ? I guess it is a kind of product. Which ? – Jérôme JEAN-CHARLES Sep 24 '10 at 1:28
Bit of a hybrid there! At the level of sets, it's just the pushout of Z <- {0} -> iZ. Or one could be a little slack and just say it's the union. I don't know what the operation 'is' at an algebraic level. – David Roberts Sep 24 '10 at 4:39

This may not be algebraic enough for you, but after some 150 years of people thinking the plane, catenoid, and the helicoid the only possible examples, Celso Costa found the Weierstrass representation for a new complete minimal surface in $R^3,$ which happened to have three ends. I will try to put the Wikipedia link. I see, if there is punctuation within the Wikipedia name, we need to click on the hyperlink icon (picture of the Earth with an arrow) and do a little extra, but then it works.

http://en.wikipedia.org/wiki/Costa's_minimal_surface

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What about a three-point compactification of the Eisenstein integers? It wouldn't be a group, but it seems fairly natural.

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What about it? What actually is it? To me it doesn't appear fairly natural, but rather daft. – Robin Chapman Sep 22 '10 at 6:42
@Robin: It's the Eisenstein integers, $\mathbb{Z}[\omega]$ with $\omega$ a third root of unity, together with three distinguished elements which, for lack of better names, I will call "$\infty$" = $1+1+\cdots$, "$\omega\infty$" = $\omega\cdot\infty$, and "$\omega^2\infty$" = $\omega^2\cdot\infty$. Conceptually, it's somewhere between $\mathbb{CP}^1$, where there is a continuum of infinite elements, and the extended reals, where there are only two. – Charles Sep 23 '10 at 0:23