Another look at recovering local homology from samples of stratified sets

Abstract

Recovering homological features of spaces from samples has become one of the central themes of topological data analysis, leading to many successful applications. Many of the results in this area focus on global homological features of a subset of a Euclidean space. In this case, homology recovery predicates on imposing well understood geometric conditions on the underlying set. Typically, these conditions guarantee that small enough neighborhoods of the set have the same homology as the set itself. Existing work on recovering local homological features of a space from samples employs similar conditions locally. However, such local geometric conditions may vary from point to point and can potentially degenerate. For instance, the size of local homology preserving neighborhoods across all points of interest may not be bounded away from zero. In this paper, we introduce more general and robust conditions for local homology recovery and show that tame homology stratified sets, including Whitney stratified sets, satisfy these conditions away from strata boundaries, thus obtaining control over the regions where local homology recovery may not be feasible. Moreover, we show that true local homology of such sets can be computed from good enough samples using Vietoris–Rips complexes.

Appendix

Proof of Lemma 3

It is known that (b)-regular stratified sets belong to a wider class of (c)-regular stratified sets introduced in Bekka (1991). The definition of (c)-regularity is somewhat technical, and we refer the interested reader to the original paper by Bekka or to the book by Pflaum (see Pflaum 2001, Section 1.4.13). It is also known that transverse union of (c)-regular stratified sets with strata $\mathcal {S}$ and $\mathcal {R}$ is again a (c)-regular stratified set with the stratification $\mathcal {S}\cup _t\mathcal {R}$. Let us show that condition (b) is satisfied for $\mathcal {S}\cup _t\mathcal {R}$. Take $Z,W\in \mathcal {S}\cup _t\mathcal {R}$, with $Z\le W$. Let $z\in Z$, and consider sequences

$$\begin{aligned} z_i,w_i\rightarrow z, \quad {\text {with}}\quad z_i\in Z, w_i\in W \end{aligned}$$

such that

$$\begin{aligned} T_{w_i}W\rightarrow \tau ,\quad l_i=\overline{w_iz_i}\rightarrow l. \end{aligned}$$

Since condition (b) holds for both $\mathcal {S}$ and $\mathcal {R}$, it holds if $Z\ne X\cap Y$ for any $X\in \mathcal {S}$, $Y\in \mathcal {R}$. Suppose

$$\begin{aligned} Z=X_1\cap Y_1,\quad X_1\in \mathcal {S},\quad Y_1\in \mathcal {R}. \end{aligned}$$

Since condition (b) holds for transverse intersections, we only need to consider the case when $W=X_2-Y_2$ or $W=Y_2-X_2$ for some $X_2\in \mathcal {S}$, $Y_2\in \mathcal {R}$. Without loss of generality, we may assume the former, $W=X_2-Y_2$. Since $Z\le W$, we have $Z\subseteq X_1\le X_2$. So, $z_i\in X_1$, $w_i\in X_2$, and since condition (b) holds for $X_1\le X_2$, we get $l\subseteq \tau$. $\square$

The following proof employs the meaning of F(a, b) from (2).

Proof of Lemma 4

It follows from the proof of Lemma 7 that for all sufficiently small $\varepsilon \ge 0$ we can find $\rho _u>\rho _l>0$ such that $(\rho ,\rho ,\varepsilon )\in A_{x}(0)$ for $\rho \in [\rho _l,\rho _u]$. Thus, $A_{x}(0\,|\,\alpha =\varepsilon )\ne \emptyset$.

Now, suppose $\rho _u>\rho _l>0$ and $\varepsilon >0$ are such that

$$\begin{aligned} (\rho _l,\rho _l,\varepsilon ) \in A_{x}(0),\quad (\rho _u,\rho _u,\varepsilon ) \in A_{x}(0). \end{aligned}$$

Take $\rho _1, \rho _2$ $\in$ $[\rho _l,\rho _u]$, $\rho _1\le \rho _2$, $\varepsilon '\in [0,\varepsilon ]$. We have the following commutative diagram:

The bottom row consists of isomorphisms, and the maps from the bottom row to the top row along each column are injective, yielding $(\rho _2,\rho _1,\varepsilon ')\in A_{x}(0)$. The diagram also implies that if $(\rho _u,\rho _l,\varepsilon )\in A_{x}(0)$ then $(\rho ,\rho ,\varepsilon ')\in A_{x}(0)$ for all $\rho \in [\rho _l,\rho _u]$, $\varepsilon '\in [0,\varepsilon ]$. Consequently, each $A_{x}(0\,|\,\alpha =\varepsilon ')$ is a right isosceles triangle with the legs parallel to the axes and the hypotenuse lying on the diagonal, and $A_{x}(0\,|\,\alpha =\varepsilon )\subseteq A_{x}(0\,|\,\alpha =\varepsilon ').$

Note that the remark at the end of the proof of Lemma 7 implies that the vertical distance between the horizontal leg of our right triangle $A_{x}(0\,|\,\alpha =\varepsilon )$ and the R-axis tends to zero as $\varepsilon \rightarrow 0$. $\square$

The following proof employs the meaning of F(a, b) from (2).

Proof of Lemma 5

Part (1). Consider diagram (6) and note that $i_*$ factors through $F(\delta , R)$ for $\delta \in (0,\varepsilon )$. Therefore, we can replace $\varepsilon$ in diagram (6) by any $\delta \in [0,\varepsilon ]$ retaining all of the properties. Hence,

$$\begin{aligned} (R,r,\alpha )\in A_{x}(\varepsilon ) \implies (R,r,\alpha )\in A_{x}(\delta ) \end{aligned}$$

for all $\delta \in [0,\varepsilon ]$.

Part (2). Assume that each of the sets contains at least two elements (otherwise its a degenerate interval). So, let

$$\begin{aligned} \begin{aligned}&\alpha _1, \alpha _2 \in A_{x}(\varepsilon | R=R', r=r'),\\&r_1, r_2 \in A_{x}(\varepsilon | R=R', \alpha =\alpha '),\\&R_1,R_2\in A_{x}(\varepsilon | r=r',\alpha =\alpha '), \end{aligned} \end{aligned}$$

with $\alpha _1<\alpha _2$, $r_1<r_2$, $R_1<R_2$, and let

$$\begin{aligned} \alpha \in (\alpha _1,\alpha _2),\quad r\in (r_1,r_2),\quad R\in (R_1,R_2). \end{aligned}$$

Inclusions yield the following commutative diagrams:

In diagram (12), $\varphi _0$ is injective and restrictions $\mathrm {im\,}{\varphi _0}\rightarrow \mathrm {im\,}{\varphi _1}$ and $\mathrm {im\,}{\varphi _0}\rightarrow \mathrm {im\,}{\varphi _3\circ \varphi _2\circ \varphi _1}$ are isomorphisms. Therefore, we must also have $\mathrm {im\,}{\phi _0}\rightarrow \mathrm {im\,}{\varphi _2\circ \varphi _1}$ is an isomorphism. Thus,

$$\begin{aligned} \alpha _1,\alpha _2\in A_{x}(\varepsilon | R=R', r=r') \implies \alpha \in A_{x}(x | R=R', r=r') \end{aligned}$$

for all $\alpha \in [\alpha _1,\alpha _2]$. An analogous analysis of diagram (13) gives the result for the set $A_{x}(\varepsilon |R=R',\alpha =\alpha ')$. In diagram (14), $\varphi _3$ is injective, therefore $\psi _2\circ \varphi _2$ is injective, and hence $\varphi _2$ is injective. And since $\mathrm {im\,}{\varphi _3}\rightarrow \mathrm {im\,}{\psi _3}$ is an isomorphism, $\mathrm {im\,}{\varphi _2}\rightarrow \mathrm {im\,}{\psi _3\circ \psi _2}$ is also an isomorphism. Thus,

$$\begin{aligned} R_1, R_2\in A_{x}(\varepsilon | r=r', \alpha =\alpha ') \implies R\in A_{x}(\varepsilon | r=r',\alpha =\alpha ') \end{aligned}$$

for all $R\in [R_1, R_2]$.

Part (3). Take $R_2>R_1\ge R_u\ge r_u\ge r_1>R_l\ge r_l$ and consider the following commutative diagram induced by inclusions.

To simplify our exposition, we shall slightly abuse notation and write $(a,b)\rightarrow (c,d)$ to indicate the homomorphism $F(a,b)\rightarrow F(c,d)$, with $[(a,b)(c,d)]=\mathrm {im\,}{( (a,b)\rightarrow (c,d) )}$ denoting the corresponding image. Note that since $(R_1,r_1,\alpha _1)\in A_{x}(\varepsilon )$, the homomorphism $(0,R_1)\rightarrow (\varepsilon ,R_1)$ is injective and $[(0,R_1)(\varepsilon ,R_1)]\rightarrow [(\varepsilon ,R_1)(\alpha _1,r_1)]$ is an isomorphism.

To show the first inclusion, we start by showing that

$$\begin{aligned} R_2\notin A_{x}(\varepsilon |r=r_1,\alpha =\alpha _1)\implies A_{x}(\varepsilon |R=R_2,\alpha =\alpha _1)=\emptyset . \end{aligned}$$

Note that the map $(0,R_2)\rightarrow (0,R_1)$ in diagram (15) cannot be an isomorphism in this case. Indeed, if it is, then

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_1)] \approx [(0,R_1)(\varepsilon ,R_1)] \approx [(0,R_2)(\varepsilon ,R_2)], \end{aligned}$$

hence

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_1)] \end{aligned}$$

is an isomorphism, implying $R_2\in A_{x}(\varepsilon |r=r_1,\alpha =\alpha _1)$, which is a contradiction. But if $(0,R_2)\rightarrow (0,R_1)$ is not an isomorphism then it follows directly from the definition of $A_{x}(\varepsilon )$ that $(R_2,r',\alpha ')\notin A_{x}(\varepsilon )$ for any $r',\alpha '$.

Now let

$$\begin{aligned} R_2\in A_{x}(\varepsilon |r=r_1,\alpha _1), \end{aligned}$$

and suppose that

$$\begin{aligned} r_l, r_u \in A_{x}(\varepsilon |R=R_1,\alpha =\alpha _1). \end{aligned}$$

Then in diagram (15) the maps

$$\begin{aligned}{}[(0,R_1)(\varepsilon ,R_1)] \rightarrow [(\varepsilon ,R_1)(\alpha _1,r_i)],\quad i\in \{l,u\}, \end{aligned}$$

as well as the map

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_1)], \end{aligned}$$

are isomorphisms. Since $(0,R_2)$ $\rightarrow$ $(0,R_1)$ is an isomorphism in this case, we have

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_1)] \approx [(0,R_1)(\varepsilon ,R_1)] \approx [(0,R_2)(\varepsilon ,R_2)], \end{aligned}$$

which implies that

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_i)],\quad i\in \{l,u\}, \end{aligned}$$

are isomorphisms. Thus,

$$\begin{aligned} r_l,r_u \in A_{x}(\varepsilon |R=R_2,\alpha =\alpha _1). \end{aligned}$$

For the second inclusion, we again start by showing

$$\begin{aligned} r_l\notin A_{x}(\varepsilon |R=R_1,\alpha =\alpha _1) \implies A_{x}(\varepsilon |r=r_l,\alpha =\alpha _1)=\emptyset . \end{aligned}$$

So, suppose that the map

$$\begin{aligned}{}[(0,R_1)(\varepsilon ,R_1)]\rightarrow [(\varepsilon ,R_1)(\alpha _1,r_l)] \end{aligned}$$

is not an isomorphism. Since

$$\begin{aligned}{}[(0,R_1)(\varepsilon ,R_1)]\rightarrow [(\varepsilon ,R_1)(\alpha _1,r_1)] \end{aligned}$$

is an isomorphism, this implies that

$$\begin{aligned}{}[(\varepsilon ,R_1)(\alpha _1,r_1)] \rightarrow [(\alpha _1,r_1)(\alpha _1,r_l)] \end{aligned}$$

is not injective. Clearly, if

$$\begin{aligned} R_2 \notin A_{x}(\varepsilon |r=r_1,\alpha =\alpha _1) \end{aligned}$$

then

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_1)] \end{aligned}$$

is not an isomorphism, implying that

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_l)] \end{aligned}$$

is not an isomorphism, i.e.

$$\begin{aligned} R_2 \notin A_{x}(\varepsilon |r=r_l,\alpha =\alpha _1). \end{aligned}$$

So, assume

$$\begin{aligned} R_2\in A_{x}(\varepsilon |r=r_1,\alpha =\alpha _1). \end{aligned}$$

Then we see that maps

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha _1,r_l)] \quad {\text {and}}\quad [(0,R_u)(\varepsilon ,R_u)] \rightarrow [(\varepsilon ,R_u)(\alpha _1,r_l)] \end{aligned}$$

are not isomorphisms. Therefore,

$$\begin{aligned} R_2, R_u \notin A_{x}(\varepsilon |r=r_l,\alpha =\alpha _1). \end{aligned}$$

Now, assuming that

$$\begin{aligned}{}[(0,R_l)(\varepsilon ,R_l)] \rightarrow [(\varepsilon ,R_l)(\alpha _1,r_l)] \end{aligned}$$

is an isomorphism would imply that

$$\begin{aligned}{}[(0,R_l)(\alpha _1,R_l)] \rightarrow [(\alpha _1,R_l)(\alpha _1,r_l)] \end{aligned}$$

is injective, which is impossible since

$$\begin{aligned}{}[(0,R_1)(\alpha _1,r_1)] \rightarrow [(\alpha _1,r_1)(\alpha _1,r_l)] \end{aligned}$$

is not injective. Thus, we also have

$$\begin{aligned} R_l \notin A_{x}(\varepsilon |r=r_l,\alpha =\alpha _1), \end{aligned}$$

and so

$$\begin{aligned} A_{x}(\varepsilon |r=r_l,\alpha =\alpha _1) = \emptyset . \end{aligned}$$

Now let

$$\begin{aligned} r_l \in A_{x}(\varepsilon |R=R_1,\alpha =\alpha _1), \end{aligned}$$

and suppose that

$$\begin{aligned} R_2, R_u \in A_{x}(\varepsilon |r=r_1,\alpha =\alpha _1). \end{aligned}$$

Then in diagram (15) the maps

$$\begin{aligned}{}[(0,R_i)(\varepsilon ,R_i)] \rightarrow [(\varepsilon ,R_i)(\alpha ,r)],\quad i\in \{u,l\}, \end{aligned}$$

as well as the maps

$$\begin{aligned} (0,R_2) \rightarrow (0,R_1)\quad {\text {and}}\quad [(0,R_1)(\varepsilon ,R_1)] \rightarrow [(\varepsilon ,R_1)(\alpha _1,r_l)], \end{aligned}$$

are isomorphisms. The latter implies that

$$\begin{aligned}{}[(\varepsilon ,R_1)(\alpha _1,r_1)] \rightarrow [(\alpha _1,r_1)(\alpha _1,r_l)] \end{aligned}$$

is injective. Therefore,

$$\begin{aligned}{}[(0,R_2)(\varepsilon ,R_2)] \rightarrow [(\varepsilon ,R_2)(\alpha ,r_1)] \quad {\text {and}}\quad [(0,R_u)(\varepsilon ,R_u)] \rightarrow [(\varepsilon ,R_u)(\alpha ,r_1)] \end{aligned}$$

are isomorphisms, and so

$$\begin{aligned} R_2, R_u \in A_{x}(\varepsilon | r=r_l, \alpha = \alpha _1). \end{aligned}$$

Part (4) follows from the above results and the fact that intersection of intervals is an interval. $\square$

Proof of Lemma 6

The first part follows immediately from the definitions and the fact that

$A_{x}([0,\varepsilon _1]\,|\, \alpha \le \beta _1) \supseteq A_{x}([0,\varepsilon _2]\,|\, \alpha \le \beta _2)$.

For part two, note that

$$\begin{aligned} 0\le \bar{\delta }_{x}(\varepsilon ,\beta )\le \sup _{\alpha '\in [\alpha _{x}^{l}(\varepsilon ), \alpha _{x}^{u}(\varepsilon )]}{ d_H(A_{x}(0\,|\, \alpha =\alpha '), A_{x}(\varepsilon \,|\, \alpha =\alpha '))}\rightarrow 0\quad {\text {as}}\quad \varepsilon \rightarrow 0. \end{aligned}$$

The other limit follows from the fact that $\alpha$-sections of $A_{x}(0|\alpha \in [0,\beta ])$ are right isosceles triangles (in the (R, r)-plane) whose lower legs descend to the R-axis as $\beta$ $\rightarrow 0$ (see proofs of Lemmas 7, 4 for details).

Part three now follows from parts one and two. Indeed, weak seemliness implies that

$$\begin{aligned} A_{x}([0,\varepsilon ]\,|\, \alpha \le \beta ) \end{aligned}$$

has non-empty interior for small enough $\varepsilon$ and $\beta \in [\alpha _{x}^{l}(\varepsilon ), \alpha _{x}^{u}(\varepsilon )]$. Hence,

$$\begin{aligned} \tau _{x}(\varepsilon ,\beta )>0. \end{aligned}$$

Then part two implies that by reducing $\varepsilon$ we can achieve

$$\begin{aligned} \tau _{x}(\varepsilon ',\beta )>\gamma , \end{aligned}$$

where

$$\begin{aligned} \varepsilon ' = \alpha _{x}^{l}(g(0,\varepsilon )),\quad \beta = g(\varepsilon ',\varepsilon ),\quad \gamma =f(0,\varepsilon )+f(\varepsilon ',\varepsilon ). \end{aligned}$$

Hence, $\varepsilon \in E_{x}(f,g)$.

The claim that $E_{x}(f,g)$ is an interval follows from the monotonicity of $\tau _{x}(\varepsilon ,\beta )$, since it implies that if $\varepsilon _2\in E_{x}(f,g)$ and $\varepsilon _1\le \varepsilon _2$, then $\varepsilon _1\in E_{x}(f,g)$. $\square$

Proof of Lemma 7

We shall prove the statement of the lemma for any compact and non-empty $L\subseteq X$, $X\in \mathcal {S}$. It is enough to show that $\alpha$-sections $A_{L}(0\,|\, \alpha =\varepsilon )\ne \emptyset$ for all sufficiently small $\varepsilon >0$. The rest of the statement follows from Lemma 4 and the fact that any intersection of right isosceles triangles with legs parallel to the axes and the hypotenuse lying on the diagonal is another such triangle.

Let

$$\begin{aligned} W_1(x)&= \{\rho>0\,|\quad \forall \; \rho '\in (0,\rho ]\; H(K, K-B_{\rho '}(x)) \approx H(K, K-\{x\})\},\\ W_2(x)&= \{\rho>0\,|\quad \forall \; \rho '\in (0,\rho ]\; D_{\rho '}(x)\quad {\text { is a topological ball}}\},\\ W_3(x)&= \left\{ \begin{aligned}&\{\rho >0\,|\quad \forall \; \rho '\in (0,\rho ]\; S_{\rho '}(x)\pitchfork K\},&\quad K \;{\text { is a Whitney stratified set}}\\&\mathbb {R},&\quad {\text {otherwise.}} \end{aligned} \right. \end{aligned}$$

Define

$$\begin{aligned} \bar{\rho }(x) = \sup {(W_1(x)\cap W_2(x)\cap W_3(x))}. \end{aligned}$$

The fact that $\mathbb {X}$ locally strongly convex, the definition of homology stratification, and properties of Whitney stratified sets imply $\bar{\rho }(x)>0$ for any $x\in K$. We shall show that $\bar{\rho }(L)$ $=$ $\inf _{x\in L}{\bar{\rho }(x)}>0$. Suppose the opposite. Then we can find a sequence $x_n\in L$ such that $\bar{\rho }(x_n)\rightarrow 0$. Due to compactness of L, we can assume without loss of generality that $x_n$ is convergent. Let $\lim _{n\rightarrow \infty }{x_n} = \hat{x}\in L$. For any $\rho$ $\in$ $(0,\bar{\rho }(\hat{x}))$ we have $x_n$ $\in$ $B_{\rho }(\hat{x})$ for all sufficiently large n. Take any $\rho '$ $\in$ $(0,\rho -d(\hat{x},x_n))$. Then we have induced homomorphisms:

$$\begin{aligned} H(K, K-B_{\rho }(\hat{x})) \rightarrow H(K, K-B_{\rho '}(x_n)) \rightarrow H(K, K-\{x_n\}). \end{aligned}$$

Conditions for homology stratification imply that if $\rho$ is sufficiently small then all of the above homomorphisms are isomorphisms. In particular,

$$\begin{aligned} H(K, K-B_{\rho '}(x_n)) \rightarrow H(K, K-\{x_n\}) \end{aligned}$$

is an isomorphism. But we can have $\rho '>\bar{\rho }(x_n)$ for large n. Contradiction. It follows that $A_{L}(0)\ne \emptyset$ since contains triples of the form $(\rho , \rho ',0)$, where $\rho ,\rho '$ $\in$ $(0,\bar{\rho }(L))$, $\rho \ge \rho '$.

Let U be a small enough neighborhood of K so that there is a retraction $\pi :U\rightarrow K$. Take $\rho _u>0$, $\beta >0$. For any $x\in K$ and any $\rho \in [0,\rho _u]$, the preimage

$$\begin{aligned} U_{\rho }(x) = \pi ^{-1}(K-D_{\rho }(x)) \end{aligned}$$

is an open set containing $K-B_{\rho +\beta }(x)$. Due to compactness of the latter, we can find $\varepsilon (x,\rho ,\beta )>0$ such that

$$\begin{aligned} D_{\varepsilon (x,\rho ,\beta )}(K) - B_{\rho +\beta }(x) \subseteq U_{\rho }(x). \end{aligned}$$

We claim that $\varepsilon (x,\rho ,\beta )$ can be chosen so that

$$\begin{aligned} \bar{\varepsilon }(\beta ) = \inf \{\varepsilon (x,\rho ,\beta )\,|\, x\in L,\rho \in [0,\rho _u]\}>0. \end{aligned}$$

Suppose the opposite. Then we can find $x_n\in K$, $\rho _n\in [0,\rho _u]$, such that $\varepsilon (x_n,\rho _n,\beta )$ $\rightarrow$ 0. Due to compactness of K and $[0,\rho _u]$, we can assume $x_n$ $\rightarrow$ $\hat{x}$ $\in$ K, $\rho _n$ $\rightarrow$ $\hat{\rho }$ $\in$ $[0,\rho _u]$. Taking $\alpha >0$ sufficiently small, we get

$$\begin{aligned} U_{\hat{\rho }+\alpha }(\hat{x}) \supseteq D_{\varepsilon (\hat{x}, \hat{\rho },\beta )}(K) - B_{\hat{\rho }+\beta -\alpha }(\hat{x}). \end{aligned}$$

For all sufficiently large n, we have

$$\begin{aligned} \varepsilon (\hat{x}, \hat{\rho },\beta ) > \varepsilon (x_n,\rho _n,\beta ),\quad B_{\hat{\rho }+\beta -\alpha }(\hat{x}) \subseteq B_{\hat{\rho }+\beta }(x_n),\quad D_{\hat{\rho }+\alpha }(\hat{x}) \supseteq D_{\hat{\rho }}(x_n). \end{aligned}$$

Hence,

$$\begin{aligned} U_{\hat{\rho }}(x_n)\supseteq U_{\hat{\rho }+\alpha }(\hat{x})\supseteq D_{\varepsilon (\hat{x},\hat{\rho },\beta )}(K)-B_{\hat{\rho }+\beta -\alpha }(\hat{x})\supseteq D_{\varepsilon (\hat{x},\hat{\rho },\beta )}(K)-B_{\hat{\rho }+\beta }(x_n). \end{aligned}$$

This shows that $\varepsilon (x_n,\rho _n,\beta )$ could have been chosen as large as $\varepsilon (\hat{x},\hat{\rho },\beta )$. Contradiction.

Now, suppose $\rho _u$ $\in$ $(0,\bar{\rho }(L))$, $\beta$ $\in$ $(0,\bar{\rho }(L)-\rho _u)$. Take

$$\begin{aligned} \rho \in [0,\rho _u],\quad \rho ' \in [\rho +\beta ,\bar{\rho }(L)),\quad \varepsilon \in (0,\bar{\varepsilon }(\beta )),\quad x\in L. \end{aligned}$$

The image $\pi (D_{\varepsilon }(K)-B_{\rho '}(x))$ is a compact set inside $K-D_{\rho }(x)$. Hence,

$$\begin{aligned} \pi (D_{\varepsilon }(K)-B_{\rho '}(x))\subseteq K-B_{\rho ''}(x) \end{aligned}$$

for some $\rho ''>\rho$. Consider

Since $\pi _*\circ i_*$ is an isomorphism, $i_*$ must be injective, yielding $(\rho ',\rho ',\varepsilon )\in A_{L}(0)$. Hence, $A_{L}(0\,|\,\alpha =\varepsilon )\ne \emptyset$.

It is worth pointing out that the aforementioned argument implies that no matter how small $\rho '>0$ is, we can find $\varepsilon >0$ such that $(\rho ',\rho ',\varepsilon )\in A_{L}(0)$. $\square$

Proof of Lemma 8

For simplicity, we shall use sub-indexes ij instead of $X_{ij}^{\varvec{w}}$ and $\varvec{w}$ instead of $K^{\varvec{w}}$.

The definition of seemliness gives us, for each $X_{ij}^{\varvec{w}}$, functions $\alpha _{ij}^{l}$, $\alpha _{ij}^{u}:$ $[0,\varepsilon _{ij}]\rightarrow \mathbb {R}_+$. Let $\varepsilon _{\varvec{w}}$ $=$ $\min \{\varepsilon _{ij}\}>0$. Then we have functions

$$\begin{aligned} \alpha _{\varvec{w}}^{l} = \max \{\alpha _{ij}^{l}\} \quad {\text {and}}\quad \bar{\alpha }_{\varvec{w}}^{u} = \min \{\alpha _{ij}^{u}\} \end{aligned}$$

defined on $[0,\varepsilon _{\varvec{w}}]$.

It follows from the proof of Lemma 7 that $\alpha$-sections of $A_{\varvec{w}}(0)$, which is the intersection of all $A_{ij}(0)$, are non-empty for sufficiently small $\alpha$, say, $\alpha \in [0,\bar{\alpha }]$, and have the structure described in the lemma. We may reduce $\varepsilon _{\varvec{w}}$, if necessary, to make sure that $\alpha _{\varvec{w}}^{l}(\varepsilon _{\varvec{w}})\le \bar{\alpha }$, and define $\alpha _{\varvec{w}}^{u}$ $=$ $\min \{\bar{\alpha }_{\varvec{w}}^{u}, \bar{\alpha }\}$.

The properties of $\alpha$-sections from Lemma 5 imply that if

$$\begin{aligned} \begin{aligned}&D_H(A_{ij}(0\,|\,\alpha =\alpha '), A_{ij}(\varepsilon \,|\,\alpha =\alpha '))< a,\\&D_H(A_{i'j'}(0\,|\,\alpha =\alpha '), A_{i'j'}(\varepsilon \,|\,\alpha =\alpha ')) < b, \end{aligned} \end{aligned}$$

then

$$\begin{aligned} D_H(A_{ij}(0\,|\,\alpha&=\alpha ')\cap A_{i'j'}(0\,|\,\alpha =\alpha '), A_{ij}(\varepsilon \,|\,\alpha \\&=\alpha ')\cap A_{i'j'}(\varepsilon \,|\,\alpha =\alpha ')) < \max \{a,b\}. \end{aligned}$$

It then follows that

$$\begin{aligned} \sup _{\alpha '\in [\alpha _{\varvec{w}}^{l}(\varepsilon ), \alpha _{\varvec{w}}^{u}(\varepsilon )]}{d_H(A_{\varvec{w}}(0\,|\, \alpha =\alpha '), A_{\varvec{w}}(\varepsilon \,|\, \alpha =\alpha '))}\rightarrow 0\quad {\text {as}}\quad \varepsilon \rightarrow 0, \end{aligned}$$

which proves the lemma. $\square$