# Weak topology on CAT(0) spaces

Alexander Lytchak  and  Anton Petrunin

###### Abstract.

We analyze weak convergence on $\mathrm{CAT}(0)$ spaces and the existence and
properties of corresponding weak topologies.

###### Key words and phrases:

Hadamard space, weak convergence, convex subsets

###### 2010 Mathematics Subject Classification:

53C20, 53C21, 53C23

##  1\. Introduction

Weak convergence and coarse topologies in $\mathrm{CAT}(0)$ spaces have
appeared in relation to very different problems and settings in the last
years, see [Jos94, Mon06, KP08, Bac13, Kel14, Str16, BDL17, GN20] and the
survey [Bac18] for an overview. On the other hand, some related fundamental
questions have remained open. This note aims to close some of these gaps.

######  Definition 1.1.

A bounded sequence $(x_{n})$ in a $\mathrm{CAT}(0)$ space $X$ _converges
weakly_ to a point $x$ if for any compact geodesic $c$ starting at $x$, the
closest-point projections $Proj_{c}(x_{n})$ of $x_{n}$ to $c$ converge to $x$.

This notion of convergence (also known as $\Delta$-convergence), introduced in
[Jos94], generalizes weak convergence in Hilbert spaces. It can be defined in
many other natural ways and is suitable for questions concerning the existence
of fixed points and gradient flows, see [Bac18]. The weak convergence
generalizes verbatim to convergence of nets and satisfies natural compactness
and separation properties.

We begin by resolving the question asked by William Kirk and Bancha Panyanak
in [KP08, Question 1] and discussed, for instance, in [Bac14, Bac18, Kel14,
DST16]. The question concerns the existence of a weak topology inducing the
weak convergence. Somewhat surprisingly, the answer is different for sequences
and for general nets. In the case of sequences, the answer is always
affirmative and the proof is general nonsense, not involving geometry:

######  Theorem 1.2.

Let $X$ be a $\mathrm{CAT}(0)$ space. There exists a unique topology
$\mathcal{T}_{\Delta}$ on $X$ with the following two properties:

  * •

A sequence $(x_{n})$ converges in $X$ with respect to $\mathcal{T}_{\Delta}$
to a point $x$ if and only if the sequence is bounded and converges to $x$
weakly.

  * •

The topology $\mathcal{T}_{\Delta}$ is sequential.

Recall, that a topology $\mathcal{T}$ is called _sequential_ if a subset is
$\mathcal{T}$-closed whenever it contains any $\mathcal{T}$-limit point of any
sequence of its elements.

This topology $\mathcal{T}_{\Delta}$, which we want to call the _weak
topology_ , has the following additional properties; see Proposition 3.1,
Corollary 5.3: $\mathcal{T}_{\Delta}$ is sequentially Hausdorff; any
metrically closed, bounded, convex subset of $X$ is
$\mathcal{T}_{\Delta}$-closed, $\mathcal{T}_{\Delta}$-sequentially compact and
$\mathcal{T}_{\Delta}$-compact. However,

######  Proposition 1.3.

There exists a bounded, separable, two-dimensional $\mathrm{CAT}(0)$
simplicial complex $X$ such that $\mathcal{T}_{\Delta}$ is not Hausdorff.

Together with Proposition 1.3, the next theorem implies that, in general,
there is no topology on a $\mathrm{CAT}(0)$ space which induces the weak
convergence of _nets_ :

######  Theorem 1.4.

Let $X$ be a $\mathrm{CAT}(0)$ space and let $\mathcal{T}_{\Delta}$ be the
weak topology defined in Theorem 1.2. For a topology $\mathcal{T}$ on $X$ the
following two conditions are equivalent:

  * •

A bounded net $(x_{\alpha})$ converges to a point $x\in X$ weakly if and only
if $(x_{\alpha})$ converges to $x$ with respect to $\mathcal{T}$.

  * •

The restriction of $\mathcal{T}$ to any closed ball in $X$ is Hausdorff and
coincides with $\mathcal{T}_{\Delta}$.

We discuss $\mathcal{T}_{\Delta}$ in some examples and relate this topology to
another coarse topology, the _coconvex_ topology introduced by Nicolas Monod
in [Mon06]. This coconvex topology $\mathcal{T}_{co}$ on a $\mathrm{CAT}(0)$
space $X$ is defined as the coarsest topology $\mathcal{T}$ on $X$ for which
all metrically closed, convex subsets are $\mathcal{T}$-closed.

Every metrically closed, bounded convex subset of $X$ is
$\mathcal{T}_{co}$-compact and $\mathcal{T}_{co}$-sequentially compact, see
Section 6. The weak topology $\mathcal{T}_{\Delta}$ is finer than the coconvex
topology $\mathcal{T}_{co}$ (Proposition 3.1); these topologies can be
different even for bounded $\mathrm{CAT}(0)$ spaces $X$ (Lemma 4.1). The
topologies $\mathcal{T}_{co}$ and $\mathcal{T}_{\Delta}$ coincide on all
bounded subsets of $X$ if and only if the topology $\mathcal{T}_{co}$ is
sequential and sequentially Hausdorff on bounded convex subsets. Whenever
$\mathcal{T}_{co}$ is Hausdorff on bounded subsets, the topologies
$\mathcal{T}_{co}$ and $\mathcal{T}_{\Delta}$ coincide on bounded subsets.

Whenever the $\mathrm{CAT}(0)$ space $X$ is locally compact, the metric
topology $\mathcal{T}_{metric}$ coincides with $\mathcal{T}_{\Delta}$. On the
other hand, for smooth 3-dimensional Riemannian $\mathrm{CAT}(-1)$ manifolds
or symmetric spaces of higher rank, the coconvex topology $\mathcal{T}_{co}$
can be non-Hausdorff and not first countable, as we will observe in Section 6.
The failure of the Hausdorff property for symmetric spaces has been expected
in [Mon06], a first explicitly confirmed failure of the Hausdorff property for
some $\mathrm{CAT}(0)$ space seems to be the example of the Euclidean cone
over a Hilbert space provided by Martin Kell in [Kel14].

On the other hand, $\mathcal{T}_{co}$ is Hausdorff (and therefore coincides
with $\mathcal{T}_{\Delta}$ and induces the weak convergence of nets) in some
geometric cases:

######  Proposition 1.5.

The topology $\mathcal{T}_{co}$ is Hausdorff in the following cases:

  1. (1)

$X$ is homeomorphic to the plane.

  2. (2)

$X$ is a Riemannian manifold with pinched negative curvature.

  3. (3)

$X$ is a finite-dimensional cubical complex.

The answer we provide to the second point above is a direct consequence of the
construction of convex hulls in manifolds with pinched negative curvature due
to Michael Anderson. While the main construction of [And83] works without
changes in infinite dimensions, it seems not to be sufficient to answer
another question from [Mon06]:

######  Problem 1.6.

Is the coconvex topology $\mathcal{T}_{co}$ Hausdorff on the infinite-
dimensional complex projective space $X=CH^{\infty}$?

Despite Lemma 4.1 and the examples by Alano Ancona [Anc94], see Example 6.4
below, we do not know the answer to the following:

######  Problem 1.7.

Find an example of bounded $\mathrm{CAT}(0)$ spaces for which
$\mathcal{T}_{\Delta}$ is Hausdorff but different from $\mathcal{T}_{co}$.

A natural question is whether for the class of non-locally compact
$\mathrm{CAT}(0)$ spaces appearing in most applications, as in [Mon06, Str16,
BDL17, Cla13, CR13], the weak topology is Hausdorff, at least when restricted
to bounded subsets. Most of the examples are subsumed by or related to the
example in the following question (we refer to [Mon06] for the definition and
properties of the spaces of $L^{2}$-maps):

######  Problem 1.8.

Let $\Omega$ be a probability space and $X$ a locally compact
$\mathrm{CAT}(0)$ space. What are the separation properties of the weak and
the coconvex topologies on the space of $L^{2}$-maps
$\mathcal{L}^{2}(\Omega,X)$?

Also, the following question seems to be very natural in view of the somewhat
cumbersome formulation of Theorem 1.4:

######  Problem 1.9.

If the restriction of the weak topology $\mathcal{T}_{\Delta}$ on any bounded
subset is Hausdorff, does it have to be a Hausdorff topology on $X$?

The paper arose in an attempt to better understand the behavior of convex
subsets and convex hulls in $\mathrm{CAT}(0)$ spaces. The non-Hausdorff
properties of $\mathcal{T}_{co}$ should be related to Gromov’s question:

######  Problem 1.10.

Is the closed convex hull of a compact subset of any $\mathrm{CAT}(0)$ space
compact?

The paper is structured as follows. In Section 3 we recall some basic
properties of the weak convergence and provide a rather straightforward proof
of Theorem 1.2. In Section 4 we provide the example verifying Proposition 1.3.
In Section 5 we prove Theorem 1.4. The main technical point in the proof is a
$\mathrm{CAT}(0)$-version of the theorem of Eberlein–Smulian in functional
analysis, saying that a bounded subset is weakly closed if and only if it is
weakly sequentially closed (Proposition 5.1). In the final Section 6, we
discuss the relations with the coconvex topology.

### Acknowledgments

We would like to thank Tamas Darvas for explaining to us the problem of the
existence of the weak topology on $\mathrm{CAT}(0)$ spaces, to Nicolas Monod
for his interest and helpful exchange about convex subsets of
$\mathrm{CAT}(0)$ spaces and to Miroslav Bačák for helpful comments on a
preliminary version of the paper. Alexander Lytchak was partially supported by
the DFG grant SPP 2026. Anton Petrunin was partially supported by NSF grant
DMS-2005279.

##  2\. Preliminaries

###  2.1. CAT(0)

We assume familiarity with the geometry of $\mathrm{CAT}(0)$ spaces and refer
to [BH99], [AKP19a], [AKP19b]. All $\mathrm{CAT}(0)$ spaces here are by
definition complete and geodesic.

By $d(x,y)=d_{X}(x,y)$ we denote the distance in a metric space $X$. By
$B_{r}(x)$ we denote the closed ball of radius $r$ around the point $x$.

Any bounded subset $A$ in a $\mathrm{CAT}(0)$ space $X$ has a unique
circumcenter $x\in X$ such that for some $r=r(A)\in\mathbb{R}$, the
_circumradius_ of $A$, we have $A\subset B_{r}(x)$ but $A\not\subset
B_{r}(y)$, for any other point $y\neq x$ [BH99].

###  2.2. General topology

We refer to [Eng89] for a detailed explanation of the notions below.

A _directed set_ $I$ is a partially ordered set, such that for any pair
$\alpha_{1},\alpha_{2}\in I$ there exist $\alpha$ with $\alpha\geq\alpha_{1}$
and $\alpha\geq\alpha_{2}$.

A net $(x_{\alpha})$ in a set $X$ is given by a map $\alpha\to x_{\alpha}$
from a directed set $I$ to $X$. We will mostly suppress the directed set $I$
since it will not play any special role.

In a topological space $X$, a net $(x_{\alpha})$ converges to a point $x$ if
for any neighborhood $U$ of $x$ there exists some $\alpha_{0}$ such that, for
all $\alpha\geq\alpha_{0}$, the elements $x_{\alpha}$ are contained in $U$.

In a topological space, convergence of nets can be used as the right
generalization of convergence of sequences from the theory of metric spaces.
For instance, a topological space is Hausdorff if and only if any net can
converge to at most one point in $X$. A topological space is compact if and
only if any net in $X$ has a converging subnet. The closure of a subset
$A\subset X$ consists of all limit points $x$ of all nets $(x_{\alpha})$ with
$x_{\alpha}\in A$.

Replacing in the above statement general nets by sequences, we obtain the
following properties of spaces which will appear below.

A topological space $X$ is called _sequentially Hausdorff_ if any sequence in
$X$ has at most one limit point. Any Hausdorff space is sequentially Hausdorff
but the opposite does not hold.

A topological space $X$ is called _sequentially compact_ if any sequence in
$X$ has a convergent subsequence. A compact space does not need to be
sequentially compact and a sequentially compact space does not need to be
compact.

###  2.3. Basics on weak convergence

Let $X$ be a $\mathrm{CAT}(0)$ space. We stick to the definition of weak
convergence given in the introduction and refer to [Bac18] for other
descriptions and for the explanations and references of the following
properties frequently used below:

Any bounded net in $X$ has at most one weak limit point in $X$. Any subnet of
a weakly converging net converges weakly to the same limit point. Any bounded
sequence (net) has a weakly converging subsequence (subnet).

##  3\. Weak convergence of sequences

In this section we provide the rather straightforward:

###### Proof of Theorem 1.2.

Define the topology $\mathcal{T}_{\Delta}$ as follows. We say that a subset
$A\subset X$ is $\mathcal{T}_{\Delta}$-closed, if, for any bounded sequence
$x_{n}\in A$ weakly converging to a point $x\in X$, we have $x\in A$.

By definition, the empty set and the whole set are
$\mathcal{T}_{\Delta}$-closed. Moreover, any intersection of
$\mathcal{T}_{\Delta}$-closed subsets $A_{\alpha}$ is
$\mathcal{T}_{\Delta}$-closed.

Finally, if $A_{1},...,A_{m}$ are $\mathcal{T}_{\Delta}$-closed and $(x_{n})$
is a bounded sequence in $A_{1}\cup....\cup A_{m}$ weakly converging to $x$,
then we find a subsequence of $(x_{n})$ contained in one of the $A_{i}$. This
subsequence also weakly converges to $x$, therefore $x\in A_{i}$. Hence
$A_{1}\cup...\cup A_{m}$ is $\mathcal{T}_{\Delta}$-closed.

Altogether, this shows that the family of all $\mathcal{T}_{\Delta}$-closed
sets is the family of closed sets of a topology, which we will denote by
$\mathcal{T}_{\Delta}$.

We claim that a sequence $(x_{n})$ in $X$ converges to a point $x$ with
respect to $\mathcal{T}_{\Delta}$ if and only if $(x_{n})$ is bounded and
converges to $x$ weakly.

Firstly, let $(x_{n})$ be bounded and weakly converge to $x$. If $(x_{n})$
does not $\mathcal{T}_{\Delta}$-converge to $x$, we would find a
$\mathcal{T}_{\Delta}$-open subset $U$ containing $x$ and a subsequence
$(x_{m_{n}})$ contained in the complement $A\mathrel{:=}X\setminus U$.
However, $(x_{m_{n}})$ also converges to $x$ weakly, hence, by the definition
of $\mathcal{T}_{\Delta}$-closed subsets, we infer $x\in A$, a contradiction.

On the other hand, let a sequence $(x_{n})$ converge in the
$\mathcal{T}_{\Delta}$-topology to $x$. If $(x_{n})$ is not bounded, we could
find a subsequence $(x_{m_{n}})$ such that $d(x_{1},x_{m_{n}})\to\infty$. Then
the countable set $\\{x_{m_{n}}\\}$ is $\mathcal{T}_{\Delta}$-closed. Hence,
$(x_{m_{n}})$ does not $\mathcal{T}_{\Delta}$-converge to $x$. Therefore,
$(x_{n})$ must be bounded.

Assume that $x_{n}$ does not converge weakly to $x$. Then we find a
subsequence $(x_{m_{n}})$ of $(x_{n})$ which converges weakly to some point
$y\neq x$. Moreover, deleting finitely many elements from the sequence, we may
assume that $x_{m_{n}}$ is not equal to $x$ for all $n$. Then the union $A$ of
all $x_{m_{n}}$ and the point $y$ is $\mathcal{T}_{\Delta}$-closed. Thus, the
complement of $A$ is a $\mathcal{T}_{\Delta}$-open neighborhood of $x$, which
does not contain all but finitely many elements of the sequence $(x_{n})$.
This contradiction proves that $(x_{n})$ weakly converges to $x$ and finishes
the proof of the claim.

The claim and the definition of $\mathcal{T}_{\Delta}$ imply that a subset $A$
of $X$ is $\mathcal{T}_{\Delta}$-closed if every $\mathcal{T}_{\Delta}$-limit
$x\in X$ of a sequence of points in $A$ is contained in $A$. This means that
$\mathcal{T}_{\Delta}$ is sequential.

We have verified the required properties of $\mathcal{T}_{\Delta}$. Let
$\mathcal{T}$ be another sequential topology on $X$, for which a sequence
$(x_{n})$ converges to $x$ if and only if $(x_{n})$ is bounded and weakly
converges to $x$. Then, for $\mathcal{T}$ and $\mathcal{T}_{\Delta}$ the
convergence of sequences coincide. Since both topologies are sequential, this
implies that the properties of being closed with respect to $\mathcal{T}$ and
$\mathcal{T}_{\Delta}$ coincide. Hence, $\mathcal{T}=\mathcal{T}_{\Delta}$. ∎

Basic properties of the weak topology $\mathcal{T}_{\Delta}$ are direct
consequence of the definition and the corresponding properties of weak
convergence:

######  Proposition 3.1.

The weak topology $\mathcal{T}_{\Delta}$ on a $\mathrm{CAT}(0)$ space $X$ is
finer than the coconvex topology and coarser than the metric topology:

$\mathcal{T}_{co}\subset\mathcal{T}_{\Delta}\subset\mathcal{T}_{metric}.$

The topology $\mathcal{T}_{\Delta}$ is sequentially Hausdorff. Any metrically
closed, bounded, convex subset $C\subset X$ is
$\mathcal{T}_{\Delta}$-sequentially compact.

The less trivial statement that any closed, bounded, convex subset is
$\mathcal{T}_{\Delta}$-compact will be derived later in Corollary 5.3.

We finish the section with two simple examples. The first example is a direct
consequence of the definition and the theorem of Hopf–Rinow:

######  Example 3.2.

Assume that the $\mathrm{CAT}(0)$ space $X$ is locally compact. Then
$\mathcal{T}_{\Delta}$ coincides with the metric topology.

The second example is a special case of the fact that the weak convergence as
defined above corresponds to the usual weak convergence in the case of Hilbert
spaces, [Bac18] and Theorem of Eberlien-Smulian, [Whi67], in the case of
Hilbert spaces, saying that a subset is compact in the weak topology if and
only if it is sequentially compact.

######  Example 3.3.

For a Hilbert space $X$, the topology $\mathcal{T}_{\Delta}$ coincides with
the weak topology of the Hilbert space and with $\mathcal{T}_{co}$.

##  4\. Example

We are going to show that $\mathcal{T}_{\Delta}$ can be non-Hausdorff:

_Proof of Proposition 1.3._ Let $Y_{1}$ be a countable family of intervals
$[0,\frac{\pi}{4}]$ glued together at the common boundary point $0$. Fix an
endpoint $b$ among the countably many endpoints of the tree $Y_{1}$. Choose a
countable family of isometric copies of $Y_{1}$ and glue all of them together
by identifying the chosen ”endpoints” $b$ with each other.

The arising space $Y$ is a tree with a special point $p$ (the point at which
all subtrees isometric to $Y_{1}$ are glued together). Point $p$ is the unique
circumcenter of the simplicial tree $Y$. The tree has countably many branches
at $p$ and every point at distance $\frac{\pi}{4}$ from $p$. There are no
other branching points in $Y$; all edges of the tree $Y$ have length
$\frac{\pi}{4}$.

We denote by $E$ the set of endpoints of the tree $Y$ and by $B$ the set of
the branching points different from $p$ (thus the $\frac{\pi}{4}$-sphere
around $p$). Any pair of different points of $B$ lie at distance
$\frac{\pi}{2}$ from each other. Any pair of different points in $E$ either
are at distance $\pi$ and have $p$ as the midpoint or are at distance
$\frac{\pi}{2}$ and have a point from $B$ as their midpoint.

Let $\hat{X}$ denote the Euclidean cone over $Y$. We identify $Y$ with the
unit sphere around the tip $o$ in $\hat{X}$. For a point $y\in Y$ and a number
$\lambda\geq 0$, we denote by $\lambda\cdot y$ the point in the cone $\hat{X}$
on the radial ray in the direction of $y$ at distance $\lambda$ from the
vertex $o$.

For any edge $I$ of $Y$ with endpoints $y_{1},y_{2}$ consider the triangle
$S_{I}$ defined by the points $o,2\cdot y_{1},2\cdot y_{2}$ in $\hat{X}$. The
union of all such triangles is a closed convex subset $X$ of $\hat{X}$. This
subset $X$ is bounded and contains the unit ball $B_{1}(o)$. Moreover, $X$ is
a $2$-dimensional simplicial complex with countably many simplices.

We are going to verify that the points $o$ and $\frac{1}{2}\cdot p$ are not
separated in the weak topology $\mathcal{T}_{\Delta}$ on space $X$.

Firstly, for any pair of different points in $E\subset Y\subset X$ the unique
geodesic in $X$ connecting them either has its midpoint in $o$ (if the points
are at distance $\pi$ in $Y$) or it has its midpoints in
$\frac{1}{\sqrt{2}}\cdot b$ for the unique midpoint $b\in B$ of the
corresponding geodesic in $Y$.

Given any sequence $(x_{n})$ of elements in $E\subset Y\subset X$ with
pairwise distance $\pi$ in $Y$, we see that the convex hull of $\\{x_{n}\\}$
is the union of the geodesic segments $[o,x_{n}]$, thus a tree with a unique
vertex in $o$. In this case, the sequence $(x_{n})$ converges weakly to $o$.

Given any sequence $x_{n}$ of pairwise different elements in $E$ with pairwise
distance $\frac{\pi}{2}$ in $Y$, the convex hull of $\\{x_{n}\\}$ is again a
tree with a unique vertex $\frac{1}{\sqrt{2}}\cdot b$, the common midpoint of
any pair of different points in the sequence $(x_{n})$. Thus, $(x_{n})$ weakly
converges to $\frac{1}{\sqrt{2}}\cdot b$.

Similarly, for any sequence of different point $b_{n}\in B\subset Y\subset X$,
the sequence $(b_{n})$ weakly converges in $X$ to the point
$\frac{1}{\sqrt{2}}\cdot p$. Thus, by rescaling, the sequence
$\frac{1}{\sqrt{2}}\cdot b_{n}$ converges weakly to $\frac{1}{2}\cdot p$.

Assume that $o$ and $\frac{1}{2}\cdot p$ can be separated in
$\mathcal{T}_{\Delta}$. Thus, we find $\mathcal{T}_{\Delta}$-closed subsets
$C_{1}$ and $C_{2}$ such that $o\notin C_{1}$, $\frac{1}{2}\cdot p\notin
C_{2}$ and $C_{1}\cup C_{2}=X$.

By above, $C_{1}$ cannot contain infinitely many points of $E$, which have in
$Y$ pairwise distance $\pi$.

Thus, for all but finitely many branch-points $b\in B\subset Y$ all points in
$E$ at distance $\frac{\pi}{4}$ from $b$ are contained in $C_{2}$. By above,
for any such $b$ we must have $\frac{1}{\sqrt{2}}\cdot b\in B$. Since we have
infinitely many such points, we conclude $\frac{1}{2}\cdot p\in C_{2}$, in
contradiction to our assumption.

Thus, we have verified that $(X,\mathcal{T}_{\Delta})$ is not Hausdorff. ∎

The provided example implies that $\mathcal{T}_{co}$ and
$\mathcal{T}_{\Delta}$ may be different:

######  Lemma 4.1.

The weak topology $\mathcal{T}_{\Delta}$ and coconvex topology
$\mathcal{T}_{co}$ do not coincide on the bounded $\mathrm{CAT}(0)$ space $X$
constructed above.

###### Proof.

Consider the set

$A=E\cup\tfrac{1}{\sqrt{2}}\cdot B\cup\\{\tfrac{1}{2}\cdot p\\}\cup\\{o\\}$

which has appeared above. As explained in the proof of Proposition 1.3 above,
the set $A$ is $\mathcal{T}_{\Delta}$-closed.

We are going to prove that $\frac{1}{4}\cdot p$ is contained in the
$\mathcal{T}_{co}$ closure of $A$. Assuming the contrary, we find finitely
many convex, metrically closed subsets $C_{1},...,C_{n}$ in $X$ which cover
$A$ and do not contain $\frac{1}{4}\cdot p$.

For any $b\in B$, consider the set $E^{b}$ of points in $E$ which are at
distance $\frac{\pi}{4}$ from $b$. Then a counting argument implies that at
least one of the sets $C_{i}$ contains at least 2 points in any of the sets
$E^{b_{1}},E^{b_{2}}$, for different $b_{1},b_{2}\in B$. Then this convex set
$C_{i}$ contains the origin $o$ (as the midpoint of a point in $E^{b_{1}}$ and
$E^{b_{2}}$), the points $\frac{1}{\sqrt{2}}\cdot b_{i}$ and therefore their
midpoint $\frac{1}{2}\cdot p$. Hence, $C_{i}$ also contains the whole geodesic
$[o,\frac{1}{2}\cdot p]$ and, therefore, $\frac{1}{4}\cdot p\in C_{i}$, in
contradiction to our assumption.

Thus, the set $A$ is not $\mathcal{T}_{co}$-closed, finishing the proof. ∎

##  5\. Compactness

The following result can be seen as an analog of the theorem of
Eberlein–Smulian in functional analysis. Unlike Theorem 1.2, here the
$\mathrm{CAT}(0)$ geometry plays an important role several times:

######  Proposition 5.1.

Let $(x_{\alpha})$ be a bounded net in a $\mathrm{CAT}(0)$ space $X$ weakly
converging to a point $x$. Then there exists a sequence
$x_{\alpha_{1}},x_{\alpha_{2}},\dots$ of elements of the net weakly converging
to $x$.

###### Proof.

Replacing the net by a subnet we may assume that the net
$r_{\alpha}\mathrel{:=}d(x_{\alpha},x)$ of real numbers converges to some
$r\geq 0$. If $r=0$, we find some $x_{\alpha_{i}}$ such that
$\lim_{i\to\infty}r_{\alpha_{i}}=0$. Thus, the sequence $x_{\alpha_{i}}$
converges to $x$ in the metric topology, and, therefore, also weakly. Thus, we
may assume $r>0$ and, after rescaling, $r=1$.

We choose inductively $\alpha_{k}\in I$, for $k=1,2,...$, starting with an
arbitrary $\alpha_{1}$. Let the elements $\alpha_{1}\leq...\leq\alpha_{k}$ in
$I$ be already chosen.

For any non-empty subset $S\subset\\{1,...,k\\}$, denote by $m_{S}$ the unique
circumcenter $m_{S}$ of the finite set $\\{x_{\alpha_{i}},i\in S\\}$. Since
the net $(x_{\alpha})$ converges weakly to $x$ and $(r_{\alpha})$ converges to
$1$, we find some $\alpha_{k+1}\geq\alpha_{k}$ with the two following
properties, for any $\alpha\geq\alpha_{k+1}$:

1) $|r_{\alpha}-1|\leq 2^{-k-1}$.

2) For all nonempty $S\subset\\{1,...,k\\}$ the projection
$Proj_{c}(x_{\alpha})$ of $x_{\alpha}$ onto the geodesic $c=[xm_{S}]$ has
distance at most $2^{-k-1}$ from $x$.

Note that any subsequence of the sequence $(x_{\alpha_{i}})$ has also the
properties (1) and (2). We claim that the so-defined sequence
$(x_{\alpha_{i}})$ converges to $x$ weakly. The proof of the claim relies only
on the strict convexity of the squared distance functions and is rather
straightforward. For the convenience of the reader, we present the somewhat
lengthy details.

Assuming the contrary and replacing the sequence by a subsequence we may
assume that the sequence converges weakly to a point $z\neq x$. Set
$\delta\mathrel{:=}d(z,x)$. Choosing yet another subsequence we may assume
that $s_{\alpha_{i}}\mathrel{:=}d(x_{\alpha_{i}},z)$ converge to some $s\geq
0$, for $i\to\infty$.

We set $\varepsilon\leq\frac{\delta^{2}}{10}$ and find some $i_{0}$ such that
$(1-2^{-i_{0}-1})^{2}>1-\varepsilon$ and such that, for all $i\geq i_{0}$,

$|r_{\alpha_{i}}^{2}-1|<\varepsilon\;;\;|s_{\alpha_{i}}^{2}-s^{2}|<\varepsilon.$

Using the weak convergence of $(x_{\alpha_{i}})$ to $z$ and $\mathrm{CAT}(0)$
comparison, we may assume in addition, that for all $i\geq i_{0}$

$d^{2}(x_{\alpha_{i}},x)-d^{2}(x_{\alpha_{i}},z)^{2}\geq\delta^{2}-\varepsilon=9\varepsilon.$

For $j=1,2...$ we consider the point $p_{j}\mathrel{:=}x_{\alpha_{i_{0}+j}}$.
By above, the circumradius $t$ of the countable set $\\{p_{j}\\}$ satisfies

$t^{2}<1-\tfrac{1}{2}\delta^{2}=1-5\varepsilon\,.$

Denote by $0\leq t_{k}\leq t$ the circumradius of the set
$\\{p_{1},...,p_{k}\\}$. We claim that there exists some positive $\rho>0$,
such that $t^{2}_{k+1}-t^{2}_{k}>\rho$ for all $k\geq 1$. Since the sequence
$(t_{k})$ is bounded above by $t$, this would provide a contradiction and
finish the proof.

In order to prove the claim, consider the circumcenter $m_{k}$ of the subset
$p_{1},...,p_{k}$. Thus, $m_{k}$ is the point at which the $2$-convex
function,

$f(y)\mathrel{:=}\max_{1\leq i\leq k}d^{2}(y,p_{i})$

assumes its unique minimum $t_{k}^{2}$. By the $2$-convexity, we deduce

$f(m_{k+1})\geq t_{k}^{2}+d^{2}(m_{k},m_{k+1}).$

On the other hand, $f(m_{k+1})\leq t^{2}_{k+1}$, hence

$t_{k+1}^{2}\geq t_{k}^{2}+d^{2}(m_{k},m_{k+1}).$

By construction of the sequence $x_{\alpha_{i}}$, we have

$d^{2}(p_{k+1},m_{k})\geq(1-2^{-i_{0}-1})^{2}>1-\varepsilon.$

Thus, by the triangle inequality and the fact

$d^{2}(p_{k+1},m_{k+1})\leq t_{k+1}^{2}\leq t^{2}\leq 1-5\varepsilon\,$

we obtain some positive lower bound $\rho>0$ on $d^{2}(m_{k},m_{k+1})$. This
finishes the proof of the claim and of the proposition. ∎

As a consequence, we derive:

######  Lemma 5.2.

If a bounded net $(x_{\alpha})$ in $X$ converges to the point $x$ weakly then
$(x_{\alpha})$ converges to $x$ with respect to the
$\mathcal{T}_{\Delta}$-topology.

###### Proof.

Assume the contrary. Then, replacing the net by a subnet, we find a
$\mathcal{T}_{\Delta}$-open neighborhood $U$ of $x$ which does not contain any
$x_{\alpha}$. Using Proposition 5.1 we find a sequence
$x_{\alpha_{1}},....,x_{\alpha_{k}},...$ of elements of the net converging
weakly to $x$. Then $x$ is contained in the $\mathcal{T}_{\Delta}$-closed set
$X\setminus U$ which contains all elements of the sequence. This contradicts
the definition of $\mathcal{T}_{\Delta}$-closed sets. ∎

Since any bounded net has weakly convergent subnets, we infer:

######  Corollary 5.3.

Every bounded $\mathcal{T}_{\Delta}$-closed subset $A$ of $X$ is
$\mathcal{T}_{\Delta}$-compact.

Now we provide:

###### Proof of Theorem 1.4.

Let $\mathcal{T}$ be a topology on $X$, such that a bounded net $(x_{\alpha})$
weakly converges to $x$ if and only if this net $\mathcal{T}$-converges to
$x$. Since any net has at most one weak limit point and since the Hausdorff
property can be recognized by the uniqueness of limit points of nets, we
deduce that any bounded subset of $X$ is Hausdorff with respect to
$\mathcal{T}$.

Let $A$ be a bounded subset of $X$. By definition, $A$ is $\mathcal{T}$-closed
if and only if it contains all weak limit points of any net $(x_{\alpha})$ of
elements in $A$. From Proposition 5.1, this happens if and only if $A$
contains all weak limit points of any _sequence_ of elements in $A$. Thus, if
and only if $A$ is $\mathcal{T}_{\Delta}$-closed. We infer, that $\mathcal{T}$
coincides with $\mathcal{T}_{\Delta}$ on bounded subsets.

Assume, on the other hand, that the weak topology $\mathcal{T}_{\Delta}$ is
Hausdorff on any ball in $X$. We claim that a bounded net $(x_{\alpha})$
converges weakly to $x$ if and only if $(x_{\alpha})$ converges to $x$ with
respect to $\mathcal{T}_{\Delta}$.

Due to Lemma 5.2, the only if conclusion always holds. On the other hand,
assume that $(x_{\alpha})$ converges to $x$ with respect to
$\mathcal{T}_{\Delta}$ but does not weakly converge to $x$. Replacing
$(x_{\alpha})$ by a subnet we may assume that $(x_{\alpha})$ weakly converges
to another point $y$. Due to Lemma 5.2, this implies that the net
$(x_{\alpha})$ converges to the point $y$ with respect to the topology
$\mathcal{T}_{\Delta}$. But this contradicts the assumption that
$\mathcal{T}_{\Delta}$ is Hausdorff on the bounded ball which contains the net
$(x_{\alpha})$.

This proves the ”if”-direction and finishes the proof of the theorem. ∎

######  Remark 5.4.

Using the considerations above, it is not difficult to prove another form of
Theorem 1.4. Namely, the topology $\mathcal{T}_{\Delta}$ is Hausdorff on any
bounded subset of $X$ (and thus weak convergence of bounded nets is equivalent
to the $\mathcal{T}_{\Delta}$-convergence) if and only if the topology
$\mathcal{T}_{\Delta}$ is _Frechet–Urysohn_ on any bounded set. Recall, that a
topology $\mathcal{T}$ is called Frechet–Urysohn, if the closure of any set
$A$ in this topology is the set of all $\mathcal{T}$-limit points in $X$ of
all _sequences_ contained in $A$.

##  6\. Coconvex topology

The coconvex topology $\mathcal{T}_{co}$ is coarser than
$\mathcal{T}_{\Delta}$, Proposition 3.1. Thus, convergence of sequences (nets)
with respect to $\mathcal{T}_{\Delta}$ implies convergence with respect to
$\mathcal{T}_{co}$. This immediately implies that any bounded,
$\mathcal{T}_{co}$-closed set is $\mathcal{T}_{co}$-compact and
$\mathcal{T}_{co}$-sequentially compact.

######  Proposition 6.1.

The topologies $\mathcal{T}_{co}$ and $\mathcal{T}_{\Delta}$ coincide on all
bounded subsets of a $\mathrm{CAT}(0)$ space $X$ if and only if the topology
$\mathcal{T}_{co}$ is sequential and sequentially Hausdorff on every closed
ball $B_{r}(x)$ in $X$. This happens if $B_{r}(x)$ is
$\mathcal{T}_{co}$-Hausdorff.

###### Proof.

We may replace $X$ by a ball $B_{r}(x)$ and assume that $X$ is bounded. The
only if statement follows from Proposition 3.1.

On the other hand, assume that $\mathcal{T}_{co}$ is sequential and
sequentially Hausdorff on the bounded $\mathrm{CAT}(0)$ space $X$. Due to
Proposition 3.1, the identity map
$Id\colon(X,\mathcal{T}_{\Delta})\to(X,\mathcal{T}_{co})$ is continuous. In
order to prove that the inverse
$Id\colon(X,\mathcal{T}_{co})\to(X,\mathcal{T}_{\Delta})$ is continuous,
consider a $\mathcal{T}_{\Delta}$-closed subset $A$ and assume that $A$ is not
$\mathcal{T}_{co}$-closed. Since $\mathcal{T}_{co}$ is sequential, we find a
sequence $(x_{n})$ in $A$ which $\mathcal{T}_{co}$-converges to a point $x\in
X\setminus A$. Using that $X$ is $\mathcal{T}_{\Delta}$-sequentially compact,
we may replace $(x_{n})$ by a subsequence and assume that $(x_{n})$ converges
to some point $y$ in $X$ with respect to $\mathcal{T}_{\Delta}$. Then, by
Proposition 3.1, the sequence converges to $y$ also with respect to
$\mathcal{T}_{co}$. The assumption that $\mathcal{T}_{co}$ is sequentially
Hausdorff gives us $x=y$. Since $A$ is $\mathcal{T}_{\Delta}$-closed, we
deduce $x=y\in A$. This contradiction implies that
$Id\colon(X,\mathcal{T}_{co})\to(X,\mathcal{T}_{\Delta})$ is continuous, hence
$\mathcal{T}_{\Delta}=\mathcal{T}_{co}$.

Finally, if $\mathcal{T}_{co}$ is Hausdorff on the bounded $\mathrm{CAT}(0)$
space $X$, then the compactness of $\mathcal{T}_{\Delta}$ and the continuity
of the identity map $Id\colon(X,\mathcal{T}_{\Delta})\to(X,\mathcal{T}_{co})$
imply $\mathcal{T}_{\Delta}=\mathcal{T}_{co}$. ∎

In the proof of Proposition 1.5 below, we assume more knowledge of non-
positive curvature than in the rest of this paper. We refer to [LN19] for
properties of geodesically complete $\mathrm{CAT}(0)$ spaces, to [Sch19] for
properties of cubical complexes, and to [And83] and [Bor92] for manifolds of
pinched negative curvature.

###### Proof of Proposition 1.5.

Assume first that $X$ is homeomorphic to the plane $\mathbb{R}^{2}$. Then each
geodesic $\gamma\colon[a,b]\to X$ extends to a geodesic
$\hat{\gamma}\colon\mathbb{R}\to X$, [BH99]. Moreover, by Jordan’s theorem,
$\hat{\gamma}$ divides $X$ into two connected components both having
$\hat{\gamma}$ as their boundaries. The closures of the connected components
are convex. Thus, the open components are $\mathcal{T}_{co}$ open.

In order to prove that $\mathcal{T}_{co}$ is Hausdorff it suffices to find,
for any pair of points $x,y$, some geodesic $\gamma\colon\mathbb{R}\to X$,
such that $x$ and $y$ are in different components of $X\setminus\gamma$. We
connect $x$ and $y$ by a geodesic $\eta$ and take the midpoint $m$ of $\eta$.
We find two points $p^{\pm}$ sufficiently close to $m$ which lie in different
components of $X\setminus\hat{\eta}$ for some extension of $\eta$ to an
infinite geodesic. Then consider a geodesic $\gamma\colon\mathbb{R}\to X$
which contains $p^{+}$ and $p^{-}$. The geodesic $\gamma$ intersect $\eta$
between $x$ and $y$. We infer that $x$ and $y$ lie in different components of
$X\setminus\gamma$. This finishes the proof if $X$ is homeomorphic to a plane.

Assume now that $X$ is a finite-dimensional cubical complex and choose $x,y\in
X$. Taking a sufficiently fine cubical subdivision, we may assume that the
diameter of all cubes is much smaller than the distance between $x$ and $y$.
Then the geodesic between $x$ and $y$ intersects at least one _hyperwall_ in
$X$. Any such hyperwall is a convex subsets dividing $X$ into two connected
and convex components. As above, we deduce that $x$ and $y$ are separated in
$\mathcal{T}_{co}$.

Finally, let $X$ be a Riemannian manifold with pinched negative curvature and
let $x,y\in X$ be arbitrary different points. Fix $r>d(x,y)$ and set
$B=B_{r}(x)$. By Anderson’s construction, [And83], see also [Bor92, Theorem
2.1], we find finitely many closed convex subsets $C_{i}$ in $X\setminus B$,
such that $V\mathrel{:=}X\setminus\cup_{i=1}^{m}C_{i}$ is bounded. Then $V$ is
a $\mathcal{T}_{co}$-open set containing $B$ and contained in some larger
closed ball $B^{\prime}$.

On the compact ball $B^{\prime}$ the topology $\mathcal{T}_{co}$ coincides
with the metric topology, [Mon06, Lemma 17], hence it is Hausdorff. Thus, we
find $\mathcal{T}_{co}$-open subsets $U_{1}$ and $U_{2}$ in $X$ containing $x$
and $y$, respectively, such that $U_{1}\cap U_{2}\cap B^{\prime}$ is empty.
Then $U_{1}\cap U$ and $U_{2}\cap U$ are disjoint
$\mathcal{T}_{co}$-neighborhoods of $x$ and $y$ in $X$. ∎

######  Remark 6.2.

As the proof and the reference to [Bor92] shows, in condition (2) one can
replace the pinching by the assumption that the quotient of the minimal and
maximal curvature in the ball $B_{r}(x_{0})$ around some chosen point $x_{0}$
is at most $2^{\lambda r}$ for some $\lambda\in\mathbb{R}$. Moreover, a closer
look at the proof shows that under the assumptions (1) or (2), the topology
$\mathcal{T}_{co}$ coincides with the metric topology on all of $X$.

We discuss finally two examples showing that the coconvex topology can be
quite strange even for rather regular spaces. Below we denote for a locally
compact $\mathrm{CAT}(0)$ space $X$ by $X^{\infty}$ its boundary at infinity
with its cone topology, [BH99]. Recall that $X^{\infty}$ is compact.

######  Lemma 6.3.

Let $X$ be a locally compact $\mathrm{CAT}(0)$ space. Assume that $X$ is not
bounded and for any closed convex subset $A$ of $X$ different from $X$, the
boundary at infinity $A^{\infty}$ is nowhere dense in $X^{\infty}$. Then the
coconvex topology $\mathcal{T}_{co}$ on $X$ is non-Hausdorff and not first-
countable.

###### Proof.

Since $X^{\infty}$ is a compact space, it is not a countable union of nowhere
dense subsets, by Baire’s theorem.

Therefore, by our assumption, $X$ is not a finite union of closed convex
subsets different from $X$. Thus, any finite intersection of non-empty
$\mathcal{T}_{co}$-open subsets is non-empty. In particular,
$\mathcal{T}_{co}$ is not Hausdorff.

Assume now that $\mathcal{T}_{co}$ is first-countable on $X$, fix an arbitrary
$x\in X$ and a $\mathcal{T}_{co}$-fundamental system of its open neighborhoods
$U_{1},...,U_{n},...$. By definition of $\mathcal{T}_{co}$, we may assume that
each $U_{i}$ is the complement of a finite union of closed convex subsets
$K_{i}^{j}$, not containing the point $x$. Hence, the union of the boundaries
at infinity

$\cup_{i,j}(K_{i}^{j})^{\infty}\subset X^{\infty}$

is not all of $X^{\infty}$. Consider an arbitrary point $z\in X^{\infty}$ not
contained in this union and a ray $\gamma$ in $X$ with endpoint $z\in
X^{\infty}$, such that $x$ is not on $\gamma$. Then $X\setminus\gamma$ is a
$\mathcal{T}_{co}$-open neighborhood of $x$ which does not contain any of the
set $U_{i}$. This contradiction shows that $\mathcal{T}_{co}$ is not first-
countable. ∎

The first example directly follows from Lemma 6.3 above and [Anc94, Theorem B,
Corollary C]:

######  Example 6.4.

There exists a smooth $3$-dimensional $\mathrm{CAT}(-1)$ Riemannian manifold
$X$ for which the coconvex topology $\mathcal{T}_{co}$ is not Hausdorff and
not first-countable.

In the final example, we use some facts about geometry of spherical buildings
arising as the boundary at infinity of symmetric space with their
corresponding _Tits-metric_ , see [KL06], [KL97], [KLP18]. The following
result might be known to specialists, accordingly to Nicolas Monod it was
known to Bruce Kleiner many years ago.

######  Proposition 6.5.

Let $X$ be an irreducible, non-positively curved symmetric space of rank at
least two. Then $X$ satisfies the assumptions, and, therefore, the conclusions
of Lemma 6.3.

###### Proof.

Assume the contrary and consider any closed convex subset $A$ of $X$ such that
the boundary at infinity $A^{\infty}$ of $A$ has non-empty interior in the
$(n-1)$-dimensional sphere $X^{\infty}$; here $n$ is the dimension of $X$.

Thus, in the cone topology, $A^{\infty}$ has dimension $n-1$. Therefore, there
are no totally geodesic symmetric spaces $Y\subsetneq X$ with
$A^{\infty}\subset Y^{\infty}$. On the other hand, if $A^{\infty}=X^{\infty}$
then $A=X$. Thus, we may assume $A^{\infty}\neq X^{\infty}$. Applying [KL06,
Theorem 3.1], we deduce that $A^{\infty}$ is not a _sub-building_ of the
_spherical building_ $X^{\infty}$.

Since $A^{\infty}$ contains an open subset in the cone topology, we find a
non-empty subset $O$ of $A^{\infty}$, open in the cone topology and consisting
of _regular_ points only. If, for some $p\in O$, we find an antipode $q\in
A^{\infty}$ (with respect to the _Tits-distance_) then $A^{\infty}$ contains a
spherical apartment (the boundary of a maximal flat in $X$), as the convex
hull in the Tits-metric of $q$ and a Tits-ball around $p$. By [BL06, Theorem
1.1], this would imply that $A^{\infty}$ is a sub-building, in contradiction
to the statements above. Thus, for no $p\in O$ and $q\in A^{\infty}$ the Tits-
distance between $p$ and $q$ equals $\pi$.

We are going to construct a pair of antipodes $p\in O$ and $q\in A^{\infty}$
and achieve a contradiction. We start with an arbitrary point $p\in O$.

Let $G$ be the isometry group of $X$ (and of $X^{\infty}$) and denote by
$\Delta$ the spherical Coxeter chamber $X^{\infty}/G$ of the spherical
building $X^{\infty}$. Let $\mathcal{P}:X^{\infty}\to\Delta$ be the canonical
projection. Denote by $I:\Delta\to\Delta$ the isometry of the Coxeter chamber
induced by the action of $-Id$ on any apartment of $X^{\infty}$. The map $I$
is an involution, which is the identity map if and only if the Coxeter group
$W$ of $X^{\infty}$ has a non-trivial center (note, that this is the case for
all Weyl groups, which are not of type $A_{m}$, $E_{6}$ or $D_{2m+1}$, see
[Hum72, p.71]).

Consider the orbit $L:=G\cdot p=\mathcal{P}^{-1}(\mathcal{P}(p))\subset
X^{\infty}$. Any element $p^{\prime}\in L$ is contained in a unique Coxeter
chamber $\Delta_{p^{\prime}}$. Consider the set $L^{op}_{p}$ of all elements
$p^{\prime}$ in $L$ which are in an _opposite_ Coxeter to $p$, thus such that
the Coxeter chamber $\Delta_{p^{\prime}}$ through $p^{\prime}$ contains an
antipode of $p$. Then $L^{op}_{p}$ is open and dense in the manifold $G\cdot
p$, see, for instance, [KLP18]. Thus, we find an element $p^{\prime}\in O\cap
L^{op}_{p}$.

If the isometry $I:\Delta\to\Delta$ is the identity (see the discussion
above), then $p^{\prime}$ is an antipode of $p$ and we are done. If $I$ is not
the identity, then looking at an apartment through $p$ and $p^{\prime}$ we
deduce that the Tits-geodesic between $p$ and $p^{\prime}$ contains a point
$q$ which is projected by $\mathcal{P}$ onto $I(p)$. Then $L$ contains all
antipodes of $q$. By convexity, $q\in A^{\infty}$. As above, the set
$L^{op}_{q}\cap O$ of elements in $O$ contained in a chamber opposite to $q$
is not empty. For any such element $p^{\prime}\in L^{op}_{q}\cap O$, the
distance between $q$ and $p^{\prime}$ is $\pi$.

Thus, in both cases we have found a pair of antipodes $p\in O$ and $q\in
A^{\infty}$, finishing the proof. ∎

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