Vertices of large depth

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Component labels and annular depth attaches to a closed set $A$ and a group element $g$ the depth $D(A,g)$, the largest number of nested annuli between $A$ and $g$. This lesson studies the vertices of the graph $K$ that have depth at least $m$, and the components into which they fall. Its results are the following.

If the depths $D(A,g_n)$ tend to infinity, then, along a subsequence, the translates $g_n^{-1}A$ shrink to one point $q$ of the continuum $L$ (Lemma 2.1). The depth measured from such a translate is the sum of two depths, with an error that is bounded (Lemma 3.1).
The vertices of depth at least $m$ form a set $I_m$. Away from $A$, the vertices outside $I_m$ do not accumulate on $L$, so there is a component label $Q_m$ on $L\setminus A$ (Section 4).
Let $g_n$ have depth at least $m_n$, with $m_n\to\infty$. Then any two points $z,z'$ of $L$ other than $q$ have translates $g_nz$, $g_nz'$ with the same label $Q_{m_n+1}$, for all large $n$ (Theorem 6.1).

These statements are the preparation for the proof that $L$ is locally connected, in The continuum is locally connected. The lesson follows the first half of Section 9 of [OpenAI-S].

1. The setting

We use the data $Y=H\cup L$, $K$, $I$ and the properties (S1) to (S7) of the lesson on component labels and annular depth, the set $\mathcal A$ of annuli of Theorem 3.2 there, and the depth $D$ of Section 4 there. The identity $1$ of $H$ is a point of $Y$. A sequence $(h_n)$ in $H$ tends to infinity if every element of $H$ occurs in it only finitely often; by (S4), $\rho(h_nw,h_n)\to0$ for such a sequence and every $w\in I$.

Throughout the lesson, $A$ is a closed subset of $Y$ with $1\in A,\qquad L\not\subseteq A .$ The following facts and notations are used.

(D1) $D(gF,gv)=D(F,v)$ for $g\in H$. If $F$ contains a point of $H$ and $v\in H$, then $D(F,v)$ is finite. In particular $D(A,g)$ is finite for every $g\in H$. (Theorem 4.1(1) and (2) of that lesson.)
(D2) If $F$ is closed, $x\in L\setminus F$ and $N\ge1$, then $D(F,v)\ge N$ for all $v$ in some neighbourhood of $x$ in $Y$. (Theorem 4.1(3).)
(D3) In a chain $\sigma_1<\dots<\sigma_n$ of annuli the negative sides increase, the positive sides decrease, and the annuli are pairwise different (Lemma 3.1 of that lesson): $\begin{gathered} \sigma_1^-\subseteq\dots\subseteq\sigma_n^-,\\ \sigma_1^+\supseteq\dots\supseteq\sigma_n^+ . \end{gathered}$
Let $d_0$ be the distance from $1$ to $L$. It is positive, because $L$ is closed and $1\notin L$. Every side of an annulus meets $L$. So a side that contains $1$ has diameter at least $d_0$.
For $a,b>0$ let $N(a,b)$ be the number of annuli $\sigma\in\mathcal A$ with $\operatorname{diam}\sigma^-\ge a$ and $\operatorname{diam}\sigma^+\ge b$. It is finite, by Theorem 3.2(1) of that lesson with $\delta$ the smaller of $a$ and $b$.
A sequence $(F_n)$ of subsets of $Y$ shrinks to the point $q\in Y$ if for every $r>0$ the set $F_n$ lies in the ball $B(q,r)$ for all but finitely many $n$. Every subsequence then shrinks to $q$ as well.

Lemma 1.1. Let $p\in I$, and let $(a_n)$ be a sequence in $H$ such that the vertices $a_np$ are pairwise distinct and converge to a point $x$ of $Y$. Then $x\in L$ and $a_n\to x$.

Proof. A limit of pairwise distinct points is not an isolated point, so $x\in L$ by (S1). The elements $a_n$ are pairwise distinct, so $(a_n)$ tends to infinity, and $\rho(a_np,a_n)\to0$ by (S4). $\blacksquare$

2. Translates that shrink to a point

Lemma 2.1. Let $(g_n)$ be a sequence in $H$ with $D(A,g_n)\to\infty$. Then $(g_n)$ has a subsequence for which the sets $g_n^{-1}A$ shrink to a point $q$ of $L$.

Proof. Let $F_n=g_n^{-1}A$. By (D1), $D(F_n,1)=D(A,g_n).$ The diameters tend to zero. Let $\varepsilon>0$ and let $\sigma_1<\dots<\sigma_\ell$ be a chain from $F_n$ to $\{1\}$. By (D3), every positive side $\sigma_i^+$ contains $1$, so its diameter is at least $d_0$; every negative side $\sigma_i^-$ contains $F_n$; and the $\ell$ annuli are pairwise different. If $\ell>N(\varepsilon,d_0)$, then not all of them have a negative side of diameter at least $\varepsilon$, so $\operatorname{diam}F_n<\varepsilon$. Since $D(F_n,1)\to\infty$, such chains exist for all large $n$. Hence $\operatorname{diam}F_n\to0$.

The limit point. For every $g\in H$ the number $D(A,g)$ is finite, so $g$ occurs only finitely often among the $g_n$. Pass to a subsequence in which the elements $g_n^{-1}$ are pairwise distinct and converge to a point $q$ of the compact space $Y$. Then $q\in L$, because a limit of pairwise distinct points is not isolated. Since $1\in A$, the point $g_n^{-1}$ lies in $F_n$. For $r>0$ and all large $n$ we have $\rho(g_n^{-1},q)<r/2$ and $\operatorname{diam}F_n<r/2$, so $F_n\subseteq B(q,r)$. $\blacksquare$

3. A comparison of depths

In the situation of Lemma 2.1, a chain from $F_n$ to a group element $b$ far from $q$ can be built from two chains: one from $F_n$ to $1$, and one from $1$ to $b$. Only a bounded number of annuli is lost.

Lemma 3.1. Let $(g_n)$ be a sequence in $H$ with $D(A,g_n)\to\infty$ such that the sets $F_n=g_n^{-1}A$ shrink to a point $q\in Y$. For $r>0$ let $C_r=N(r/2,d_0)+N(d_0,r)+2 .$ Then there is $n_r$ such that for all $n\ge n_r$ and all $b\in H$ with $\rho(b,q)\ge3r$, $D(F_n,b)\ge D(F_n,1)+D(1,b)-C_r .$

Proof. Let $N_-=N(r/2,d_0)$ and $N_+=N(d_0,r)$. Choose $n_r$ such that for $n\ge n_r$ $\begin{gathered} F_n\subseteq B(q,r/2),\\ D(F_n,1)\ge N_-+2 . \end{gathered}$ Fix $n\ge n_r$ and $b\in H$ with $\rho(b,q)\ge3r$. The set $F_n$ contains the group element $g_n^{-1}$, so $\ell=D(F_n,1)$ and $t=D(1,b)$ are finite, by (D1). Let $\begin{gathered} U=B(q,r),\\ V=\{y\in Y\mid\rho(y,q)>2r\}. \end{gathered}$ These are disjoint open sets, and $b\in V$.

The first chain. Let $\sigma_1<\dots<\sigma_\ell$ be a chain from $F_n$ to $\{1\}$. Suppose $\sigma_j^-\not\subseteq U$. The side $\sigma_j^-$ contains a point of $F_n$, at distance less than $r/2$ from $q$, and a point at distance at least $r$ from $q$; so its diameter is at least $r/2$. The side $\sigma_j^+$ contains $1$, so its diameter is at least $d_0$. Since the annuli of the chain are pairwise different, there are at most $N_-$ such indices $j$. The negative sides increase along the chain, so these are the last $k$ indices, for some $k\le N_-$. Let $p=\ell-k-1\ge1 .$ Then $\sigma_{p+1}^-\subseteq U$, and $\sigma_p^+\cup\sigma_{p+1}^-=Y$ gives $Y\setminus U\subseteq\sigma_p^+$. So $\sigma_1<\dots<\sigma_p$ is a chain from $F_n$ to $Y\setminus U$, of length at least $\ell-N_--1$.

The case $t\le N_++1$. The first chain goes from $F_n$ to $\{b\}$, because $b\notin U$. So $D(F_n,b)\ge\ell-N_--1\ge\ell+t-C_r .$

The case $t\ge N_++2$. Let $\tau_1<\dots<\tau_t$ be a chain from $\{1\}$ to $\{b\}$. Suppose $\tau_j^+\not\subseteq V$. The side $\tau_j^+$ contains $b$ and a point at distance at most $2r$ from $q$, so its diameter is at least $r$. The side $\tau_j^-$ contains $1$, so its diameter is at least $d_0$. Hence there are at most $N_+$ such indices $j$. The positive sides decrease along the chain, so these are the first $k'$ indices, for some $k'\le N_+$. Let $p'=k'+2\le t .$ Then $\tau_{p'-1}^+\subseteq V$, and $\tau_{p'-1}^+\cup\tau_{p'}^-=Y$ gives $Y\setminus V\subseteq\tau_{p'}^-$. The chain $\tau_{p'}<\dots<\tau_t$ has length at least $t-N_+-1$.

Since $U$ and $V$ are disjoint, $\sigma_p^+\cup\tau_{p'}^-\supseteq(Y\setminus U)\cup(Y\setminus V)=Y ,$ that is, $\sigma_p<\tau_{p'}$. So $\sigma_1<\dots<\sigma_p<\tau_{p'}<\dots<\tau_t$ is a chain from $F_n$ to $\{b\}$. Its length is at least $\begin{aligned} &(\ell-N_--1)+(t-N_+-1)\\ &\qquad=\ell+t-C_r . \end{aligned}$ $\blacksquare$

Corollary 3.2. Let $(g_n)$, $F_n$ and $q$ be as in Lemma 3.1, and let $(b_n)$ be a sequence in $H$ that converges to a point $z\in L$ with $z\ne q$. Then $D(1,b_n)\to\infty$, and there is a number $C$ such that for all but finitely many $n$ $D(F_n,b_n)\ge D(F_n,1)+D(1,b_n)-C .$

Proof. Choose $r>0$ with $4r<\rho(q,z)$. Then $\rho(b_n,q)\ge3r$ for all large $n$, and Lemma 3.1 gives the inequality with $C=C_r$. By (D2) with $F=\{1\}$ and $x=z$, for every $N$ the elements $b_n$ with large $n$ satisfy $D(1,b_n)\ge N$. $\blacksquare$

4. Vertices of depth at least m

A vertex of $K$ is in general the image of a base vertex under many group elements, and each of these has its own depth. The following definition asks for a bound on all of them.

By (S3) the set $I$ consists of finitely many orbits of $H$. Choose one point $p_1,\dots,p_k$ in each orbit; if $I=H$, there is one orbit, and we choose $p_1=1$. For a vertex $u$ in the orbit of $p_i$ let $R(u)=\{g\in H\mid gp_i=u\},$ the set of representatives of $u$. For $g\in H$ we have $R(gu)=gR(u)$.

For an integer $m\ge0$ let $I_m$ be the set of all vertices $u\in I\setminus A$ such that $D(A,g)\ge m$ for every $g\in R(u)$. So $I\setminus A=I_0\supseteq I_1\supseteq I_2\supseteq\cdots .$ We use the notation of Section 2 of the lesson on component labels and annular depth: $D'$ is the set of limits of sequences of pairwise distinct points of $D$, and $\Omega_T=L\setminus(I\setminus T)'$.

Lemma 4.1.

For every $m\ge0$, no point of $L\setminus A$ is the limit of a sequence of pairwise distinct vertices in $I\setminus I_m$. So $L\setminus A\subseteq\Omega_{I_m}$.
For every finite set $E\subseteq I$ there is $m$ with $E\cap I_m=\varnothing$.

Proof. (1) Suppose pairwise distinct vertices $u_n\in I\setminus I_m$ converge to $x\in L\setminus A$. Since $A$ is closed, $u_n\notin A$ for large $n$; for these $n$ there is $a_n\in R(u_n)$ with $D(A,a_n)<m$, by the definition of $I_m$. Pass to a subsequence in which all $u_n$ lie in the orbit of one $p_i$, so that $u_n=a_np_i$. By Lemma 1.1, $a_n\to x$. By (D2), $D(A,a_n)\ge m$ for large $n$. This is a contradiction.

(2) For every $u\in E\setminus A$ choose one $a_u\in R(u)$. If $m$ is larger than the finitely many numbers $D(A,a_u)$, no point of $E\setminus A$ lies in $I_m$; and no point of $A$ lies in $I_m$. $\blacksquare$

By Lemma 4.1(1) and Theorem 2.1 of the lesson on component labels and annular depth, applied to the set $T=I_m$ of allowed vertices, there is for every $m\ge0$ a map $Q_m\colon L\setminus A\to\{\text{components of }K[I_m]\}$ with the following properties.

(Q1) Every $x\in L\setminus A$ is the limit of a sequence of pairwise distinct vertices in $I_m$, and all vertices of $I_m$ in some neighbourhood of $x$ in $Y$ belong to the component $Q_m(x)$.
(Q2) $Q_m$ is locally constant on the open subset $L\setminus A$ of $L$.

By (Q1), $Q_m(x)$ is the only component of $K[I_m]$ that contains a sequence of pairwise distinct vertices converging to $x$.

5. Deleted vertices after a translation

For Sections 5 and 6 we fix integers $m_n\ge0$ with $m_n\to\infty$ and elements $g_n\in H$ with $D(A,g_n)\ge m_n ,$ such that the sets $F_n=g_n^{-1}A$ shrink to a point $q\in L$. By Lemma 2.1, every pair of sequences $(m_n)$, $(g_n)$ with $m_n\to\infty$ and $D(A,g_n)\ge m_n$ has a subsequence of this kind.

The translate $g_n^{-1}(I\setminus I_{m_n+1})$ is the set of vertices $w$ whose image $g_nw$ is deleted at depth $m_n+1$. The next lemma says that away from $q$ only finitely many vertices are of this kind, and that the finite set does not depend on $n$.

Lemma 5.1. For every compact set $E\subseteq Y\setminus\{q\}$ there are a finite set $S_E\subseteq I$ and an index $n_E$ such that for all $n\ge n_E$: every vertex $w\in I\cap E$ with $g_nw\notin I_{m_n+1}$ belongs to $S_E$.

Proof. Suppose there are no such $S_E$ and $n_E$. Then there are indices $n_1<n_2<\cdots$ and pairwise distinct vertices $w_1,w_2,\dots$ in $I\cap E$ with $g_{n_j}w_j\notin I_{m_{n_j}+1} .$ Indeed, when $n_1,\dots,n_{j-1}$ and $w_1,\dots,w_{j-1}$ are chosen, the finite set $\{w_1,\dots,w_{j-1}\}$ and the index $n_{j-1}+1$ do not have the property of the lemma, which gives $n_j$ and $w_j$. Replace $(m_n)$ and $(g_n)$ by the subsequences with the indices $n_j$; the assumptions of this section still hold. So there are pairwise distinct vertices $w_n\in I\cap E$ with $g_nw_n\notin I_{m_n+1}$ for all $n$.

Pass to a subsequence such that $w_n$ converges to a point $z$ of the compact set $E$ and all $w_n$ lie in the orbit of one $p_i$. Then $z\in L$, because the $w_n$ are pairwise distinct, and $z\ne q$.

Since $E$ is compact and $q\notin E$, some ball $B(q,r)$ does not meet $E$, and $F_n\subseteq B(q,r)$ for large $n$. For these $n$, $w_n\notin F_n$, that is, $g_nw_n\notin A$. So $g_nw_n$ is a vertex outside $A$ and outside $I_{m_n+1}$, and by the definition of $I_{m_n+1}$ there is $c_n\in R(g_nw_n)$ with $D(A,c_n)\le m_n$. Let $b_n=g_n^{-1}c_n$. Then $\begin{gathered} b_np_i=w_n,\\ D(F_n,b_n)=D(A,c_n)\le m_n , \end{gathered}$ by (D1). By Lemma 1.1, $b_n\to z$. By Corollary 3.2, $D(1,b_n)\to\infty$ and $\begin{aligned} D(F_n,b_n)&\ge D(F_n,1)+D(1,b_n)-C\\ &\ge m_n+D(1,b_n)-C , \end{aligned}$ because $D(F_n,1)=D(A,g_n)\ge m_n$. For large $n$ the right side is larger than $m_n$. This is a contradiction. $\blacksquare$

6. Labels after a translation

Theorem 6.1. Let $(m_n)$, $(g_n)$ and $q$ be as in Section 5.

For every $z\in L\setminus\{q\}$ there are a neighbourhood $V_z$ of $z$ in $L$ and an index $n_z$ such that for all $n\ge n_z$ and all $w\in V_z$: $g_nw\notin A$ and $Q_{m_n+1}(g_nw)=Q_{m_n+1}(g_nz).$
For any two points $z,z'\in L\setminus\{q\}$ there is an index $n_0$ such that for all $n\ge n_0$: $g_nz\notin A$, $g_nz'\notin A$ and $Q_{m_n+1}(g_nz)=Q_{m_n+1}(g_nz').$

Proof. (1) Choose an open neighbourhood $O$ of $z$ in $Y$ and $r>0$ such that $\overline O$ does not meet $B(q,r)$. Lemma 5.1 for the compact set $\overline O$ gives a finite set $S\subseteq I$ and an index $n_1$. Let $T=(I\cap O)\setminus S .$ For $n\ge n_1$ and $t\in T$ the vertex $g_nt$ lies in $I_{m_n+1}$, by Lemma 5.1. So $g_nT\subseteq I_{m_n+1}\qquad(n\ge n_1).$ The set $I\setminus T$ is the union of the finite set $S$ and a subset of the closed set $Y\setminus O$. So no point of $O$ is a limit of pairwise distinct points of $I\setminus T$, and $O\cap L\subseteq\Omega_T$. By Theorem 2.1 of the lesson on component labels and annular depth there are a component $G$ of $K[T]$ and a neighbourhood $V_z\subseteq O\cap L$ of $z$ in $L$ such that $Q_T(w)=G$ for all $w\in V_z$.

Choose $n_z\ge n_1$ such that $F_n\subseteq B(q,r)$ for $n\ge n_z$. Fix $n\ge n_z$ and $w\in V_z$. Then $w\notin F_n$, so $g_nw\notin A$.

The map $g_n$ sends edges of $K$ to edges and $T$ into $I_{m_n+1}$. So it sends paths of $K[T]$ to paths of $K[I_{m_n+1}]$, and $g_nG$ lies in one component $G_n$ of $K[I_{m_n+1}]$. By Theorem 2.1(1) and (2) of that lesson, $w$ is the limit of a sequence $(t_j)$ of pairwise distinct vertices in $G$. The vertices $g_nt_j$ are pairwise distinct, lie in $G_n$, and converge to $g_nw$, because $g_n$ is a homeomorphism of $Y$. By the remark after (Q2), $Q_{m_n+1}(g_nw)=G_n .$ The component $G_n$ does not depend on $w\in V_z$, and $z\in V_z$. This proves (1).

(2) For $z,z'\in L\setminus\{q\}$ write $z\sim z'$ if the conclusion of (2) holds for them. By (1) with $w=z$, $z\sim z$. The relation is symmetric, and it is transitive: if the labels of $g_nz$ and $g_nz'$ agree for $n\ge n_0$ and those of $g_nz'$ and $g_nz''$ agree for $n\ge n_0'$, then those of $g_nz$ and $g_nz''$ agree for $n$ at least the larger of $n_0$ and $n_0'$. By (1), $w\sim z$ for every $w\in V_z$. So the classes of $\sim$ are open subsets of $L\setminus\{q\}$. This set is connected, because $q$ is not a cut point of $L$, by (S1). A connected space is not the union of two or more disjoint nonempty open sets, so there is one class. $\blacksquare$

The threshold $n_0$ in (2) depends on the two points. The set of labels changes with $n$, since $Q_{m_n+1}$ takes values in the components of a graph that depends on $n$; the relation $\sim$ compares labels for one $n$ at a time.

7. Exercises

7.1. In Lemma 2.1, show that $g_n^{-1}\to q$ along the subsequence. Let $A_1$ be another closed subset of $Y$ with $1\in A_1$, $L\not\subseteq A_1$ and $D(A_1,g_n)\to\infty$. Show that the sets $g_n^{-1}A_1$ shrink to the same point $q$ along that subsequence.

7.2. Show that $N(a,b)=N(b,a)$, and that $N(a_1,b_1)\le N(a,b)$ if $a_1\ge a$ and $b_1\ge b$.

7.3. Let $I=H$ and $A=\{1\}$. Show that $I_m$ is the set of all $g\in H$ with $D(1,g)\ge m$, that $H\setminus I_m$ is finite for every $m$, and that $Q_m$ is constant on $L$.

7.4. Show that every $I_m$ is infinite and that the intersection of all $I_m$ is empty.

7.5. Let $I\subseteq L$ and $u\in I$. Show that $R(u)=gR(p_i)$ for every $g\in R(u)$, where $p_i$ is the chosen point in the orbit of $u$, that $R(p_i)$ is the stabilizer of $p_i$ in $H$, and that $R(u)$ is infinite.

8. Solutions

7.1. Since $1\in A$, the point $g_n^{-1}$ lies in $g_n^{-1}A$, and these sets shrink to $q$; so $g_n^{-1}\to q$. The first part of the proof of Lemma 2.1 uses only that the set is closed, contains $1$ and has depths tending to infinity; applied to $A_1$ it shows that the diameter of $g_n^{-1}A_1$ tends to $0$. These sets contain $g_n^{-1}$, which converges to $q$ along the subsequence. So they shrink to $q$.

7.2. The reversal $(\sigma^-,\sigma^+)\mapsto(\sigma^+,\sigma^-)$ maps $\mathcal A$ onto itself and is its own inverse. It maps the annuli counted by $N(a,b)$ onto those counted by $N(b,a)$. If $a_1\ge a$ and $b_1\ge b$, every annulus counted by $N(a_1,b_1)$ is counted by $N(a,b)$.

7.3. Here $p_1=1$ and $R(g)=\{g\}$, so $I_m$ consists of the $g\in H\setminus\{1\}$ with $D(1,g)\ge m$. For $m\ge1$ the condition $D(1,g)\ge m$ excludes $g=1$ (Exercise 5.3 of the lesson on component labels and annular depth); for $m=0$ the set is $H\setminus\{1\}$, and the description holds for $m\ge1$. By Exercise 5.4 of that lesson, $D(1,g)\ge m$ for all but finitely many $g$. So $H\setminus I_m$ is finite. The map $Q_m$ is defined on $L\setminus A=L$ and locally constant; a locally constant map on a connected space is constant.

7.4. By (Q1), a point of $L\setminus A$ is the limit of a sequence of pairwise distinct vertices in $I_m$; so $I_m$ is infinite. If $u$ were in every $I_m$, Lemma 4.1(2) for $E=\{u\}$ would be contradicted.

7.5. By definition $R(p_i)=\{h\mid hp_i=p_i\}$ is the stabilizer. If $g\in R(u)$, then $g'p_i=u$ holds exactly if $g^{-1}g'$ fixes $p_i$; so $R(u)=gR(p_i)$. By (S6) the stabilizer of $p_i$ contains an element of infinite order, so it is infinite. This is why the definition of $I_m$ bounds the depth of every representative: a vertex in $L$ has infinitely many of them.

References

[OpenAI-S] OpenAI, The Singer conjecture in dimension four, OpenAI Math Release preprint, 25 September 2026, Section 9. https://github.com/openai/math/tree/main/preprints/The-Singer-conjecture-in-dimension-four-September-25-2026