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Volume 28, Issue 5 (Sep 2016)


Corrigendum to: Separable ultrametric spaces and their isometry groups

Maciej Malicki
  • Corresponding author
  • Department of Mathematics and Mathematical Economics, Warsaw School of Economics, al. Niepodleglosci 162, 02-554,Warsaw, Poland
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Published Online: 2015-11-17 | DOI: https://doi.org/10.1515/forum-2015-0050

This Erratum corrects the original online version which can be found here: https://doi.org/10.1515/forum-2011-0149

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 B0, B1 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 B0,B1 be disjoint ‘open’ balls in an ultrametric space X. Suppose that the equivalence classes they determine in EB0,EB1, respectively, are agreeable. If x0B0,x1B1, then



Suppose that d(x0,x1)>dist([x0],[x1]), and put r=d(x0,x1), C0=B<r(x0), C1=B<r(x1). Then there exists ϕIso(X) such that ϕ[C0]=C1, so EC0=EC1, and EB0,EB1EC0. However, C0C1, and B0C0, B1C1. Therefore B0,B1 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 YB 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 EB on the set SX of all d-transversals in X, defined by


Suppose that D is an equivalence class of an equivalence relation EB 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 CB. Then one of the following must hold:

  • (i)

    for every xCC , and every ball DB with DC there is a ball EB such that ED , and E[xC]=,

  • (ii)

    there is a decreasing sequence {Cn} of balls in B such that CnC , and diam(Cn)0,

  • (iii)

    there is DB with DC which is inclusion-minimal in B.


Suppose that (i) does not hold, and fix xCC and DB witnessing it. Suppose that (ii) and (iii) do not hold either. We will construct a strictly decreasing sequence {Cα}α<ω1 of balls in B. As this is obviously not possible (because B is countable), (ii) or (iii) must hold.

Put C0=D. Suppose that we have constructed Cβ for β<α. Suppose that α=γ+1. Since (iii) does not hold, Cγ is not inclusion-minimal in B, and there is EB with ECγ. Put Cα=E.

Suppose now that α is a limit ordinal. Find a cofinal sequence {Dn} in {Cβ}β<α. As (ii) does not hold, the radii of Dn 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=Br(xC), F=B<r(xC), the relation EF strictly refines the relation EF, that is, δ=(EF,EF) is a covering pair in ΔX.

Fix n. Since Dn[xC], and r<diam(Dn), there exists ϕnIso(X) such that ϕn[F]Dn, that is, EF refines EDn. In other words, δ witnesses that covering pairs coming from {Dn} are bounded from below. The boundedness condition implies that there is a ball GB with GDn for every n. Put Cα=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 CB and xCC are such for every ball DB with DC there is a ball ED such that ED, and E[xC]=. Then there is xCC such that d(xC,y)=dist([xC],[y]) for every DB with xCD, and for every yD.


Fix a maximal pairwise disjoint family BB of balls E such that EC, and E[xC]=. Fix yEE for every EB. Because B is agreeable, Lemma 1 implies that d(yE,yE)=dist([yE],[yE]) for every E,EB. As X is a W-space, Lemma 4.10 in [1] implies that there exists xC[xC] such that d(xC,yE)=dist([xC],[yE]) for EB. Clearly, xCC, and xCE for every EB.

Let DB be a ball such that xCD. Fix yD. If DC=, then Lemma 1 implies d(xC,y)=dist([xC],[y]). Otherwise DC, so there is EB such that DE or ED. In both cases xCDE, so d(y,yE)<d(y,xC). Suppose that there is x[xC] such that d(y,x)<d(y,xC). But then the ultrametric triangle inequality implies that d(yE,x)<d(yE,xC), which is impossible. Thus, d(y,xC)=dist(y,[xC]). ∎

([1, Theorem 4.13])

Let X be a W-space, and let SX, ΔX, Gδ for δΔX be defined as above. Then ΔX is plenary, Iso(X) acts transitively and faithfully on SX, and all elements of Iso(X) respect ΔX. Thus, it can be regarded as a transitive permutation group (Iso(X),SX,ΔX). Moreover,



Consider the action of Iso(X) on SX given by


for ϕIso(X), YSX.

By [1, Lemma 4.7], ΔX is a countable plenary family, and every relation in ΔX has countably many classes. By [1, Lemmas 4.10, 4.11 and 4.12], the action defined above is faithful, and (Iso(X),SX,Δ) is a transitive maximal permutation group respecting ΔX. 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

B0={CB:there exists DB such that D is inclusion-minimal in B,DC},B1={CB:there exists {Cn}B such that {Cn} is decreasing,diam(Cn)0,CnC},

and let B2=B(B0B1).

Let A0 be a selector for the family of all inclusion-minimal balls in B. Let A1 be the set of all points xX of the form {x}=Cn, where {Cn} is a sequence of balls in B with diam(Cn)0. Then A0 intersects every ball in B0, A1 intersects every ball in B1, and by Lemma 1, d(a,b)=dist([a],[b]) for all a,bA0A1.

We show how to construct A2 intersecting every element of B2 so that A=A0A1A2 intersects every element of B, and d(a,b)=dist([a],[b]) for all a,bA.

Clearly, for every ball in B2, point (i) of Lemma 2 holds. Let C0,C1, be an enumeration of all balls in B2. By Lemma 3, there is x0C0 such that d(x0,y)=dist([x0],[y]) for every DB with x0D, and for every yD. Put D0=C0. Suppose that we have constructed x0,,xn, D0,,Dn so that xiCi, and xi is chosen as in Lemma 3 for the ball Di, in. If xi0Cn+1 for some i0n, then put xn+1=xi0, Dn+1=Di0. Otherwise, fix xn+1 for Cn+1 as in Lemma 3, and put Dn+1=Cn+1.

By the construction, we have that {xn} intersects every element of B2, and d(xn,xm)=dist([xn],[xm]) for every n,m. By Lemma 1, A2={xn} is as required.

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


  • [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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