Proof [III.] Since $L_n \rightarrow L$ , by Theorem 1.8.7 in [Reference Schneider15], for all $q\in L \cap K$ , there exists $q_n \in L_n \cap K$ such that $q_n \rightarrow q$ . We denote by $\Gamma _n$ the orthogonal hyperplane to $L_n$ passing through $q_n$ , and by $\Gamma $ the hyperplane orthogonal to L passing through q. Since $L_n$ is an axis of symmetry of K, $\Gamma _n \cap K$ is centrally symmetric with center at $q_n$ . In virtue that $L_n \rightarrow L$ and $q_n \rightarrow q$ , we have $\Gamma _n \rightarrow \Gamma $ . Thus, $\Gamma _n \cap K \rightarrow \Gamma \cap K$ . From II., it follows that $\Gamma \cap K$ is centrally symmetric with center at q. Thus, L is an axis of symmetry of K.

Proof Suppose that there exists a hyperplane of symmetry $\Gamma $ of K such that $o\notin \Gamma $ . Then $K=S_{\Gamma }(K)$ . Furthermore, since $S_{\Gamma }(\Omega )\not = \Omega $ and $K\subset \Omega $ , we have $K=S_{\Gamma }(K)\subset S_{\Gamma }(\Omega )$ . Hence, $K\subset \Omega \cap S_{\Gamma }(\Omega )$ . This implies that there exists a sphere of smaller radius that contains K, contradicting the assumption that $\Omega $ is the circumsphere of K.

On the other hand, let L be an axis of symmetry of K, and assume that L does not contain o. Then, $\Omega \not =R_{L}(\Omega )$ . Since $K\subset \Omega ,$ we have that

$$\begin{align*}K=R_{L}(K)\subset R_{L}(\Omega), \end{align*}$$

because L is an axis of symmetry of K. Thus, $K\subset \Omega \cap R_{L}(\Omega )$ . Consequently, K is contained in a sphere of smaller radius than $\Omega $ , which again is a contradiction.

Proof Let $\Gamma $ be a k-axis and let $\Omega $ be a $(d-k)$ -flat such that $\Omega $ is orthogonal to $\Gamma $ and $\Omega \cap K \not =\emptyset $ . For every $(k+1)$ -flat $\Pi , \Gamma \subset \Pi $ , the section $\Pi \cap K$ is symmetric with respect to $\Gamma $ . Hence, the line segment $\Omega \cap (\Pi \cap K)$ has $\Omega \cap \Gamma $ as its midpoint. Varying $\Pi $ and $\Gamma \cap \Pi $ , it follows that $\Omega \cap K$ is centrally symmetric with center at $\Omega \cap \Gamma $ .

Proof of Theorem 1.2

We will show that all the $(k+1)$ sections of K parallel to $\Lambda $ are Euclidean balls, whose centers are situated in a $(d-(k+1))$ -flat orthogonal to $\Lambda $ passing through p, that is, K is a $(k+1)$ -body of revolution.

First, we prove that K is centrally symmetric. Recall that $\Lambda $ is a $(k+1)$ -flat, and thus $\Lambda ^{\perp }$ is $(d-(k+1))$ -flat. For every $(d-k)$ -flat $\Omega $ , $\Lambda ^{\perp } \subset \Omega $ , we consider the k-flat $\Omega ^{\perp }\subset \Lambda $ . By hypothesis, $\Omega ^{\perp }$ is a k-axis of symmetry of K. From Lemma 5.1, it follows that $\Omega \cap K$ is centrally symmetric with center at p. Varying $\Omega $ , $ \Lambda ^{\perp } \subset \Omega $ , we obtain that K is centrally symmetric with center at p.

We take a system of coordinates with p as the origin. We will show next that every $(d-k)$ -flat $\Omega $ , $\Lambda ^{\perp } \subset \Omega $ , is a $(d-k)$ -axis of symmetry of K. Observe that this fact is, in a sense, dual to our hypothesis.

Let $\Omega $ be a $(d-k)$ -flat such that $\Lambda ^{\perp } \subset \Omega $ , and let $\Gamma \subset \Lambda $ be a k-flat passing through $p\in \Gamma $ and orthogonal to $\Omega $ . We need to show that for every $(d-k+1)$ -flat $\Sigma $ containing $\Omega $ , $\Omega $ is a $(d-k)$ -plane of symmetry of $\Sigma \cap K$ . Let $\Sigma $ be a ${(d-k+1)}$ -flat with $\Omega \subset \Sigma $ and let $\Phi \subset \Sigma $ be a $(d-k)$ -flat parallel to $\Omega $ (thus, orthogonal to $\Gamma $ ) such that $ \Phi \cap K\not = \emptyset $ . We take $x\in \Phi \cap \text {bd} K$ . We claim that there exists a line $\Delta $ perpendicular to $\Omega $ , with $x\in \Delta $ and $\Delta \cap \text {bd} K=\{x,y\}$ , where $y\in (-\Phi ) \cap \text {bd} K$ , and $\Delta \subset \Sigma $ . By Lemma 5.1, the sections $\Phi \cap \text {bd} K$ and $-(\Phi \cap \text {bd} K)$ are centrally symmetric. We denote their centers by $c, -c$ , respectively. Notice that $c,-c\in \Gamma $ and

(5.1) $$ \begin{align} -(\Phi \cap \text{bd} K)=(\Phi \cap \text{bd} K)-2c. \end{align} $$

(Since $-c+(c-x)=-x\in -(\Phi \cap \text {bd} K)$ , we get $-c-(c-x)=x-2c \in -(\Phi \cap \text {bd} K)$ .) Thus, letting $L(p,c)$ be the line passing through the points $p,c$ , and defining $\Delta =x+L(p,c)$ and $y=x-2c$ , it follows from (5.1) that $y\in -(\Phi \cap \text {bd} K)$ , and we obtain what we have claimed.

Now, let $\Lambda '$ be a $(k+1)$ -plane parallel to $\Lambda $ such that $\Lambda ' \cap K\not =\emptyset $ . Consider a line $L\subset \Lambda '$ passing through $\Lambda ' \cap \Lambda ^{\perp }$ and let $\Omega $ be a $(d-k)$ -flat containing $\Lambda ^{\perp }$ such that $L=\Omega \cap \Lambda '$ . In order to prove that L is a line of symmetry of $\Lambda ' \cap K$ , we will show that, for every two-dimensional plane $\Pi \subset \Lambda '$ containing L, the section $\Pi \cap K$ has L as a line of symmetry. Let $\Pi \subset \Lambda '$ be a plane with $L\subset \Pi $ , and let $M\subset \Pi $ be a line perpendicular to L. Denote by $\Sigma $ the $(d-k+1)$ -plane defined by $\Omega $ and M. By virtue that $\Omega $ is a $(d-k)$ -axis of symmetry of K, $\Omega $ is a $(d-k)$ -plane of symmetry of $\Sigma \cap K$ . It follows from the choice of M that $\Pi \cap K$ has L as a line of symmetry (notice that $M\perp \Omega $ because $M\subset \Pi \subset \Lambda '$ and $\Lambda ^{\perp }\subset \Omega $ ). From property (A) mentioned in the introduction, we conclude that $\Lambda ' \cap K$ is a Euclidean ball with center at $\Lambda ' \cap \Lambda ^{\perp }$ .

Proof Let v be a unit vector parallel to $L'$ . For each $u\in \mathbb {S}^{k-1} \subset \Pi $ , we denote by $\alpha (u)$ the angle between the vectors u and v. Let $\delta _1$ and $\delta _2$ be the minimum and the maximum of $\alpha (\cdot )$ . Since $L'\subset \mathbb {E}^{d} \backslash \Pi $ and $L'\not =\Pi ^{\perp }$ , it follows that $\delta _1>0$ and ${\delta _2<\pi /2}$ . By virtue of the continuity of $\alpha (u)$ and the compactness of $\mathbb {S}^{k-1}$ , we have that $\alpha (\mathbb {S}^{k-1})=[\delta _1,\delta _2]$ . We denote by F the subset of irrational multiples of $\pi $ in $[\delta _1,\delta _2]$ and by E the set $\alpha ^{-1}(F)$ . Since F is dense in $[\delta _1,\delta _2]$ , the set E is dense in $\mathbb {S}^{k-1}$ . For each $u\in E$ , we denote by $L(u)$ the line through the origin with direction u and by $\Pi (u)$ the plane $\text {aff} \{u,v\}$ . Since $u\in E$ , we have that $\alpha (u)\in F$ and, consequently, $\Sigma (L(u), L')$ is a set of axes of symmetry of K in $\Pi (u),$ which is dense in $\Omega (L(u), L')$ . Thus, by Lemma 3.2 [III], every line in $\Pi (u)$ through o is an axis of symmetry of K.

Let w be a unit vector in span $\{\Pi ,L'\}$ . We will show that there exists an axis of symmetry $L_w$ of K such that $L_w=\mathrm{span}\{w\}$ . Let $e \in \mathbb {S}^{k-1}$ such that $\mathrm{span}\{e\}=\Pi \cap \mathrm{span}\{v,w\}$ . Since E is dense in $\mathbb {S}^{k-1}$ , there exists a sequence $\{e_i\}\subset E$ such that $e_i \rightarrow e$ , when $i \rightarrow \infty $ . It follows that $\Pi (e_i) \rightarrow \mathrm{span}\{v,w\}$ . Hence, we can find a sequence of lines $\{\Delta _i\}$ , $\Delta _i \subset \Pi (e_i)$ , such that $\Delta _i \rightarrow L_w=\mathrm{span}\{w\}\subset \mathrm{span}\{v,w\}$ . By the argument in the previous paragraph, every line $\Delta _i$ is an axis of symmetry of the body K. Therefore, by Lemma 3.2 [III], $L_w$ is an axis of symmetry of K.

Proof of Theorem 1.3

In virtue of Lemma 3.3, there exists a point $o\in \mathbb {E}^{d}$ such that $o\in L_n$ , for all n. We take a coordinate system having o as the origin. We assume that $V=\mathrm{span}\{L_1,L_2,\ldots \}$ has dimension k, $k\geq 2$ .

First, we consider the case when there exists a two-dimensional plane $H\subset V$ and an infinite subsequence $L_{n_i}$ of $\{L_n\}$ such that $\{L_{n_i}\} \subset H$ , for all i. By the compactness of $\mathbb {S}^{2}$ , there is an infinite subsequence of $L_{n_i}$ , which will be denote again by $L_{n_i}$ , and a line $L\subset H$ through o such that $L_{n_i} \rightarrow L$ , when $i \rightarrow \infty $ . By Lemma 3.2 [III], L is an axis of symmetry of K.

We will show that every line $\Delta \subset H$ , $o \in \Delta $ , is an axis of symmetry of K. By Lemma 3.2 [III], it is enough to find a sequence of axes of symmetry $\{\Delta _n\}\subset H$ of K such that $\Delta _n \rightarrow \Delta $ . First of all, suppose that, for all i, $\Sigma (L,L_{n_i})$ is an $f(n_i)$ -star of lines for some integer $f(n_i)$ . Since $L_{n_i} \rightarrow L$ , it follows that $f(n_i) \rightarrow \infty $ . Thus, given $\epsilon> 0$ , there exists an integer $n_i$ such that the difference between the angle of $\Delta $ and the angle of some line $\Delta _{k(n_i)}$ in $\Sigma (L,L_{n_i})$ is less than $\epsilon $ . On the other hand, if for some $n_i$ , the angles corresponding to the lines in the star $\Sigma (L,L_{n_{i}})$ determine a dense set in $[0,2\pi ]$ , we obtain immediately the existence of a sequence $\{\Delta _n\}\subset \Sigma (L,L_{n_i})$ of axes of symmetry of K such that $\Delta _n \rightarrow \Delta $ .

In particular, we have proven the case $k=2$ of the theorem. Since every line $L\subset V$ , through o, is an axis of symmetry of the body K, by Theorem 1.1, we conclude that K is a 2-body of revolution.

We consider the case in which there is no plane $H\subset V$ containing an infinite subsequence of $\{L_n\}$ . Then, letting $\Pi _n$ be the plane $\mathrm{span}\{L,L_{n}\}$ , we have that the sequence of planes $\{\Pi _n\}$ is infinite, hence there exists a limiting plane $\Pi \subset V$ for a subsequence, which we will still denote as $\Pi _{n}$ . If for some integer $n_0$ , the angles corresponding to the lines in $\Sigma (L,L_{n_0})$ are a dense set in $[0,2\pi ]$ , we have that, for each line $\Delta \subset \mathrm{span}\{L,L_{n_0}\},$ there exists a sequence $\{\Delta _n\}\subset \Sigma (L,L_{n_0})$ of axes of symmetry of K, such that $\Delta _n \rightarrow \Delta $ . Thus, by Lemma 3.2 [III], $\Delta $ is an axis of symmetry of the body K, and we obtain that all lines in $\Pi $ are axes of symmetry of K. On the other hand, if we do not have a dense set of angles, then every $\Sigma (L,L_n)$ is an $f(n)$ -star of lines, for some integer $f(n)$ . Since the sequence of planes $\{\Pi _n\}$ is infinite, the sequence $\{f(n)\}$ is also infinite. In virtue that $L_n \rightarrow L$ , it follows that $f(n) \rightarrow \infty $ . Let $\Delta \subset \Pi $ be a line, $x\in \Delta $ . Again, we will show that there exists a sequence $\{\Delta _n\}$ of axes of symmetry of K such that $\Delta _n \rightarrow \Delta $ . Since $\Pi _n \rightarrow \Pi $ , there exists a sequence of lines $\{\Gamma _n \}$ such that $\Gamma _n \subset \Pi _ n $ and $\Gamma _n \rightarrow \Delta $ , that is, given $\epsilon> 0$ , there exists $N_1$ such that if $n> N_1$ , then $\delta _p(\Gamma _n,\Delta )< \epsilon /2$ . On the other hand, since $f(n) \rightarrow \infty $ , given $\epsilon> 0$ , there exists $N_2$ such that $n> N_2$ , there exists a line of $\Sigma (L,L_n)$ , say $\Delta _{k(n)}$ , such that $\delta _p(\Delta _{k(n)}, \Gamma _n)< \epsilon /2$ . Taking $n> \max \{N_1,N_2\}$ , then $\delta _p(\Delta _{k(n)}, \Delta )\leq \delta _p(\Delta _{k(n)}, \Gamma _n)+ \delta _p( \Gamma _n, \Delta )< \epsilon /2 + \epsilon /2=\epsilon $ , that is, $\Delta _{k(n)} \rightarrow \Delta $ .

In the previous paragraph, we have proven that there exists a plane $\Pi \subset V$ such that every line in $\Pi $ , through o, is an axis of symmetry of K. Since the case $k=2$ has already been established, let $3\leq k \leq d-1$ . Since $k>2$ , there exists an integer $n_1$ such that $L_{n_1}$ is not contained in $\Pi $ and, by condition (iii) of Theorem 1.3, $L_{n_1}$ is not perpendicular to $\Pi $ . It follows from Lemma 6.1 that every line in $\mathrm{span}\{\Pi , L_{n_1}\}$ , passing through o, is an axis of symmetry of K. If $d=3$ , then $k=3$ and $\mathbb {E}^3=\mathrm{span}\{\Pi , L_{n_1}\}$ and, consequently, K is a sphere by property (A) in the introduction. If $d>3$ and $k=3$ , then, by Theorem 1.1, K is a 3-body of revolution. If $k>3$ , using Lemma 6.1 repeatedly, we can assume the existence of axes of symmetry $\{L_{n_1}, L_{n_2},\ldots ,L_{n_{(k-2)}}\} \subset \{L_n\}$ of K such that every line $L'\subset V=\mathrm{span}\{\Pi , L_{n_1}, L_{n_2},\ldots ,L_{n_{(k-2)}}\}$ through o is an axis of symmetry of K. Thus, if $k\leq d-1$ , then K is a k-body of revolution by Theorem 1.1. If $k=d$ , then K is a sphere and the proof of Theorem 1.3 is now complete.

Proof of Theorem 1.4

In virtue of the hypothesis, there exists an integer k, $1\leq k \leq d-2$ , and a k-flat $\Gamma $ such that the hyperplanes of symmetry $\Pi _i$ of K satisfy (1.1). Let $\Delta $ be a $(d-k)$ -flat orthogonal to $\Gamma $ and such that $\Delta \cap K\not = \emptyset $ . By (1.1), $\Delta $ is orthogonal to $\Pi _i, i=1,2,\ldots $ . Consequently, $\Delta \cap \Pi _i$ is a $(d-k-1)$ -flat of symmetry of $\Delta \cap K$ . We claim that for $1\leq t \leq d-k-1$ , there is no t-flat $\Psi \subset \Delta $ and no subsequence $\Delta \cap \Pi _{i_j}$ of $\Delta \cap \Pi _i$ such that $ \Psi \subset \Pi _{i_j}, j=1,2,\ldots $ Otherwise, we would have that $\text {aff}\{ \Gamma , \Psi \}\subset \Pi _{i_j}$ and since $\dim (\text {aff}\{ \Gamma , \Psi \})>k,$ we would contradict the maximality of k. By Proposition 3.4, $\Delta \cap K$ is a sphere with center at $\Gamma $ , that is, K is a $(d-k)$ -body of revolution.

Proof of Theorem 1.5

Let $\Pi $ be a hyperplane, and $p\in \Pi $ . We will show that for every hyperplane $\Gamma $ with $p\in \Gamma $ and $\Gamma \not =\Pi $ , the sections $\Pi \cap K $ and $\Gamma \cap K$ are congruent, that is, there exists an orthogonal transformation $\Omega : \mathbb {E}^{d} \rightarrow \mathbb {E}^{d}$ such that $\Omega (\Pi \cap K) =\Gamma \cap K$ . We denote by $\Delta $ the subset of $\mathbb {S}^{d-1}$ defined as follows: $u\in \Delta $ if there exists a hyperplane $\Sigma $ , $p\in \Sigma $ , whose corresponding unit normal vector is u and $\Sigma \cap K$ and $\Pi \cap K$ are congruent. We are going to show that $\Delta =\mathbb {S}^{d-1}$ . We denote by w the unit normal vector of $\Pi $ . Let $L\subset \Pi $ be a $(d-2)$ -plane, $p\in L$ . In virtue that L is a $(d-2)$ -axis of rotation of order k, there are hyperplanes $\Sigma _1$ , $\Sigma _2$ with unit normal vectors $v_1,v_2$ , respectively, such that $\Pi \cap K$ can be obtained from $\Sigma _1 \cap K$ and $\Sigma _2 \cap K$ after applying two rotations, both with axis L, one by an angle $2\pi /k$ and the other by an angle $2(k-1)\pi /k$ . Since the boundary of K is a continuous surface, varying L, always contained in $\Pi $ , we have that the collection of unit normal vector of the hyperplanes $\Sigma $ is a $(d-2)$ -sphere $\tau _{w}\subset \mathbb {S}^{d-1}$ with center at w and each vector $v\in \tau _{w}$ makes an angle $2\pi /k$ with w. It is clear that $\tau _{w}\subset \Delta $ . Applying the same argument, it follows that if v is in $\Delta $ , then $\tau _v \subset \Delta $ . Thus,

(7.1) $$ \begin{align} \bigcup_{v\in \tau_{w}} \tau_{v}\subset \Delta. \end{align} $$

We denote by $B_w$ the interior of the spherical cap with boundary $\tau _{w}$ in $\mathbb {S}^{d-1}$ . In virtue that

$$\begin{align*}B_w=\bigcup_{v\in \tau_{w}} \tau_{v}, \end{align*}$$

we get from (7.1) that $B_w \subset \Delta $ . Finally, we observe that the collection of sets $\{B_v:v\in \tau _w\}$ covers the sphere. Hence, $\Delta = \mathbb {S}^{d-1}$ , that is, any two sections of K passing through p are congruent. By the Süss–Schneider theorem (see [Reference Schneider14, Reference Süss16]), we conclude that K is a sphere.