Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have begun to learn complex analysis and have solved a few problems on stereographic projection and Riemann sphere but can't solve the problem in the subject.

Could you help please?

share|improve this question
"few" means "not many", "a few" means "several" -- the "but" seems to imply that you meant "a few", since "few" doesn't form a contrast to not being able to solve a problem. –  joriki Mar 3 '12 at 11:52
@joriki thanks, english is not my native language. –  Sergey Filkin Mar 3 '12 at 13:24
Maybe you could see the stereographic projection as a diffeo from the 3-sphere (minus a point) to $\mathbb{R}^3$, and then transport the Hopf fibration to $\mathbb{R}^3$. That still leaves one manifold homeomorphic to $\mathbb{R}$ and not to $\mathbb{S}_1$, because of the point we eliminated at the beginning. –  D. Thomine Mar 3 '12 at 14:02
I've found a way to represent $\mathbb{R}^3$ as a disjoint union of manifolds diffeomorphic to $\mathbb{S}_1$, but it is not very elegant and it makes no use of the stereographic projection. Are you interested? –  D. Thomine Mar 7 '12 at 15:30
@D.Thomine sure! –  Sergey Filkin Mar 8 '12 at 1:32
add comment

1 Answer

up vote 2 down vote accepted

Since you said you're interested in D. Thomine's answer (and s/he seems to have forgotten about it), here's an answer using no complex analysis or stereographic projection. We will use transfinite induction. We will show $\mathbb{R}^3$ can be partitioned into circles of radius $1$. This idea actually given as an exercise in Ciesielski's book Set Theory for the Working Mathematician. (In the same book, he proves that $\mathbb{R}^2$ can't be partitioned as a union of circles of positive radius.)

To that end, choose a bijection with $\mathbb{R}^3$ and $\mathfrak{c}$, the cardinality of the continuum. (Equivalently, well order $\mathbb{R}^3$ with its minimal well ordering.) The important thing about $\mathfrak{c}$ is that it is a cardinal number, meaning for any ordinal number $\alpha < \mathfrak{c}$, we have $|\alpha| < |\mathfrak{c}| = \mathfrak{c}$ where $|\cdot |$ denotes cardinality.

To begin the induction, let $r_0$ denote the "first" real number. Choose any circle $C_0$ of unit radius which goes through $r_0$.

Now, assume inductively that for all $\beta < \alpha$, we have chosen pairwise disjoint circles of radius $1$ so that all of the $r_\beta$ lie on 1. We wish to extend the induction to $r_\alpha$.

First, if $r_\alpha$ already lies on a prechosen circle, we are done. So, we may assume $r_\alpha$ does not lie on any previously chosen circle of radius 1.

Now, consider all the planes in $\mathbb{R}^3$ passing through $r_\alpha$. It is easy to see that there are $\mathfrak{c}$ such planes. Since, at this point, we have chosen at most $|\alpha|$ circles and $|\alpha|< \mathfrak{c}$, and each circle lies in one plane, there must be a plane $P_\alpha$ which doesn't contain any of our previously chosen circles (though, of course, they may intersect it in $1$ or $2$ points).

Let's just focus on $P_\alpha$ for now. Consider all the circles of radius $1$ contained in $P_\alpha$ passing through $r_\alpha$, which I'll call candidate circles. Again, a not-too-hard counting argument shows there are $\mathfrak{c}$ candidate circles. Each of our previously chosen circles intersects $P_\alpha$ in at most 2 points so there are at most $2|\alpha| = |\alpha| < \mathfrak{c}$ "bad" points in $P_\alpha$ which we must avoid. Any point in $P_\alpha$ is on at most 2 candidate circles, so all "bad" points remove at most $2|\alpha| < \mathfrak{c}$ candidate circles from consideration. But since there are $\mathfrak{c}$ candidate circles, there must be at least one left over. Let $C_\alpha$ be one of these left overs candidates.

By "candidateness", $r_\alpha \in C_\alpha$. Further, $C_\alpha$ cannot intersect any of our previously chosen circles because those intersection points would correspond to "bad" points, which we avoided. Thus, we have continued the induction.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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