Represent $\mathbb{R}^3$ as an union of disjoint circles using stereographic projection 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?
 A: 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.
