2. Preliminaries

Here we collect some necessary notions and notation for this note.

Let $(X,d)$ be a metric space. For $x \in X$ and $R \geq 0$ , denote the closed (metric) ball by $B(x,R) = \{ y\in X \;:\; d(x,y) \leq R\}$ . We say that $(X,d)$ is bounded if there exist $x\in X$ and $R \geq 0$ such that $X = B(x,R)$ , and unbounded if it is not bounded. For a subset $A \subset X$ and $R\geq 0$ , denote $\mathcal{N}_R(A) = \{x\in X\;:\; d(x,A) \leq R\}$ the closed $R$ -neighbourhood of $A$ in $X$ . For subsets $A,B \subset X$ , the Hausdorff distance between $A$ and $B$ is

\begin{equation*} d_H(A,B) = \inf \{R \geq 0\;:\; A \subseteq \mathcal {N}_R(B) \text { and } B \subseteq \mathcal {N}_R(A)\}. \end{equation*}

Recall that a path in a metric space $(X,d)$ is a continuous map $\gamma \;:\; [a,b] \to X$ . A path $\gamma$ is called rectifiable if its length

\begin{equation*} \ell (\gamma )\;:\!=\; \sup \!\left \{ \sum _{i=1}^{n} d(\gamma (t_{i-1}),\gamma (t_i)) \;:\; a=t_0\lt t_1 \lt \cdots \lt t_n=b, n\in \mathbb {N} \right \} \end{equation*}

is finite. Usually, it is convenient to change the parameter $t\in [a,b]$ to the standard arc parameter $s\in [0,\ell (\gamma )]$ as follows. Define a map $\varphi \;:\; [a,b] \rightarrow [0,\ell (\gamma )]$ by $t \mapsto s=\ell (\gamma |_{[a,t]})$ , and let $\tilde{\gamma }\;:\;[0,\ell (\gamma )] \to X$ to be the unique path satisfying $\tilde{\gamma } \circ \varphi =\gamma$ . Then, we have $\ell (\tilde{\gamma }|_{[s_1, s_2]}) = |s_1 - s_2|$ for any $0 \leq s_1 \leq s_2 \leq \ell (\gamma )$ .

Now we recall the notions of quasi-isometry and (quasi-)geodesic.

Definition 2.1. Let $(X,d_X), (Y,d_Y)$ be metric spaces and $L\geq 1,C\gt 0$ be constants. An $(L,C)$ -quasi-isometric embedding from $(X,d_X)$ to $(Y,d_Y)$ is a map $f\;:\;X\rightarrow Y$ such that for any $x,x'\in X$ , we have

\begin{equation*}L^{-1} d_X(x,x')-C\leq d_Y(f(x),f(x'))\leq Ld_X(x,x')+C.\end{equation*}

If additionally, we have $\mathcal{N}_C(f(X)) = Y$ , then $f$ is called an $(L,C)$ -quasi-isometry. In this case, we say that $(X,d_X)$ and $(Y,d_Y)$ are $(L,C)$ -quasi-isometric.

We also need the notion of ultrametric space. Recall that a metric space $(X,d)$ is called ultrametric if there exists $\delta \gt 0$ such that for any points $x,y,z \in X$ , we have

\begin{equation*} d(x,y) \leq \mathrm {max} \{d(x,z),d(y,z)\} + \delta. \end{equation*}

The following is due to Gromov:

Recall from Section 1 that for a metric space $(X,d)$ , Gromov considered another metric $d'$ on $X$ defined in (1.2) and noticed that

\begin{eqnarray*} d'(x,y) &\leq & \ln\!(1+d(x,z)+d(z,y)) \\[5pt] &\leq & \ln\!(2+2\mathrm{max} \{ d(x,z),d(y,z) \}) \\[5pt] &=& \mathrm{max} \{ d'(x,z),d'(y,z) \} + \ln 2. \end{eqnarray*}

Combining with Lemma 2.3, we obtain the following:

3. Proof of Theorem 1.2

This whole section is devoted to the proof of Theorem 1.2, which is divided into several parts.

Firstly, we would like to study the property of (quasi-)geodesics for the new metric $d'$ defined in (1.2). To simplify notation, for a path $\gamma$ in $X$ , we denote $\ell (\gamma )$ and $\ell '(\gamma )$ its length with respect to the metric $d$ and $d'$ , respectively.

We need the following lemma:

Proof. Firstly, we assume that $\gamma$ is rectifiable with respect to $d$ , that is, $\ell (\gamma )\lt \infty$ . Note that $\ln\!(1+x) \leq x$ holds for all $x \geq 0$ . Hence for any partition $a=t_0\lt t_1 \lt \cdots \lt t_n=b$ , we have

\begin{equation*} \sum _{i=1}^{n} d'(\gamma (t_{i-1}),\gamma (t_i)) \leq \sum _{i=1}^{n} d(\gamma (t_{i-1}),\gamma (t_i)) \leq \ell (\gamma ), \end{equation*}

which implies that $\ell '(\gamma ) \leq \ell (\gamma )$ . In particular, $\gamma$ is rectifiable with respect to $d'$ .

Conversely, we assume that $\gamma$ is rectifiable with respect to $d'$ , that is, $\ell '(\gamma )\lt \infty$ . Without loss of generality, we can assume that $\gamma$ is parametrised by the standard arc parameter with $a=0$ and $b=\ell '(\gamma )$ . Note that for any $\alpha \in (0,1)$ , there exists $\delta \gt 0$ such that $\alpha x \leq \ln\!(1+x)$ holds for all $x \in [0,\delta ]$ . Given a partition $a=s_0\lt s_1 \lt \cdots \lt s_m=b$ , we choose a refinement $a=t_0\lt t_1 \lt \cdots \lt t_n=b$ such that $|t_{i-1}-t_i|\lt \ln\!(1+\delta )$ holds for all $i$ . Hence, we have

\begin{equation*} d'(\gamma (t_{i-1}),\gamma (t_i)) \leq l'(\gamma \vert _{[t_{i-1},t_i]}) =|t_{i-1}-t_i| \lt \ln\!(1+\delta ), \end{equation*}

which implies that $d(\gamma (t_{i-1}),\gamma (t_i)) \lt \delta$ due to (1.2). Therefore, we obtain

\begin{equation*} \sum _{i=1}^{m} d(\gamma (s_{i-1}),\gamma (s_i)) \leq \sum _{i=1}^{n} d(\gamma (t_{i-1}),\gamma (t_i)) \leq \frac {1}{\alpha } \cdot \sum _{i=1}^{n} d'(\gamma (t_{i-1}),\gamma (t_i)) \leq \frac {1}{\alpha } \cdot \ell '(\gamma ) \end{equation*}

for all $\alpha \in (0,1)$ . Letting $\alpha \to 1$ and taking the supremum of the left hand side, we obtain that $\ell (\gamma ) \leq \ell '(\gamma )$ , which concludes the proof.

As a direct corollary, we obtain the following:

Proof. By assumption, we take two distinct points $x, y \in X$ . If $(X,d')$ is geodesic, we choose a geodesic $\gamma$ between $x$ and $y$ . In particular, $\gamma$ is rectifiable with respect to $d'$ . Hence by Lemma 3.1, we know that $\gamma$ is also rectifiable with respect to $d$ and we have

\begin{equation*} \ell '(\gamma ) = \ell (\gamma ) \geq d(x,y) \gt d'(x,y), \end{equation*}

where the last inequality follows from the assumption that $x\neq y$ . This is a contradiction to the assumption that $\gamma$ is a geodesic between $x$ and $y$ with respect to the metric $d'$ . Hence, we conclude the proof.

Moreover, with an extra hypothesis, the new metric $d'$ cannot even be quasi-geodesic.

To prove Corollary 3.3, we need the following lemma to tame quasi-geodesics. The idea is similar to [Reference Bridson and Haefliger1, Lemma III.H.1.11], but the setting is slightly different.

Careful readers might already notice that in the situation of [Reference Bridson and Haefliger1, Lemma III.H.1.11], we need to assume that the new metric $d'$ on $X$ is geodesic instead of the current setting that the original metric $d$ is geodesic. Although the proof is similar, here we also provide one for convenience to readers.

Proof of Lemma 3.4. Define $\gamma '$ to agree with $\gamma$ on $\Sigma \;:\!=\; \{a,b\} \cup (\mathbb{Z} \cap (a,b))$ , then choose geodesic segments with respect to $d$ joining the images of successive points in $\Sigma$ and define $\gamma '$ by concatenating linear reparameterisations of these geodesic segments.

Let $[t]$ denote the point of $\Sigma$ closest to $t \in [a,b]$ . Note that the $d'$ -distance of the images of successive points in $\Sigma$ is at most $L+C$ , and hence, we have

\begin{equation*} d'(\gamma '(t),\gamma '([t])) = \ln\!\left (1+ d(\gamma '(t),\gamma '([t]))\right ) \leq \ln\!(1+\exp\!(L+C)-1) \leq L+C, \end{equation*}

which implies that $d'_{\!\!H}(\mathrm{Im}(\gamma ), \mathrm{Im}(\gamma ')) \leq L+C$ . Since $\gamma$ is an $(L,C)$ -quasi-geodesic with respect to $d'$ , and $\gamma ([t]) = \gamma '([t])$ for all $t\in [a,b]$ , we have

\begin{equation*} d'(\gamma '(t),\gamma '(t')) \leq d'(\gamma '([t]),\gamma '([t'])) + 2(L+C) \leq L|[t]-[t']| + C +2(L+C) \leq L|t-t'| +3(L+C), \end{equation*}

and similarly, we have

(3.1) \begin{equation} d'(\gamma '(t),\gamma '(t')) \geq \frac{1}{L}|t-t'| -3(L+C) \end{equation}

for all $t,t' \in [a,b]$ . Hence, $\gamma '$ is an $(L,C')$ -quasi-geodesic with respect to $d'$ .

For any integers $s,s^{\prime} \in \Sigma$ with $s \leq s^{\prime}$ , Lemma 3.1 tells us that

\begin{equation*} \ell '(\gamma ' \vert _{[s,s^{\prime}]}) = \ell (\gamma ' \vert _{[s,s^{\prime}]}) = \sum _{k=s}^{s^{\prime}-1} \ell (\gamma ' \vert _{[k,k+1]}) \leq |s-s^{\prime}|(L+C). \end{equation*}

Similarly for any $s,s^{\prime} \in \Sigma$ , we have $\ell '(\gamma ' \vert _{[s,s^{\prime}]}) \leq (|s-s^{\prime}|+2)(L+C)$ . Hence for any $t,t' \in [a,b]$ , we have

\begin{equation*} \ell '(\gamma ' \vert _{[t,t']}) \leq (|[t]-[t']|+2)(L+C) + (L+C) \leq (|t-t'|+4)(L+C). \end{equation*}

Combining with inequality (3.1), we obtain that $\ell (\gamma '|_{[t,t']}) \leq k_1 d'(\gamma '(t),\gamma '(t')) + k_2$ for $k_1,k_2$ defined in (3). Hence, we conclude the proof.

Proof of Corollary 3.3. Assume that $(X,d')$ is $(L,C)$ -quasi-geodesic for some $L \geq 1$ and $C \geq 0$ . For any $x,y\in X$ , choose an $(L,C)$ -quasi-geodesic $\gamma \;:\; [0,T] \rightarrow X$ (with respect to $d'$ ) connecting them. Lemma 3.4 implies that there is a rectifiable path $\gamma '\;:\;[0,T] \rightarrow X$ which is an $(L,3L+3C)$ -quasi-geodesic (with respect to $d'$ ) connecting $x$ and $y$ . Moreover, we have $\ell '(\gamma ') \leq k_1 d'(x,y) + k_2$ for $k_1=L(L+C)$ and $k_2 =(3L(L+C)+4)(L+C)$ . Hence, Lemma 3.1 implies that

\begin{equation*} k_1 d'(x,y) + k_2 \geq \ell '(\gamma ') = \ell (\gamma ') \geq d(x,y) = \exp\!(d'(x,y)) -1. \end{equation*}

Note that $(X,d)$ is unbounded, which implies that $(X,d')$ is also unbounded. Hence taking $d'(x,y) \to \infty$ , we conclude a contradiction and finish the proof.

Next, we move to study the relation of the quasi-ball property and the bounded eccentricity property between the metric spaces $(X,d)$ and $(X,d')$ . Again to save the notation, for $x\in X$ and $r\geq 0$ , we denote $B(x,r)$ and $B'(x,r)$ the closed balls with respect to the metrics $d$ and $d'$ , respectively. For $A \subset X$ and $\delta \geq 0$ , we denote $\mathcal{N}_\delta (A)$ and $\mathcal{N}'_\delta (A)$ the closed $\delta$ -neighbourhood of $A$ with respect to the metrics $d$ and $d'$ , respectively.

Proof. Firstly, we assume that $(X,d')$ has the quasi-ball property, that is, there exists $\delta \geq 0$ such that the intersection of any two balls in $(X,d')$ is $\delta$ -close to another ball (i.e., their Hausdorff distance is bounded by $\delta$ ). Note that there exists $\alpha =\alpha (\delta ) \gt 1$ such that $x \leq \alpha \ln\!(1+x)$ for all $x \in [0,\delta ]$ . This implies that for $A,B \subseteq X$ with $A \subseteq \mathcal{N}'_\delta (B)$ , we have $A \subseteq \mathcal{N}_{\alpha \delta }(B)$ .

Given two balls $B(x,s)$ and $B(y,t)$ in $(X,d)$ , it is clear that $B(x,s) = B'(x,\ln\!(1+s))$ and $B(y,t) = B'(y,\ln\!(1+t))$ . Hence by the assumption, there exists another ball $B'(c,r)$ in $(X,d')$ such that

\begin{equation*} d'_{\!\!H}\!\left ( B'(x,\ln\!(1+s)) \cap B'(y,\ln\!(1+t)), B'(c,r) \right ) \leq \delta. \end{equation*}

It follows from the previous paragraph that in this case, we have

\begin{equation*} d_H\!\left ( B(x,s) \cap B(y,t), B(c, \exp\!(r)-1) \right ) = d_H\!\left ( B'(x,\ln\!(1+s)) \cap B'(y,\ln\!(1+t)), B'(c,r) \right ) \leq \alpha \delta. \end{equation*}

Hence, $(X,d)$ has the quasi-ball property.

The converse holds similarly, using the fact that $\ln\!(1+x) \leq x$ holds for all $x \geq 0$ .

Combining Lemma 2.4, Corollary 3.3, Lemma 3.5 and [Reference Chatterji and Niblo2, Theorem 1], we obtain the following, which concludes one part of Theorem 1.2.

Concerning the bounded eccentricity property, we have the following:

Proof. Given two balls $B'(x,\ln\!(1+r))$ and $B'(y,\ln\!(1+s))$ in $(X,d')$ , we assume that their intersection $Y$ is non-empty. Note that

\begin{equation*} Y=B'(x,\ln\!(1+r)) \cap B'(y,\ln\!(1+s)) = B(x,r)\cap B(y,s) \end{equation*}

is again an intersection of balls in $(X,d)$ . Hence by the assumption, there exist $c,c' \in X$ and $R \geq 0$ such that

\begin{equation*} B(c,R) \subseteq Y \subseteq B(c',R+\delta _0). \end{equation*}

Therefore in $(X,d')$ , we have

\begin{equation*} B'(c,\ln\!(1+R)) \subseteq Y \subseteq B'(c',\ln\!(1+R+\delta _0)). \end{equation*}

Then, the eccentricity of $Y$ in $(X,d')$ is bounded above by

\begin{equation*} \ln\!(1+R+\delta _0) - \ln\!(1+R) = \ln\!\left(1+\frac {\delta _0}{1+R}\right) \leq \ln\!(1+\delta _0) \leq \delta _0, \end{equation*}

which concludes the proof.

Remark 3.8. Readers might wonder whether the converse to Lemma 3.7 holds. Unfortunately, in general this is false. Note that if we have

\begin{equation*} B'(c,\ln\!(1+R)) \subseteq Y \subseteq B'(c',\ln\!(1+R) + \delta _0) \end{equation*}

in $(X,d')$ , then it implies that

\begin{equation*} B(c,R) \subseteq Y \subseteq B(c',\exp\!(\delta _0) \cdot (1+R)-1) \end{equation*}

in $(X,d)$ . Hence, the eccentricity of $Y$ in $(X,d)$ is bounded by $(\!\exp\!(\delta _0)-1)(1+R)$ . However, if $R$ (i.e., the radius of the ball contained in $Y$ ) does not have a uniform upper bound, then neither does the eccentricity of $Y$ .

Remark 3.8 suggests the following partial converse to Lemma 3.7:

Proof. Assume the opposite, that is, there exists $\delta _0\gt 0$ such that the eccentricity of the intersection of any two balls in $(X,d')$ is uniformly bounded by $\delta _0$ . Hence by condition (1), we know that for each $n\in \mathbb{N}$ , there exist $c_n,c'_{\!\!n} \in X$ and $r_n \geq 0$ such that

\begin{equation*} B'(c_n,\ln\!(1+r_n)) \subseteq Y_n \subseteq B'(c'_{\!\!n}, \ln\!(1+r_n)+\delta _0). \end{equation*}

Hence in $(X,d)$ , we have

\begin{equation*} B(c_n,r_n) \subseteq Y_n \subseteq B(c_n,\exp\!(\delta _0) \cdot (1+r_n)-1). \end{equation*}

By condition (2), we know that $r_n \leq M$ for all $n\in \mathbb{N}$ . Therefore, we have

\begin{equation*} \exp\!(\delta _0) \cdot (1+r_n)-1-r_n = (\!\exp\!(\delta _0)-1)(1+r_n) \leq (\!\exp\!(\delta _0)-1)(1+M), \end{equation*}

which is a contradiction to condition (3) in the assumption.

Example 3.10. Now we show that there exists a non-quasi-geodesic metric space which satisfies Gromov’s $4$ -point condition but not the bounded eccentricity property. For example, consider the Euclidean plane $\mathbb{R}^2$ equipped with the Euclidean metric $d_E$ , and let $d'_{\!\!E}$ be the metric on $\mathbb{R}^2$ defined in (1.2). Lemma 2.4 and Corollary 3.3 imply that $(\mathbb{R}^2,d'_{\!\!E})$ is not quasi-geodesic but satisfies Gromov’s $4$ -point condition.

For each $n\in \mathbb{N}$ , take $x_n=(0,0)$ and $y_n=(2n,0)$ and set

\begin{equation*} Y_n = B(x_n,n+1) \cap B(y_n,n+1). \end{equation*}

It is easy to see that in $(\mathbb{R}^2, d_E)$ , the biggest ball contained in $Y_n$ is $B((n,0),1)$ . Moreover, the diameter of $Y_n$ is

\begin{equation*} d\!\left ((n,\sqrt {2n+1}),(n,-\sqrt {2n+1})\right ) = 2\sqrt {2n+1}, \end{equation*}

which implies that the eccentricity of $Y_n$ cannot be uniformly bounded. Therefore applying Lemma 3.9, we conclude the result.

Proof of Theorem 1.2. Combining Corollary 3.6 and Example 3.10, we conclude the proof for Theorem 1.2.

Recall that in [Reference Wenger5] the following weaker form of the bounded eccentricity property was considered: For a metric space $(X,d)$ , there exist $\lambda \gt 0$ and $\delta \gt 0$ such that for any two balls $B_1, B_2 \subseteq X$ with non-empty intersection there exist $z,z' \in X$ and $r \geq 0$ such that

\begin{equation*} B(z,r) \subseteq B_1 \cap B_2 \subseteq B(z', \lambda r+ \delta ). \end{equation*}

Wenger showed in [Reference Wenger5] that this condition also implies Gromov’s hyperbolicity for geodesic metric spaces.

We remark that this weaker form of the bounded eccentricity property does not hold for the space $(\mathbb{R}^2,d'_{\!\!E})$ constructed in Example 3.10 either.

4. Miscellaneous comments

This section derives from comments by Indira Chatterji, who reminds us that our example is quasi-isometric to a horosphere in the hyperbolic space $\mathbb{H}^3$ .

More precisely, we consider the half-space model $\mathbb{H}^3 = \{ (x,t)\;:\; x\in \mathbb{R}^2, t \gt 0 \}$ for the $3$ -dimensional hyperbolic space. Note that horospheres centred at $\infty$ are precisely the horizontal planes $H_k = \{(x,t) \in \mathbb{H}^3\;:\; t = k \}$ with $k \gt 0$ . Endow each $H_k$ with the subspace metric $d_k$ induced by the hyperbolic metric on $\mathbb{H}^3$ . Direct calculations show that

\begin{equation*} d_k\!\left ( (x,k),(y,k) \right ) = 2 \ln\!\left (\sqrt {1+ \frac {d_E(x,y)^2}{4k^2}} + \frac {d_E(x,y)}{2k}\right ), \end{equation*}

where $x,y \in \mathbb{R}^2$ and $d_E$ is the Euclidean metric on $\mathbb{R}^2$ .

The following lemma shows that the space constructed in Example 3.10 is quasi-isometric to the horospheres $H_k$ .

Proof. Consider the map $f\;:\;\mathbb{R}^2 \rightarrow S_k$ by $f(x) = (x,k)$ . Note that

\begin{equation*} 1 + \frac {1}{2k} d_E(x,y) \leq \sqrt {1+ \frac {d_E(x,y)^2}{4k^2}} + \frac {d_E(x,y)}{2k} \leq 1 + \frac {1}{k} d_E(x,y). \end{equation*}

Since $k \geq 1$ , we have

\begin{equation*} d_k((x,k),(y,k)) \leq 2\ln\!\left ( 1+\frac {1}{k} d_E(x,y) \right ) \leq 2\ln\!(1+d_E(x,y)) = 2d'_{\!\!E}(x,y), \end{equation*}

and

\begin{equation*} d_k((x,k),(y,k)) \geq 2\ln\!\left ( 1+\frac {1}{2k} d_E(x,y) \right ) \geq 2d'_{\!\!E}(x,y) - 2\ln\!(2k), \end{equation*}

which concludes the proof.

Recall that $\mathbb{H}^3$ is a hyperbolic geodesic metric space, and hence, it satisfies the quasi-ball property and the bounded eccentricity property according to [Reference Chatterji and Niblo2]. However as shown in Corollary 3.6 and Example 3.10, the space $(\mathbb{R}^2,2d_E')$ does not satisfy either of these two properties for $d_E'$ defined in (1.2). This implies that the quasi-ball property cannot be preserved either by taking subspaces or quasi-isometries (even for $(1,C)$ -quasi-isometries), and the same holds for the bounded eccentricity property. We guess that both of these properties should be preserved under $(1,C)$ -quasi-isometries, which leads to the failure of the permanence by taking subspaces. Unfortunately, meanwhile we cannot exclude the suspicion of taking quasi-isometries. Therefore, we pose the following question to end this note.