Fractal tiles and quasidisks

Abstract

We show that the interiors of a large class of planar disklike self-similar tiles are quasidisks. We also show that in general, the interior of a disklike self-affine tile may not be a John domain and thus may not be a quasidisk.

Appendix

Lemma 5.1

In Case (b) of Lemma 4.5, let the smaller halves of $(T^1)^\circ {\setminus }[a,a']$ and $(T^1)^\circ {\setminus }[b',b]$ be $V^a$ and $V^b$ respectively. Then either $V^a\subseteq V^b$ or $V^b\subseteq V^a$. Also, $V^a$ and $V^b$ are attached to $[a,a']$ and $[b',b]$ respectively on different sides.

Proof

Let $U^a, U^b$ be the bigger halves of $(T^1)^\circ {\setminus } [a,a']$ and $(T^1)^\circ {\setminus } [b',b]$, respectively. Notice that as $(a,a')\cap (b',b)=\emptyset$, some half of $(T^1)^\circ {\setminus } [a,a']$ is contained in some half of $(T^1)^\circ {\setminus } [b',b]$. A bigger half cannot be contained in a smaller half, as by Lemma 4.2(a), $\mathrm{diam}\,U^\alpha \ge 3m/4>\mathrm{diam}\,V^\beta$ for $\{\alpha ,\beta \}=\{a,b\}$.

Claim 1. It is impossible for a smaller half to be contained in a bigger half either.

To see this, suppose $V^a \subset U^b$. Since $U^b\cap V^b=\emptyset$, we have $V^a\cap V^b=\emptyset$. However, as $a\in T^1\cap T''$ for some $T''\notin \mathcal {P}$, and $a'\in T^1\cap T'''$ for some $T'''\in \mathcal {P}{\setminus }\{T^1\}$, we conclude from Lemma 4.3 that there are vertices $p_a\in \partial V^a{\setminus } (a,a')$ and similarly $p_b\in \partial V^b{\setminus } (b',b)$. Since by our choice of $\theta$,

$$\begin{aligned} \vert p_a-p_b \vert \le \vert p_a-a \vert + \vert a-b \vert +\vert b-p_b \vert \le \mathrm{diam} V^a + \vert a-b \vert + \mathrm{diam} V^b < m, \end{aligned}$$

we have $p_a=p_b=:p$. Hence $\overline{V^a}\cap \overline{V^b}\ne \emptyset$. $\overline{V^a}\cap \overline{V^b}$ cannot contain more than one point. The reason is as follows. Suppose $\overline{V^a}\cap \overline{V^b}$ contains a nondegenerate arc $\widehat{q_1q_2}$. Then $\widehat{q_1q_2}{\setminus }\{q_1,q_2\}\subset (T^1)^\circ$, a contradiction. Suppose it contains two points $q_1, q_2$, but no point in the open subarc $\widehat{q_1q_2}{\setminus }\{q_1,q_2\}$ of $\widehat{aa'}\subset \partial V^a{\setminus } (a,a')$. Let $I_2=\widehat{q_1q_2}\subset \widehat{bb'}\subset \partial V^b{\setminus }(b',b)$ and $I_1=\widehat{q_1q_2}\subset \widehat{aa'}\subset \partial V^a{\setminus }(a,a')$. Then $I_1\cup I_2$ is a topological circle enclosing a region $S$ with $\mathrm{diam}\,S\le \mathrm{diam}\,I_1+\mathrm{diam}\,I_2\le \mathrm{diam}\,V^a+\mathrm{diam}\,V^b<m/2$. Hence by Lemma 4.2 (a) (i), there exists $T'\in \mathcal T$ with $(T')^\circ \cap S\ne \emptyset$ and $(T')^\circ \cap (\mathbb R^2{\setminus }\overline{S})\ne \emptyset$. Since $(T')^\circ$ is a topological disk, some point on $\partial S=I_1\cup I_2$ belongs to $(T')^\circ$, a contradiction. Hence $\overline{V^a}\cap \overline{V^b}$ is a singleton.

Suppose that $\overline{V^a}\cap \overline{V^b}=\{p\}$. If $p\ne a'$ or $b'$, then $p$ is a cut-point of some open neighborhood of $p$ in $T^1$, again contradictory to the disklikeness of $T^1$. Suppose $p=a'$ or $b'$. Let $\widehat{ap}$ and $\widehat{pb}$ be the arcs in $\partial V^a{\setminus } (a,a')$ and $\partial V^b{\setminus } (b',b)$ respectively. As the arc $\widehat{ap}\cup \widehat{pb}\subset \partial T^1$, and $\mathrm{diam}\,(\widehat{ap}\cup \widehat{pb}) \le \mathrm{diam}\,\overline{V^a} + \mathrm{diam}\,\overline{V^b}< m/2$ with $a,b\in \partial P$, Lemma 4.4 implies that $\widehat{ap}\cup \widehat{pb}\subset \partial P$. This is contradictory to $p\in \{a',b'\}\subset (a,b)\subset P^\circ$. Claim 1 is proved.

We assume in the following that $V^a\subset V^b$. Next, we show that $V^a, V^b$ are attached to the cross-cuts $[a,a']$ and $[b',b]$ on opposite sides. Suppose otherwise, and assume that both of them are attached to the cross-cuts from above. Notice that $V^a\subset V^b$ implies that $a,a'\in \partial V^b{\setminus }[b',b]$, and hence the arc $\widehat{a'a}$ of $\partial V^a{\setminus } (a,a')$ is also contained in $\partial V^b{\setminus } (b',b)$. In the following, let $\widehat{ba}, \widehat{ab'}$ be the arcs (or curves as specified) contained in $\partial V^b{\setminus } (b',b)$.

Claim 2. $a'\in \widehat{ab}$.

The reason is as follows. As we are supposing that $V^b$ is attached to $[b',b]$ from above, an upper half disk $B^+(x,\epsilon )\subset V^b$ is to the immediate left of the curve $[b',b]$. It follows that $V^b$ is to the left of the curve $\gamma \,{:=}\,[b',b]\cup \widehat{ba}\cup \widehat{ab'}$ $=\partial V^b$ and hence $\gamma$ is positively oriented. Similarly, by the supposition that $V^a$ is attached to $[a,a']$ from above, the curve $\widetilde{\gamma }\,{:=}\,[a,a']\cup \widehat{a'a}=\partial V^a$ is positively oriented. It follows that the points $b',b, a',a$, in that order, are located counterclockwise along $\partial V^b$ and hence along $\partial T^1$ (see Fig. 16). Hence the arc $\widehat{ab}\subset \partial V^b{\setminus } (b',b)$ contains $a'$, proving Claim 2.

As $\mathrm{diam}(\widehat{ab})\le \mathrm{diam}\,V^b <m/4$, $\widehat{ab}\subset \partial T^1$ and $\{a,b\}\subset \partial P$, Lemma 4.4 gives $\widehat{ab}\subset \partial P$. This contradicts the fact that $a'\in (a,b)\subset P^\circ$, and completes the proof of the lemma. $\square$

Fig. 16

Since $V^a\subset V^b$ and the curves $[b',b]\cup \widehat{ba}\cup \widehat{ab'}$ and $[a,a']\cup \widehat{a'a}$ are positively oriented, the points $b',b,a',a$, in that order, are arranged counterclockwise as in a. The ordering $b',b,a,a'$ as shown in b is impossible as $[a,a']\cup \widehat{a'a}$ is not positively oriented

Lemma 5.2

The point $p_a$ in Step 1 of the proof of Lemma 4.8 is a vertex.

Proof

If $p_a=a$, then $p_a\in T^1\cap T^h\cap T'$ for some $T'\notin \mathcal P$ and thus $p_a$ is a vertex.

Suppose $p_a\ne a$ and $p_a$ were not a vertex. Then there is a $\delta >0$ such that $B_\delta (p_a)\subset T^1\cup T^h$. Let $x$ be the first point along the curve $\widehat{aa'}$ in $\widehat{aa'}\cap \overline{B_\delta (p_a)}$. Notice that one hits $x$ before hitting $p_a$. As $x\in \overline{B_\delta (p_a)}\subset T^1\cup T^h$ and $x\in \widehat{aa'}\subset \partial T^1$, we have $x\in T^1\cap T^h$. Together with $a\in T^1\cap T'$ with $T'\notin \mathcal P$, the intermediate value theorem implies that there exists a vertex $p^*$ on $\widehat{ax}\subsetneqq \widehat{aa'}\subset \partial V^1$. On the other hand, as $b\in T^h\cap T''$ for some $T''\notin \mathcal {P}$, and $b'\in T^h\cap T'''$ for some $T'''\in \mathcal {P}$, Lemma 4.3 again gives a vertex $p^{**}\in \partial V^h{\setminus } [b',b]$. Since $|p^*-p^{**}|\le |p^*-a|+|a-b|+|b-p^{**}|<m$, $p^*=p^{**}$. Hence $p^*\in \overline{V^1}\cap \overline{V^h}$, contradicting $p_a$ being the first point along $\widehat{aa'}$ in $\overline{V^1}\cap \overline{V^h}$. $\square$

Lemma 5.3

In Step 2 of the proof of Lemma 4.8, $\overline{V^1}\cap \overline{V^h}=\{a\}$ or $\{b\}$.

Proof

Let $\widehat{a'p}, \widehat{ap}$ be arcs contained in $\partial V^1{\setminus } (a,a')$, and $\widehat{b'p}, \widehat{bp}$ be arcs in $\partial V^h{\setminus } (b',b)$.

Claim Under the conditions of Case (c) and the results of Step 1, we have the following three possibilities:

(1$\,{}^\circ$) $[a',b']\cup \widehat{b'p}\cup \widehat{pa'}$ contains a simple closed arc enclosing $a$ or $b$,

(2$\,{}^\circ$) $(\widehat{ap}\cup \widehat{pb})\cap [a',b']\ne \emptyset$, or

(3$\,{}^\circ$) $a=p$ or $b=p$.

See Fig. 17(A)–(I).

Fig. 17

Illustrations of the proof of Lemma 5.3. Only (e) (i.e., (E)), or with $a$ and $b$ interchanged, can occur, up to homotopic equivalence of $[a',b']$ relative to $\{a',b'\}$ among curves avoiding $[a,a'),(b',b]$ and $\widehat{ab}(=\!\widehat{pb})$. In a–c, and the degenerate cases (h) and (i) of c, $a\ne p$, while in e–g, $p=a$. a–i are topologically equivalent to A–I: straightening $\widehat{ap}\cup \widehat{pb}$ and $\widehat{a'p}\cup \widehat{pb'}$ in the latter gives the former. If $p\ne a$, then obviously the relative position of $[a',b']$ and various points can only be as in a–c (including (h) and (i)), or with $[a',b']$ winding about $a$ or $b$ in an even more complicated manner as indicated in d. Consequently either $[a',b']\cup \widehat{b'p}\cup \widehat{pa'}$ contains a simple closed arc enclosing $a$ or $b$ (a, b, d), or $[a',b']$ cuts through $\widehat{ap}\cup \widehat{pb}$ (c), proving the claim in Lemma 5.3. However the former is absurd as $a,b \in \partial P$, while the latter is absurd by Lemma 4.4 (which implies that $(\widehat{ap}\cup \widehat{pb})\cap [a',b']=\emptyset$). These reasons also exclude (f) and (g). Thus $p=a$ (or $b$) and only (e) (E) can occur, up to the equivalence mentioned at the very beginning of this caption

Proof of the claim

We start from the assumptions and and draw simpler figures topologically equivalent to the actual configurations of $V^1,V^h, a,a',b,b',p$ etc. that reveal the desired conclusion. As $V^1$ is attached to $[a,a']$ from below, it contains some lower half disk $B_\epsilon ^-(x)$ with $x\in (a,a')$, which lies on the left of the curve $[a',a]$. Hence the simple closed curve $[a',a]\cup \widehat{ap}\cup \widehat{pa'}$ is positively oriented. Similarly, that $V^h$ is attached to $[b',b]$ from above implies that the simple closed curve $[b',b]\cup \widehat{bp}\cup \widehat{pb'}$ is positively oriented. Hence the relative positions of $V^1, V^h, a,a', p, b, b'$ are as shown in Fig. 17a–i, corresponding to actual configurations in Fig. 17A–I. We use the same symbols $[a',b']$, $\widehat{ap}$ etc. to denote the corresponding curves in (a)–(i) and (A)–(I). (a)–(i) are probably simpler as $[a',b']$ is the only varying element there. In Fig. (a)–(c) with (h) and (i) as degenerate cases of (c), $p\ne a,b$ while (e)–(g) are degenerate cases of (c), with $a=p$. One can enumerate all the possible configurations by drawing $[a',b']$ subject to the constraint that it avoids $[a,a']$, $[b',b]$ and has no self-intersections.

If $p\ne a$ and $p\ne b$, it can be easily seen that the relative positions of $[a',b']$ and various points can only be as in Fig. (a)–(c) (including (h) and (i)) or with $[a',b']$ winding about $a$ or $b$ in an even more complicated way as indicated in (d). Consequently $[a',b']\cup \widehat{b'p}\cup \widehat{pa'}$ contains a simple closed arc enclosing $a$ or $b$ (Fig. (a, b, d)), or $[a',b']$ cuts through (intersects in an odd number of times with) $\widehat{ap}\cup \widehat{pb}$ (Fig. (c)), or $a=p$ or $b=p$. This completes the proof of the claim.

Assume (1${}^\circ$) holds. Then since $[a',b']\cup \widehat{b'p}\cup \widehat{pa'}\subset P$ and $P$ is disklike, $a$ (or $b$) would be an interior point of $P$, a contradiction. If (2$\,{}^\circ$) holds, then the arc $\widehat{ap}\cup \widehat{pb}$ contains a point in $[a',b']\subset (a,b)\subset P^\circ$. However, as $\widehat{ap}\cup \widehat{pb}\subset \partial T^1 \cup \partial T^h$ with $\mathrm{diam}(\widehat{ap}\cup \widehat{pb})<m$ and $\{a,b\}\subset \partial P$, Lemma 4.4 implies that it is contained in $\partial P$, a contradiction. Hence (3$\,{}^\circ$) holds and our assertion is proved. $\square$

Lemma 5.4

In Step 1 of the proof of Lemma 4.9, $p\notin \{a',b'\}$. Furthermore, $\overline{V^1}\cap \overline{V^h}$ is equal to $\{p\}$, $\widehat{pa'}\subset \partial V^1{\setminus } (a,a')$ or $\widehat{pb'}\subset \partial V^h{\setminus } (b',b)$.

Proof

We first prove that $p\notin \{a',b'\}$. Suppose otherwise that $p=a'\ne b'$; the cases $p=b'\ne a'$ and $p=a'= b'$ are similar. Let $\widehat{aa'}$ be the arc in $\partial V^1{\setminus } (a,a')$ and $\widehat{a'b}\subset \partial V^h{\setminus } (b',b)$. Then $\widehat{aa'}\cup \widehat{a'b}\subset \partial T^1\cup \partial T^h$ with diameter bounded above by $\mathrm{diam}\,V^1+\mathrm{diam}\,V^h < m/2$, and is hence contained in $\partial P$ by Lemma 4.4, contradicting $a'\in (a,b)\subset P^\circ$. Thus $p\notin \{a',b'\}$.

Suppose $\overline{V^1}\cap \overline{V^h}$ contains more than one point. We claim that either $\widehat{pa'}\subset \widehat{pb'}$ or $\widehat{pb'}\subset \widehat{pa'}$, which will imply that $\overline{V^1}\cap \overline{V^h}=\widehat{pa'}$ or $\widehat{pb'}$.

Suppose there is an $x\in \widehat{pa'}$ that is not in $\widehat{pb'}$. Let $I$ be the maximal relatively open subarc of $\widehat{pa'}$ containing $x$ that is not in $\widehat{pb'}$. First consider the case that its end-points $q_1,q_2\in \widehat{pa'}\cap \widehat{pb'}$. Let the arcs $I_1\,{:=}\,\widehat{q_1q_2}\subset \widehat{pa'}$ and $I_2\,{:=}\,\widehat{q_1q_2}\subset \widehat{pb'}$. Then $I_1\cup I_2$ is a topological circle of diameter bounded by $\mathrm{diam}\,\overline{V^1}+\mathrm{diam}\,\overline{V^h}<m/2$, enclosing a region $S$ with diameter $<m/2$. Notice that $I_1\cup I_2$ consists of boundary points of tiles in $\mathcal T$. Now by Lemma 4.2(a) (i), there is a $T'\in \mathcal {T}$ with $(T')^\circ \cap S\ne \emptyset$ and $(T')^\circ \cap (\mathbb {R}^2{\setminus } \overline{S})\ne \emptyset$, a contradiction (see Fig. 18a).

Now consider the case that $I=\widehat{qa'}{\setminus }\{q\}$ for some $q\in \widehat{pa'}$. As we are assuming that $\overline{V^1}\cap \overline{V^h}$ contains more than one point, $q\ne p$. Let $\widehat{pq}$ be the arc contained in $\widehat{pa'}$. If $\widehat{pq}\not \subset \widehat{pa'}\cap \widehat{pb'}$, then the argument in the previous paragraph gives a contradiction. Hence $\widehat{pq}\subset \widehat{pa'} \cap \widehat{pb'}\subset \overline{V^1}\cap \overline{V^h}$.

We will show that $q=b'$ and hence $\overline{V^1}\cap \overline{V^h}=\widehat{pb'}$ and our claim will be proved. Suppose $q\ne b'$ (i.e., $\widehat{qb'}\subset \widehat{bb'}$ is nondegenerate). We will show that $q$ is a vertex, which results in a contradiction as its distance to the vertex $p$ is $|p-q|\le \mathrm{diam}\overline{V^1}<m/4$.

As $q\in \overline{V^h}{\setminus } [b',b]$, there is a $\delta >0$ such that $B_\delta (q)\cap (T^h{\setminus }\overline{V^h})=\emptyset$. Then $(\widehat{qa'}{\setminus }\{q\})\cap B_\delta (q)\cap (T^h{\setminus }\overline{V^h})=\emptyset$. Together with the fact that $(\widehat{qa'}{\setminus }\{q\})\cap \overline{V^h}=\emptyset$, we get $(\widehat{qa'}{\setminus }\{q\})\cap B_\delta (q)\cap T^h=\emptyset$. Since $(\widehat{qa'}{\setminus }\{q\})\cap B_\delta (q)$ consists of boundary points of $T^1$, there exists a $T'\in \mathcal T{\setminus }\{T^1,T^h\}$ and a sequence $\{x_i\}\subset (\widehat{qa'}{\setminus }\{q\})\cap B_\delta (q)\cap T'$ with $x_i\rightarrow q$. Hence $q\in T^1\cap T^h\cap T'$ is a vertex. The proof is completed. $\square$

Fig. 18

a Since $\mathrm{diam}\,S<m/2$, there exists $T'\in \mathcal T$ with $(T')^\circ \cap S\ne \emptyset$ and $(T')^\circ \cap (\mathbb {R}^2{\setminus }\overline{S})\ne \emptyset$, a contradiction. b The thickened curve $(\widehat{qa'}{\setminus }\{q\})\cap B_{\delta }(q)$ is in $\partial T^1$, and has empty intersection with $\overline{V^h}$ and $T^h{\setminus }\overline{V^h}$ and hence $T^h$. Thus there exists a sequence of points in there converging to $q$, lying in $T^1\cap T'$ for some $T'\in \mathcal T{\setminus }\{T^1,T^h\}$. Hence $q\in T^1\cap T^h\cap T'$ is a vertex