Krzysztof Krupiǹski pointed out to me that there is a mistake in the statement of [1, Lemma 4.9], and a gap in the proof of [1, Theorem 4.13]. In this corrigendum, the right version of the lemma and a correct proof of the theorem are presented. We use most of the definitions and notation conventions from [1] without explicitly formulating them.

First of all, in the statement of Lemma 4.9 there should be an additional assumption that balls ${B}_{0}$, ${B}_{1}$ are disjoint. Then the lemma is true, its original proof goes through without any changes, and it still can be used in the proof of Theorem 4.13, which is the only place in [1], where Lemma 4.9 is applied.

#### ([1, Lemma 4.9])

*Let ${B}_{\mathrm{0}}\mathrm{,}{B}_{\mathrm{1}}$ be disjoint ‘open’ balls in an ultrametric space X. Suppose that the equivalence classes they determine in ${E}_{{B}_{\mathrm{0}}}\mathrm{,}{E}_{{B}_{\mathrm{1}}}$, respectively, are agreeable. If ${x}_{\mathrm{0}}\mathrm{\in}{B}_{\mathrm{0}}\mathrm{,}{x}_{\mathrm{1}}\mathrm{\in}{B}_{\mathrm{1}}$, then*

$d({x}_{0},{x}_{1})=dist([{x}_{0}],[{x}_{1}]).$

#### Proof.

Suppose that $d({x}_{0},{x}_{1})>dist([{x}_{0}],[{x}_{1}])$, and put $r=d({x}_{0},{x}_{1})$, ${C}_{0}={B}_{<r}({x}_{0})$, ${C}_{1}={B}_{<r}({x}_{1})$. Then there exists $\varphi \in Iso(X)$ such that $\varphi [{C}_{0}]={C}_{1}$, so ${E}_{{C}_{0}}={E}_{{C}_{1}}$, and ${E}_{{B}_{0}},{E}_{{B}_{1}}\le {E}_{{C}_{0}}$. However, ${C}_{0}\ne {C}_{1}$, and ${B}_{0}\le {C}_{0}$, ${B}_{1}\le {C}_{1}$. Therefore ${B}_{0},{B}_{1}$ cannot be agreeable. ∎

In the proof of Theorem 4.13, it is claimed that if *B* is an agreeable family satisfying the boundedness condition, then, for every chain *L* of balls coming from equivalence classes in *B*, either *L* is bounded from below in *B*, or it must contain a sequence of balls with diameters converging to 0. This is not true in general, so the presented construction of a *d*-transversal $Y\in \bigcap B$ is not quite correct. The remaining part of the corrigendum is devoted to a correct proof of this theorem.

Let *X* be a *W*-space with metric *d*. Recall that every ball *B* in *X* gives rise to an equivalence relation ${E}_{B}$ on the set ${S}_{X}$ of all *d*-transversals in *X*, defined by

${Y}_{0}\cong {Y}_{1}(\mathrm{mod}{E}_{B})\iff \exists \varphi \in Iso(X)({Y}_{0}\cap \varphi [B]\ne \mathrm{\varnothing}\mathrm{and}{Y}_{1}\cap \varphi [B]\ne \mathrm{\varnothing}).$

Suppose that *D* is an equivalence class of an equivalence relation ${E}_{B}$ defined as above by an ‘open’ ball *B*. Then *D* determines a unique ball in *X*, so we will identify such equivalence classes with the balls determined by them. In particular, by an agreeable family of balls (or a family of balls satisfying the boundedness condition), we will mean a family of balls determined by an agreeable family of equivalence classes (or a family of classes satisfying the boundedness condition.)

In order to prove the theorem, we need two auxiliary results.

*Let be X a W-space, and let B be an agreeable family of balls in X which satisfies the boundedness condition. Let $C\mathrm{\in}B$. Then one of the following must hold:*

- (i)
*for every*${x}_{C}^{\prime}\in C$*, and every ball*$D\in B$*with*$D\subseteq C$*there is a ball*$E\in B$*such that*$E\subseteq D$*, and*$E\cap [{x}_{C}^{\prime}]=\mathrm{\varnothing}$, - (ii)
*there is a decreasing sequence*$\{{C}_{n}\}$*of balls in**B**such that*${C}_{n}\subseteq C$*, and*$diam({C}_{n})\to 0$, - (iii)
*there is*$D\in B$*with*$D\subseteq C$*which is inclusion-minimal in**B*.

#### Proof.

Suppose that (i) does not hold, and fix ${x}_{C}^{\prime}\in C$ and $D\in B$ witnessing it. Suppose that (ii) and (iii) do not hold either. We will construct a strictly decreasing sequence ${\{{C}_{\alpha}\}}_{\alpha <{\omega}_{1}}$ of balls in *B*. As this is obviously not possible (because *B* is countable), (ii) or (iii) must hold.

Put ${C}_{0}=D$. Suppose that we have constructed ${C}_{\beta}$ for $\beta <\alpha $. Suppose that $\alpha =\gamma +1$. Since (iii) does not hold, ${C}_{\gamma}$ is not inclusion-minimal in *B*, and there is $E\in B$ with $E\u228a{C}_{\gamma}$. Put ${C}_{\alpha}=E$.

Suppose now that α is a limit ordinal. Find a cofinal sequence $\{{D}_{n}\}$ in ${\{{C}_{\beta}\}}_{\beta <\alpha}$. As (ii) does not hold, the radii of ${D}_{n}$ are bounded from below by some $r>0$. Observe that since *X* is locally non-rigid, *r* can be chosen so that, for the balls $F={B}_{\le r}({x}_{C}^{\prime})$, ${F}^{\prime}={B}_{<r}({x}_{C}^{\prime})$, the relation ${E}_{{F}^{\prime}}$ strictly refines the relation ${E}_{F}$, that is, $\delta =({E}_{F},{E}_{{F}^{\prime}})$ is a covering pair in ${\mathrm{\Delta}}_{X}$.

Fix *n*. Since ${D}_{n}\cap [{x}_{C}^{\prime}]\ne \mathrm{\varnothing}$, and $r<diam({D}_{n})$, there exists ${\varphi}_{n}\in Iso(X)$ such that ${\varphi}_{n}[{F}^{\prime}]\subseteq {D}_{n}$, that is, ${E}_{{F}^{\prime}}$ refines ${E}_{{D}_{n}}$. In other words, δ witnesses that covering pairs coming from $\{{D}_{n}\}$ are bounded from below. The boundedness condition implies that there is a ball $G\in B$ with $G\subseteq {D}_{n}$ for every *n*. Put ${C}_{\alpha}=G$. This finishes the inductive construction.
∎

*Let X be a W-space, and let B be an agreeable family of balls in X. Suppose that $C\mathrm{\in}B$ and ${x}_{C}^{\mathrm{\prime}}\mathrm{\in}C$ are such for every ball $D\mathrm{\in}B$ with $D\mathrm{\subseteq}C$ there is a ball $E\mathrm{\in}D$ such that $E\mathrm{\subseteq}D$, and $E\mathrm{\cap}\mathrm{[}{x}_{C}^{\mathrm{\prime}}\mathrm{]}\mathrm{=}\mathrm{\varnothing}$. Then there is ${x}_{C}\mathrm{\in}C$ such that $d\mathit{}\mathrm{(}{x}_{C}\mathrm{,}y\mathrm{)}\mathrm{=}\mathrm{dist}\mathit{}\mathrm{(}\mathrm{[}{x}_{C}\mathrm{]}\mathrm{,}\mathrm{[}y\mathrm{]}\mathrm{)}$ for every $D\mathrm{\in}B$ with ${x}_{C}\mathrm{\notin}D$, and for every $y\mathrm{\in}D$.
*

#### Proof.

Fix a maximal pairwise disjoint family ${B}^{\prime}\subseteq B$ of balls *E* such that $E\subseteq C$, and $E\cap [{x}_{C}^{\prime}]=\mathrm{\varnothing}$. Fix ${y}_{E}\in E$ for every $E\in {B}^{\prime}$. Because *B* is agreeable, Lemma 1 implies that $d({y}_{E},{y}_{{E}^{\prime}})=dist([{y}_{E}],[{y}_{{E}^{\prime}}])$ for every $E,{E}^{\prime}\in {B}^{\prime}$.
As *X* is a *W*-space, Lemma 4.10 in [1] implies that there exists ${x}_{C}\in [{x}_{C}^{\prime}]$ such that $d({x}_{C},{y}_{E})=dist([{x}_{C}],[{y}_{E}])$ for $E\in {B}^{\prime}$. Clearly, ${x}_{C}\in C$, and ${x}_{C}\notin E$ for every $E\in {B}^{\prime}$.

Let $D\in B$ be a ball such that ${x}_{C}\notin D$. Fix $y\in D$. If $D\cap C=\mathrm{\varnothing}$, then Lemma 1 implies $d({x}_{C},y)=dist([{x}_{C}],[y])$. Otherwise $D\subseteq C$, so there is $E\in {B}^{\prime}$ such that $D\subseteq E$ or $E\subseteq D$. In both cases ${x}_{C}\notin D\cup E$, so $d(y,{y}_{E})<d(y,{x}_{C})$. Suppose that there is $x\in [{x}_{C}]$ such that $d(y,x)<d(y,{x}_{C})$. But then the ultrametric triangle inequality implies that $d({y}_{E},x)<d({y}_{E},{x}_{C})$, which is impossible. Thus, $d(y,{x}_{C})=dist(y,[{x}_{C}])$. ∎

#### ([1, Theorem 4.13])

*Let X be a W-space, and let ${S}_{X}$, ${\mathrm{\Delta}}_{X}$, ${G}_{\delta}$ for $\delta \mathrm{\in}{\mathrm{\Delta}}_{X}$ be defined as above. Then ${\mathrm{\Delta}}_{X}$ is plenary, $\mathrm{Iso}\mathit{}\mathrm{(}X\mathrm{)}$ acts transitively and faithfully on ${S}_{X}$, and all elements of $\mathrm{Iso}\mathit{}\mathrm{(}X\mathrm{)}$ respect ${\mathrm{\Delta}}_{X}$. Thus, it can be regarded as a transitive permutation group $\mathrm{(}\mathrm{Iso}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{,}{S}_{X}\mathrm{,}{\mathrm{\Delta}}_{X}\mathrm{)}$. Moreover,*

$Iso(X)\cong W{r}_{\delta \in {\mathrm{\Delta}}_{X}}{G}_{\delta}.$

#### Proof.

Consider the action of $Iso(X)$ on ${S}_{X}$ given by

$\varphi .Y=\varphi [Y]$

for $\varphi \in Iso(X)$, $Y\in {S}_{X}$.

By [1, Lemma 4.7], ${\mathrm{\Delta}}_{X}$ is a countable plenary family, and every relation in ${\mathrm{\Delta}}_{X}$ has countably many classes. By [1, Lemmas 4.10, 4.11 and 4.12], the action defined above is faithful, and $(Iso(X),{S}_{X},\mathrm{\Delta})$ is a transitive maximal permutation group respecting ${\mathrm{\Delta}}_{X}$. Therefore, to prove the theorem, we only need to check condition (3) of Theorem 3.5 in [1].

Suppose that *B* is an agreeable family of balls in *X* which satisfies the boundedness condition. Let

$\begin{array}{cc}\hfill {B}_{0}& =\{C\in B:\text{there exists}D\in B\text{such that}D\text{is inclusion-minimal in}B,D\subseteq C\},\hfill \\ \hfill {B}_{1}& =\{C\in B:\text{there exists}\{{C}_{n}\}\subseteq B\text{such that}\{{C}_{n}\}\text{is decreasing},diam({C}_{n})\to 0,{C}_{n}\subseteq C\},\hfill \end{array}$

and let ${B}_{2}=B\setminus ({B}_{0}\cup {B}_{1})$.

Let ${A}_{0}$ be a selector for the family of all inclusion-minimal balls in *B*. Let ${A}_{1}$ be the set of all points $x\in X$ of the form $\{x\}=\bigcap {C}_{n}$, where $\{{C}_{n}\}$ is a sequence of balls in *B* with $diam({C}_{n})\to 0$. Then ${A}_{0}$ intersects every ball in ${B}_{0}$, ${A}_{1}$ intersects every ball in ${B}_{1}$, and by Lemma 1, $d(a,b)=dist([a],[b])$ for all $a,b\in {A}_{0}\cup {A}_{1}$.

We show how to construct ${A}_{2}$ intersecting every element of ${B}_{2}$ so that $A={A}_{0}\cup {A}_{1}\cup {A}_{2}$ intersects every element of *B*, and $d(a,b)=dist([a],[b])$ for all $a,b\in A$.

Clearly, for every ball in ${B}_{2}$, point (i) of Lemma 2 holds. Let ${C}_{0},{C}_{1},\mathrm{\dots}$ be an enumeration of all balls in ${B}_{2}$. By Lemma 3, there is ${x}_{0}\in {C}_{0}$ such that $d({x}_{0},y)=dist([{x}_{0}],[y])$ for every $D\in B$ with ${x}_{0}\notin D$, and for every $y\in D$. Put ${D}_{0}={C}_{0}$. Suppose that we have constructed ${x}_{0},\mathrm{\dots},{x}_{n}$, ${D}_{0},\mathrm{\dots},{D}_{n}$ so that ${x}_{i}\in {C}_{i}$, and ${x}_{i}$ is chosen as in Lemma 3 for the ball ${D}_{i}$, $i\le n$. If ${x}_{{i}_{0}}\in {C}_{n+1}$ for some ${i}_{0}\le n$, then put ${x}_{n+1}={x}_{{i}_{0}}$, ${D}_{n+1}={D}_{{i}_{0}}$. Otherwise, fix ${x}_{n+1}$ for ${C}_{n+1}$ as in Lemma 3, and put ${D}_{n+1}={C}_{n+1}$.

By the construction, we have that $\{{x}_{n}\}$ intersects every element of ${B}_{2}$, and $d({x}_{n},{x}_{m})=dist([{x}_{n}],[{x}_{m}])$ for every $n,m$. By Lemma 1, ${A}_{2}=\{{x}_{n}\}$ is as required.

Now, by [1, Lemma 4.10], there exists a *d*-transversal *Y* such that $A\subseteq Y$. Obviously, *A* intersects every element of *B*.
∎

## References

- [1]
Malicki M., Separable ultrametric spaces and their isometry groups, Forum Math. 26 (2014), 1663–1683. Google Scholar

## About the article

**Received**: 2015-03-19

**Revised**: 2015-10-03

**Published Online**: 2015-11-17

**Published in Print**: 2016-09-01

**Citation Information: **Forum Mathematicum, ISSN (Online) 1435-5337, ISSN (Print) 0933-7741, DOI: https://doi.org/10.1515/forum-2015-0050.

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