Prove that $S$ is a sphere.

Question

Let $S\subset {\mathbb{R}}^3$ with the following properties:

$1.$ For any line $l$:

$|l\cap S|=2$ or $|l\cap S|=1$ or $|l\cap S|=0$

$2.$ For any plane $P$

$P\cap S=\text{circle}$ or $|P\cap S|=1$ or $|P\cap S|=0$

Prove that $S$ is a sphere.

Answer 1

Assume $S$ contains more than one points. Let's say $p_1, p_2 \in S$. Let $P$ be any plane containing $p_1$ and $p_2$, then $C = S \cap P$ is a circle. Choose a coordinate system such that $C$ becomes the unit circle on the $xy$-plane. i.e.

$$C = S \cap P = \{\;(x,y,z) \in \mathbb{R}^3 : x^2+y^2 = 1, z = 0\;\}$$

Now consider the intersection of $S$ with the $xz$-plane: $$C' = S \cap \{\;(x,y,z) \in \mathbb{R}^3 : y = 0\;\}$$

$C'$ contains two points $( \pm 1, 0, 0 )$ from $C$, so $C'$ itself is a circle. It is clear $C'$ intersect the $z$-axis in two points. Let $u_1 = (0,0,d_1)$ and $u_2 = (0,0,d_2)$ be the two intersections and $v = (0,0,\frac{d_1+d_2}{2})$ be the center of $C'$.

Now consider the collection of planes $P_{\theta} : \theta \in [0,\pi)$ containing the $z$ axis. $P_{\theta}$ is obtained from the $xz$-plane by rotating it with respect to the $z$-axis for an angle $\theta$. It is clear $P_{\theta} \cap S$ contains at least 4 points from $S$:

$$u_i = (0,0,d_i)\quad\text{ and }\quad \pm (\cos\theta,\sin\theta,0) \in C$$

So $P_\theta \cap S$ is again a circle. Since three points on a plane determine the circle and using symmetry, we see $P_\theta \cap S$ share the same center $v$ with $P_0 \cap S$. When $\theta$ varies from $0$ to $\pi$, the circles $P_\theta \cap S$ trace out a sphere centered at $v$.

Conclusion: If $S$ contains more than one points, then it is a non-degenerate sphere.

Thoughts

The above discussion is interpreting the circle as a geometric circle in Euclidean geometry. Let's see what can we salvage if we reinterpret the word circle in topological setting. i.e replace the concept circle by a simple Jordan curve.

The existence of $C$ and $C'$ doesn't seems to have any issue. By a suitable choice of coordinate, we can still assume $C$ and $C'$ contains $(\pm 1,0,0)$ and $C'$ continue to intersect the $z$-axis at two points $(0,0,d_1), (0,0,d_2)$ because the $z$-axis split the $xz$-plane in two components. It is also clear the point $v = (0,0,\frac{d_1+d_2}{2})$ lies in the bounded component in the $xz$-plane associated with the Jordan curve $C'$.

One consequence of this is any ray in $xz$-plane starting at $v$ intersect $C'$ and hence $S$ at one and only one point. As we change $\theta$ and following the intersection of $P_\theta$ with $S$, this will set up a bijection $\varphi$ between the unit sphere $S^2$ (direction of rays starting at $v$) and points on $S$. $\varphi^{-1}$ is clearly continuous. It seems to me $\varphi$ is also continuous but without further assumption, I can't justify this part.

To me, $S$ seems to be a topological sphere but I don't have any idea how to prove that.