Circumcenters of bounded sets and fixed sets of isometries in complete CAT(0) spaces

For every $x\in X$ and $y\in Y$, the triangle inequality gives $d(x,y)\le d(x,x_0)+d(x_0,y)\le d(x,x_0)+R$. The nonempty set of distances defining $r_Y(x)$ is therefore bounded above, so [F8] gives its finite supremum and $r_Y(x)\le d(x,x_0)+R$. If $Y$ is finite, the nonempty finite image $\{d(x,y):y\in Y\}$ has a maximum by [F9], so its supremum is attained for every $x$. Also $\mathrm{\mid }d(x,y)-d(x^{\mathrm{\prime }},y)\mathrm{\mid }\le d(x,x^{\mathrm{\prime }})$ for every $y\in Y$, so taking suprema in both directions gives $\mathrm{\mid }{r}_Y(x)-r_Y(x^{\mathrm{\prime }})\mathrm{\mid }\le d(x,x^{\mathrm{\prime }})$ (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric, Upper bound, least upper bound, and strict upper bound); hence $r_Y$ is continuous (Continuity of a map between metric spaces, at a point and globally, in the $\epsilon$-$\delta$ form). The range $\{r_Y(x):x\in X\}$ is nonempty because $Y\mathrm{\ne }\mathrm{\varnothing }$ implies $X\mathrm{\ne }\mathrm{\varnothing }$, and it is bounded below by $0$, so [F8] gives $a:={\inf }_X{r}_Y$ and $0\le a\le r_Y(x_0)$. For fixed $x$ and $r>0$, $Y\subseteq \bar{B}(x,r)$ iff $r_Y(x)\le r$ (Open ball, closed ball and sphere in a metric space); thus $r_Y(x)$ is the infimum of the admissible positive radii at $x$, since every such radius is at least $r_Y(x)$ and every $r_Y(x)+\epsilon$ ($\epsilon >0$) is admissible. Let $\mathcal{R}:=\{r>0:\mathrm{\exists }x\in X,Y\subseteq \bar{B}(x,r)\}$. For every $r\in \mathcal{R}$, $a\le r_Y(x)\le r$ for its witnessing $x$, so $a$ is a lower bound of $\mathcal{R}$. Conversely, for $\epsilon >0$, [F8] gives $x$ with $r_Y(x)<a+\epsilon \mathrm{/}2$; then $r:=r_Y(x)+\epsilon \mathrm{/}2>0$ is admissible and $r<a+\epsilon$. Hence $\inf \mathcal{R}=a$, the radius of $Y$.

Assume the Axiom of Choice. For each $n\in \mathbb{N}$, the set $A_n:=\{x\in X:r_Y(x)<a+1\mathrm{/}(n+1)\}$ is nonempty by the infimum approximation property [F8]; a choice function for $(A_n{)}_{n\in \mathbb{N}}$ yields a sequence $(x_n{)}_{n\in \mathbb{N}}$ with $r_Y(x_n)<a+1\mathrm{/}(n+1)$ for every $n$.

Midpoint inequality and uniqueness (clause (6)). Let $y,y^{\mathrm{\prime }}\in X$, let $m$ be a midpoint of a geodesic segment $[y,y^{\mathrm{\prime }}]$ and let $z\in X$; [F3] gives the stated inequality. For any $q\in X$ put $r:=r_Y(q)$. The nonnegative distances $d(z,q)$, $z\in Y$, have supremum $r$, and their squares have supremum $r^2$: if $r=0$ they all vanish; if $r>0$, for any $\epsilon >0$ use [F8] to find $z$ with $r-\delta <d(z,q)\le r$, where $\delta :=\epsilon \mathrm{/}(2r+1)$, and then $0\le r^2-d(z,q{)}^2<2r\delta <\epsilon$. Now let $c,c^{\mathrm{\prime }}$ be minimizers with value $a$, and let $m$ be the midpoint of their unique geodesic segment, which exists by [F1] and [F2]. For every $z\in Y$, [F3] gives $d(c,c^{\mathrm{\prime }}{)}^2\le 2(d(z,c{)}^2+d(z,c^{\mathrm{\prime }}{)}^2)-4d(z,m{)}^2\le 4a^2-4d(z,m{)}^2$. The infimum over $z\in Y$ of the last right-hand side is $4a^2-4r_Y(m{)}^2$ by the square-supremum fact just proved; since $r_Y(m)\ge a$, this gives $d(c,c^{\mathrm{\prime }}{)}^2\le 0$. Thus $c=c^{\mathrm{\prime }}$, so any minimizer is unique.

Fixed sets (clause (3), first part). Let $\phi$ be an isometry of $X$ (Isometry, isometric embedding, and the subspace metric on a subset). If $\phi (x)\mathrm{\ne }x$, put $\delta :=d(\phi (x),x)\mathrm{/}3>0$. For every $x^{\mathrm{\prime }}$ with $d(x,x^{\mathrm{\prime }})<\delta$, the reverse triangle inequality and isometry property give $d(\phi (x^{\mathrm{\prime }}),x^{\mathrm{\prime }})\ge d(\phi (x),x)-d(\phi (x),\phi (x^{\mathrm{\prime }}))-d(x,x^{\mathrm{\prime }})=d(\phi (x),x)-2d(x,x^{\mathrm{\prime }})>0$. Thus a ball about each point outside $\mathrm{Fix}(\phi )$ lies in its complement, which is open by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement; hence the fixed set is closed. The same argument applies to $C={\bigcap }_i\mathrm{Fix}({\phi }_i)$: if $x\mathrm{\notin }C$, some $i$ has ${\phi }_i(x)\mathrm{\ne }x$, and the ball just constructed avoids $C$; if the family is empty then $C=X$. Now if $x,y\in \mathrm{Fix}(\phi )$ and $\gamma$ is a geodesic segment from $x$ to $y$ (Geodesics and geodesic metric spaces), then $\phi \circ \gamma$ is another geodesic from $x$ to $y$, so it equals $\gamma$ by uniqueness [F2]; hence $\mathrm{Fix}(\phi )$ is convex. Every common fixed set is an intersection of convex sets and is therefore convex.

Isometric invariance (clause (2), first part). Let $\phi$ be an isometry of $X$ with $\phi (Y)=Y$ (Isometry, isometric embedding, and the subspace metric on a subset). Then ${\phi }^{-1}(Y)=Y$, so for every $x\in X$ the substitution $z:={\phi }^{-1}(y)$ gives $r_Y(\phi (x))=\sup \{d(\phi (x),y):y\in Y\}=\sup \{d(x,{\phi }^{-1}(y)):y\in Y\}=\sup \{d(x,z):z\in Y\}=r_Y(x)$; hence $\phi$ maps the set of minimizers of $r_Y$ onto itself.

Proper spaces are complete. Let $(z_n{)}_{n\in \mathbb{N}}$ be a Cauchy sequence in a proper space $X$ (Cauchy sequence in a metric space). For each $k\ge 1$, the Cauchy condition makes the set of indices $N$ satisfying $d(z_m,z_n)\le 1\mathrm{/}k$ for all $m,n\ge N$ nonempty; let $N_k$ be its least member, which exists by [F11] and requires no choice. Then $N_k\le N_j$ for $k\le j$. Every closed ball $\bar{B}(q,r)$ is closed: if $d(q,x)>r$, the radius $\delta :=(d(q,x)-r)\mathrm{/}2>0$ gives $d(q,x^{\mathrm{\prime }})\ge d(q,x)-d(x,x^{\mathrm{\prime }})>r$ whenever $d(x,x^{\mathrm{\prime }})<\delta$, so the complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space). Hence the closed balls $G_k:=\bar{B}(z_{N_k},1\mathrm{/}k)$ and $E:=\bar{B}(z_{N_1},2)$ are closed and bounded, thus compact by properness [F6]. Each $G_k$ lies in $E$: $d(z_{N_k},z_{N_1})\le 1$, so $d(x,z_{N_1})\le d(x,z_{N_k})+d(z_{N_k},z_{N_1})\le 2$ for $x\in G_k$. The sets $G_k\cap E=\{x\in E:d(z_{N_k},x)\le 1\mathrm{/}k\}$ are closed in $(E,d)$ by the same ball argument and have the finite intersection property: the empty finite intersection is $E$, which contains $z_{N_1}$, and for any nonempty finite subfamily, if $K$ is its largest index then $z_{N_K}\in G_k\cap E$ for every index $k\le K$. By [F5] applied in $(E,d)$ there is $p\in {\bigcap }_{k\ge 1}{G}_k$. For $n\ge N_k$, $d(p,z_n)\le d(p,z_{N_k})+d(z_{N_k},z_n)\le 2\mathrm{/}k$; given any rational $\epsilon >0$, [F10] supplies $k$ with $2\mathrm{/}k<\epsilon$, so $z_n\to p$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$) and $X$ is complete (Complete metric space: every Cauchy sequence converges in the space).

Proper case, compactness (clause (5), first part). Assume now that $X$ is proper, and fix $x_0\in X$. For any real $b$, the sublevel set $\{x:r_Y(x)\le b\}$ is closed: if $r_Y(x)>b$, continuity from step 1.1 gives $\delta >0$ such that $d(x,x^{\mathrm{\prime }})<\delta$ implies $\mathrm{\mid }{r}_Y(x^{\mathrm{\prime }})-r_Y(x)\mathrm{\mid }<(r_Y(x)-b)\mathrm{/}2$, and then $r_Y(x^{\mathrm{\prime }})>b$; its complement is therefore open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). In particular $K:=\{x\in X:r_Y(x)\le r_Y(x_0)\}$ is closed. It is bounded: for a fixed $y_0\in Y$, every $x\in K$ satisfies $d(x,y_0)\le r_Y(x)\le r_Y(x_0)$, so $K\subseteq \bar{B}(y_0,r_Y(x_0)+1)$ (Open ball, closed ball and sphere in a metric space). Hence $K$ is compact by properness [F6]. If $r_Y(x_0)=a$, then $x_0\in K$ and ${\inf }_K{r}_Y=a$. If $r_Y(x_0)>a$, then for every $\epsilon >0$ put $\eta :=\min \{\epsilon ,r_Y(x_0)-a\}>0$; by the infimum approximation property [F8] there is $x\in X$ with $r_Y(x)<a+\eta \le r_Y(x_0)$ and $r_Y(x)<a+\epsilon$, so $x\in K$. Thus ${\inf }_K{r}_Y\le a+\epsilon$ for every $\epsilon >0$, while $a\le {\inf }_K{r}_Y$ since $K\subseteq X$; hence $a={\inf }_K{r}_Y$.

The minimizing sequence is Cauchy (clause (1), first part). For $m,n\in \mathbb{N}$, let $m_{mn}$ be the midpoint of the unique geodesic from $x_m$ to $x_n$ ([F1], [F2]). Applying [F3] to each $z\in Y$ and using the square-supremum fact from step 1.3 gives $r_Y(m_{mn}{)}^2\le \frac{1}{2}(r_Y(x_m{)}^2+r_Y(x_n{)}^2)-\frac{1}{4}d(x_m,x_n{)}^2$. Hence $d(x_m,x_n{)}^2\le 2(r_Y(x_m{)}^2+r_Y(x_n{)}^2)-4r_Y(m_{mn}{)}^2\le 2((a+1\mathrm{/}(m+1){)}^2+(a+1\mathrm{/}(n+1){)}^2)-4a^2$, since $r_Y(m_{mn})\ge a$ and step 1.2 bounds the selected radii. The final expression tends to $0$ as $m,n\to \mathrm{\infty }$ by [F10], so $(x_n)$ is Cauchy (Cauchy sequence in a metric space).

The fixed set is CAT(0) and carries a contraction (clause (3), second part). Let $C\mathrm{\ne }\mathrm{\varnothing }$ be the common fixed set of a family of isometries. By convexity from step 1.4, the unique geodesic of $X$ between two points of $C$ lies in $C$ (Comparison triangles, the CAT(0) and CAT(1) inequalities, local CAT, local geodesics and round circles, Geodesics and geodesic metric spaces), so $C$ with the induced metric is geodesic and inherits the CAT(0) comparison inequality; if $X$ is complete, then $C$ is closed by step 1.4 and complete by [F4]. For $p\in C$, define $H_t(x)$ to be the point at fraction $t$ of the unique geodesic from $p$ to $x$. Convexity makes $H_t(x)\in C$, and $H_0$ is constant at $p$ while $H_1={\mathrm{id}}_C$. For $x,y\in C$ and $s,t\in [0,1]$, [F2] gives $d(H_t(x),H_t(y))\le td(x,y)$ and the geodesic parametrization gives $d(H_t(y),H_s(y))=\mathrm{\mid }t-s\mathrm{\mid }d(p,y)$; therefore $d(H_t(x),H_s(y))\le td(x,y)+\mathrm{\mid }t-s\mathrm{\mid }d(p,y)$, which proves joint continuity at every $(x,t)$. Thus $H$ is a homotopy from the constant map at $p$ to the identity, and $C$ is contractible by A nonempty space is contractible if and only if its identity map is nullhomotopic and Nullhomotopic maps and contractible spaces.

Proper case, the center without Choice (clause (5), second part). In the notation of step 2.1, the sets $K_n:=\{x\in K:r_Y(x)\le a+1\mathrm{/}n\}$, $n\ge 1$, are closed subsets of compact $K$ by the sublevel-set argument of step 2.1, and are nested. Each is nonempty because $a={\inf }_K{r}_Y$; the empty finite intersection is $K$, which contains $x_0$, and any nonempty finite intersection is the set with the largest index in it. Thus $(K_n)$ has the finite intersection property (Finite intersection property). By [F5] there is $c\in {\bigcap }_{n\ge 1}{K}_n$. If $r_Y(c)>a$, choose $n$ with $1\mathrm{/}n<r_Y(c)-a$ using [F10]; then $r_Y(c)\le a+1\mathrm{/}n<r_Y(c)$, a contradiction. Since $r_Y(c)\ge a$, we get $r_Y(c)=a$, and step 1.3 gives uniqueness. The family $(K_n)$ is defined by a formula, so this proper-space argument uses no choice principle.

Attainment in the complete case (clause (1), second part). If $X$ is complete, the Cauchy sequence $(x_n)$ from step 2.2 converges to some $c\in X$ (Complete metric space: every Cauchy sequence converges in the space, Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$); continuity of $r_Y$ from step 1.1 gives $r_Y(c)={\lim }_n{r}_Y(x_n)=a$, because $a\le r_Y(x_n)<a+1\mathrm{/}(n+1)$ and [F10] makes the error tend to $0$. Thus the infimum is attained.

Clause (1) concluded. Assume $X$ complete; let $c$ be the point of step 3.2 and $a$ the infimum. Then $d(c,y)\le r_Y(c)=a$ for every $y\in Y$; step 1.1 identifies $a$ with the radius of $Y$. If $Y$ is finite, its nonempty finite image under $y\mapsto d(c,y)$ has a maximum by [F9], so the supremum defining $r_Y(c)$ is that maximum; uniqueness follows from step 1.3. This proves (1).

Clauses (2) and (4) concluded. Let $X$ be complete and let $c$ be the center of $Y$ as in step 4.1, and let $\phi$ be an isometry with $\phi (Y)=Y$; by step 1.5 the isometry $\phi$ permutes the minimizers of $r_Y$, and since $c$ is the unique minimizer by step 4.1, $\phi (c)=c$. Hence a group $G$ of isometries with bounded orbit $Y=G{x}_0$ has $\mathrm{Fix}(G)\mathrm{\ne }\mathrm{\varnothing }$, because every $g\in G$ satisfies $g(Y)=Y$ and fixes $c$. A finite group has a finite nonempty orbit, and its finite set of distances from any point has a maximum by [F9], so that orbit is bounded and it too has a fixed point. By step 2.3 the set $C=\mathrm{Fix}(G)$ is closed, convex, complete and CAT(0) in the induced metric, and contractible; it is a geodesic space whose points are joined by the geodesic segments of $X$ lying in $C$, unique by [F2]. This proves (2) and (4).

Clause (5) concluded, and the proof. Let $X$ be proper. By steps 2.1 and 3.1 the center of every nonempty bounded $Y\subseteq X$ exists and is unique, produced by the finite intersection property and not by the sequence of step 1.2, so no Choice is used; by step 1.6 a proper space is complete. Hence the assertions of (1) hold for $X$ with no use of Choice, by the argument of step 4.1 with the center of step 3.1 in place of the attained minimizer of step 3.2, and the assertions of (2), (3) and (4) follow by the same steps 1.4, 1.5 and 2.3, none of which uses Choice: the only use of the Axiom of Choice in this proof is the extraction of the minimizing sequence in step 1.2, which is needed only when $X$ is complete but not proper. Thus the conclusions of (1)-(4) hold for proper $X$ without Choice.