# Why is one “$\infty$” number enough for complex numbers?

Can anyone give me a rigorous explanation, why one needs only one number "$\infty$", when dealing with complex numbers, instead of 2 numbers $+\infty, \ -\infty$ like in the case, when dealing with real numbers?

I was told, that by adjoining a point $\{\infty\}$ to $\mathbb{C}$, the newly obtained set becomes a compact one (with respect to the euclidean topology, when $\mathbb{C}$ is viewed as $\mathbb{C}=\mathbb{R}^2$, I presume, though I am not sure), so I would assume, that in the case of the reals, just using one "$\infty$" wouldn't suffices to make it compact ? (Please note, that my knowledge of topology is very limited.)

Are there also other reasons for just using just one "$\infty$"?

What would happen/would it make sense, if we decided to use multiple $\infty$-type numbers, when dealung with complex numbers?

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Well, for one thing, you would want to add one $\infty$ for every direction (that is, one for every real number), not merely two as in the reals... – Arturo Magidin Apr 5 '11 at 19:14
en.wikipedia.org/wiki/Riemann_sphere may also be helpful. – JavaMan Apr 5 '11 at 19:21
For some purposes (for example, in studying the real projective plane), one adds a single "point at infinity" to the ordinary $x$-axis. – André Nicolas Apr 5 '11 at 19:22
You certainly can add just one $\infty$, which is in some sense at both ends simultaneously, to the real line. The resulting object is called the real projective line. – Chris Eagle Apr 5 '11 at 19:24

There are two main types of infinity used in single-variable complex analysis. But first, let me rephrase the question. When you talk about "types of infinity" what you're really talking about is different useful compactifications for the purpose of complex analysis. The reason why one settles for the one-point compactification of the complex plane is it gives you a complex manifold -- the Riemann sphere. So it allows for convienient descriptions of things like Moebius transformations and meromorphic functions.

But other natural compactifications appear in complex analysis, the primary one being the Poincare disc. The core idea is to consider "infinities" to consist of asymptotic directions of curves "heading off to infinity". You can apply this to the complex plane itself, but it's standard to apply it to $\{ z \in \mathbb C : |z|<1 \}$, the open disc. From this perspective, "infinity" is all the points of the form $\{z \in \mathbb C : |z|=1\}$ and the Poincare disc is $\{ z \in \mathbb C : |z|\leq 1 \}$. This is a natural setting for hyperbolic geometry, if your goal is to put it into the language of complex analysis.

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This is kind of misleading. Although the open unit disc and the complex plane are isomorphic as manifolds, they are NOT isomorphic as complex manifolds. (For example, the open disc has bounded non-constant holomorphic functions, and the complex plane doesn't.) The Riemmann sphere is the ONLY compact connected complex manifold which contains the complex plane. – David Speyer Apr 5 '11 at 19:25
@David: I find your interpretation of my post more confusing than my post. But I'll rephrase it a little. – Ryan Budney Apr 5 '11 at 20:54
It's even worse: if a (connected) Riemann surface contains all of C, then it is C itself or the Riemann sphere (that's a straightforward application of the uniformisation theorem). – PseudoNeo Apr 5 '11 at 20:59

There is such a thing as a "one-point compactification" of $\mathbb{R}$; you end up with something that is "like" a circle (where "like" can be made very precise) with a point corresponding to $\infty$.

This is in exactly analogy to viewing the complex numbers with $\infty$ added as a sphere: you do it by stereographic projection. Take a unit circle and put it on the plane so that its center is at $(0,1)$. It is tangent to the $x$-axis at $(0,0)$. To each real number $r$ there corresponds one and only one point in this circle, obtained by taking the line through $(r,0)$ and $(0,1)$, and identifying $(r,0)$ with the second point of intersection of the line with the circle (the first point being $(0,1)$). Every point on the circle except for $(0,1)$ itself corresponds to a real number, so you can think of the point $(0,1)$ as corresponding to "the point at infinity". In fact, this is a natural way of constructing the real projective line.

The counterpart of this is that you can "add infinities" to the complex plane the same way that we "add infinities" to the real line, one for every direction. In the real line, we have the "positive direction" and the "negative direction", which lead to a $+\infty$ and a $-\infty$, respectively. In the complex plane, you would want to add an "infinity" in every direction, so you would have to add an $+\infty_m$ for every real number $m$ that corresponds to the direction from $0$ to $1+mi$, and a $-\infty_m$ corresponding to the direction from $0$ to $-1-mi$ (in the "opposite direction" of $+\infty_m$); plus a $+\infty_v$ corresponding to the direction from $0$ to $i$, and yet another $-\infty_v$ for the direction from $0$ to $-i$. Doing this essentially gives you a closed disc.

Yet a third way is to consider adding a point for every slope, adding an $\infty_m$ corresponding to a line of slope $m$, plus an $\infty_v$ for the vertical lines on the complex plane. In this case, what you get is essentially the real projective plane.

You can do any of the three, but each gives a different kind of structure. Just like the extended reals (with $+\infty$ and $-\infty$) is (usually) the "right" setting to do a lot of calculus, rather than trying to do it in the projective real line, so the Riemann sphere (obtained by adding a single $\infty$ to $\mathbb{C}$) is (usually) the "right" setting to do complex analysis, rather than trying to do it in the closed disc or the projective plane. That is, you "can" any of them, but the "one $\infty$" completion of the complex numbers is more useful for analysis (and in other settings) than the "one $\infty$ per direction" completion or the "one $\infty$ per slope" completion. If you are doing other things (like hyperbolic geometry or projective geometry), then one of the other completions may be more useful.

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I'd say the natural meaning of "an infinity in every direction" would give you an infinity for every angle in $[0, 2 \pi )$. Thus for example there are two imaginary infinities, one along the positive imaginary axis corresponding to the angle $\pi /2$ and one along the negative imaginary axis for the angle $3\pi /2$. This is how you get something homeomorphic to the closed disc. To get the projective plane, you have to identify the infinity in one direction with the infinity in the opposite direction, just as you do to get the projective real line. – Chris Eagle Apr 5 '11 at 19:58
@Chris: Fair enough... That's three ways, then. (-: – Arturo Magidin Apr 5 '11 at 20:04
@Arturo: and we haven't even counted Stone–Čech yet! – Chris Eagle Apr 5 '11 at 20:10
Arturo: Please see my comment on Ryan Budney's answer. The Poincare disc is not a compactification of $\mathbb{C}$ as a complex manifold. – David Speyer Apr 5 '11 at 20:11
I don't think Arturo or I have made any claims of the sort you're asserting, David. IMO you've got too narrow a reading of these posts. – Ryan Budney Apr 5 '11 at 21:01

This is a result of point-set topology. There is a one-point compactification for any locally compact Hausdorff space. I put the definitions you need (for a terse but rigorous run at it) down below as a starter for some words in case you care to go into details. Don't worry if it scares you! If you study mathematics, this will all become clear in time. I just felt like writing. This is all in the following notes; you should go there for a fully fledged proof of the result (p. 60, 3.7.1): http://folk.uio.no/rognes/kurs/mat4500h10/topology.pdf

The statement of the theorem is as follows:

One-point compactification: Let $X$ be a locally compact Hausdorff space, and $Y=X\cup\{\infty\}$. Give $Y$ the topology consisting of the open sets of $X$ together with, for each compact subset $K$ of $X$, sets of the form $Y-K$. Then $Y$ is a compact Hausdorff space. Proof is in the notes above.

Topological space: A topology on a set $X$ is a collection $T$ of subsets of $X$ called open sets such that 1) $T$ contains $X$ and the empty set $\emptyset$, 2) $T$ is closed under arbitrary unions, and 3) $T$ is closed under finite intersections. We say that a subset of $X$ is closed if its compliment is an open set (note that $X$ and $\emptyset$ are both open and closed subsets of $X$). We call the pair $(X,T)$ a topological space, and often denote it by just $X$ and think of the topology $T$ as given.

Here closed under finite intersections means if you take a finite collection of open sets, their intersection is an open set. Similarily for closure under arbitrary unions. This is very different from what we called a closed subset of $X$.

Neighbourhood: If $x\in X$ we say that a neighbourhood of $x$ is any subset of $X$ containing an open set containing $x$. (Beware that some authors define a neighbourhood of $x$ simply as an open set containing $x$; we call these open neighbourhoods.)

Hausdorff: We say that a space $X$ is Hausdorff if for each pair of distinct points $x,y$, there exist disjoint open neighbourhoods of $x$ and $y$.

Compactness: An open covering of $X$ is a collection of open subsets of $X$ such that their union is all of $X$ (note that with these definitions $T$ is the biggest possible open covering of $X$). A space $X$ is called compact if each open covering has a finite open subcovering.

Subspaces: We will want to talk about compact subsets $K$ of $X$, and we can by viewing them as subspaces of $X$ in the subspace topology. That is the set $K$ together with the topology induced by $X$; open sets of $K$ are sets of the form $U\cap K$ with $U$ open in $X$.

Compactness, as you might have guessed, is a very nice property. It is supposed to capture our intuition of being closed, small and bounded. In fact, any compact subset of a Hausdorff space is closed in the technical sense. Also, any compact subset of a metric space has a well-defined diameter. It also gives a finiteness condition that can help us out in all sorts of ways, and it has some nice formal properties: the product of an arbitrary family of compact spaces (with the product topology) is compact (Tychonoff); the image under a continuous map of a compact space, is compact.

A closed and bounded interval of the real line is compact, and (to some, more importantly) all spheres are compact!

Local compactness: We say that a space $X$ is locally compact at $x\in X$ if there is a compact neighbourhood of $x$ (that is, there is a compact subset $K\subseteq X$ and an open subset $V\subseteq K$, with $x\in V\subseteq K\subseteq X$).

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I'd love it if that downvote came with a comment. – Eivind Dahl Aug 19 '11 at 16:05
For the other downvote : You may also compactify $\mathbb R$ by one-point compactification to get the circle, and thus "one" infinity. If you want two infinities for $\mathbb R$, then use Stone-Cech one. Your answer is not bad, and consistent, but does not answer the question at all, at least from my point of view. – Student May 1 '12 at 11:13
I see your point of view as well; there are apparently many angles on a somewhat open question. My approach was to point out that the dichotomy between the case of $\mathbb{R}$ and $\mathbb{C}$ implied by the word 'instead' could be considered somewhat artificial o_O – Eivind Dahl May 2 '12 at 10:43
By the way, you could consider a closed 2-dimensional disc a compactification of $\mathbf{C}$ with a full circle worth of infinities. – Eivind Dahl Oct 3 '14 at 12:03

Adjoining a single $\infty$ is particularly nice for $\mathbb C$ because you get the Riemann sphere, and meromorphic functions on $\mathbb C$ extend to maps of this sphere to itself. But sometimes you do want to talk about, say, sequences going to $\infty$ in a particular direction.

The one-point compactification of $\mathbb R$ works perfectly well, in fact there is a one-point compactification of any locally compact, noncompact Hausdorff space.

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The real numbers also have a $1$-point compactification, the real projective line $\mathbb{RP}^1$, which is homeomorphic to a circle. The plane has other compactifications including $\mathbb{RP}^2$ which has an added $\mathbb{RP}^1$ at infinity.

The $1$-point compactification is particularly natural for the the plane because $\mathbb{R}^2$ (and $\mathbb{C}$) has one topological end. If $S$ is compact, $\mathbb{R}^2 - S$ has $1$ unbounded connected component. By contrast, $\mathbb{R}$ has $2$ ends. It is also nice that you can extend the manifold, smooth, and complex structures from $\mathbb{C}$ to the Riemann sphere.

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I think you'll find very helpful the introductory exposition on points at infinity, projective closure, compactifications, modifications, etc. in Ch. 7, Points at Infinity, in: H. Behnke and H. Grauert: Fundamentals of Mathematics, vol. III, MIT Press, 1983.

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Its like for real numbers u need specfiy in (x,y) form where x varies from -inf to +inf and y varies from -inf to +inf.But if in complex plane u can express each point as (r,theta) where r varies from 0 to inf and theta vaires from 0 to 2pi

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