Proof Suppose $F(y)=\{y\}$ is degenerate. We show that Y is zero-dimensional at y. Let V be an arbitrary relatively open subset of Y that contains y. Pick an open subset W of X such that $W \cap Y = V$ . Since $F(y) = \{y\} \subset V$ , there is by compactness an element $U \in \mathcal U_y$ such that $\overline {U} \subset W$ . But then $y \in U \cap Y \subset V$ , and $U \cap Y$ is clopen in Y.

Now suppose that $y\in Y$ and $z\in Y\setminus \{y\}$ . Pick by total disconnectedness of Y clopen subsets $C_0$ and $C_1 = Y \setminus C_0$ of Y such that $y \in C_0$ and $z \in C_1$ . There are disjoint open subsets $V_0$ and $V_1$ of X such that $V_0 \cap Y = C_0$ and $V_1 \cap Y = C_1$ . Observe that the boundary of $V_0$ is contained in $X \setminus Y$ . Since $V_1 \cap V_0 = \varnothing $ , this shows that $z \notin F(y).$ Hence, $F(y)\cap Y = \{y\}$ .

It remains to prove $F(y)$ is connected. Just suppose $F(y) = A \cup B$ where A and B are disjoint closed sets with $y \in A$ . Let U and V be disjoint open neighborhoods of A and B in X, respectively. By compactness, there exists $W \in \mathcal U_y$ such that $\overline W \subset U \cup V$ . Then $W \cap U \in \mathcal U_y$ , so $B = \varnothing .$

Proof We may assume that S is the unit circle in the complex plane $ \mathbb C$ , and

$$ \begin{align*}a_1=1,a_2={\mathrm{i}} , a_3=-1,a_4=-{\mathrm{i}} .\end{align*} $$

Let $\gamma \subset \mathbb C\setminus (A_1\cup S\cup A_3)$ be any arc from $b_2$ to $b_4$ . We will show that $\gamma $ intersects K. This will imply that $b_2$ and $b_4$ are in different components of $\mathbb C\setminus (A_1\cup S\cup A_3\cup K)$ , as all such components are path-connected.

In $\gamma \cup A_2\cup A_4$ , there is an arc $\hat \gamma $ intersecting S only at endpoints $\pm {\mathrm {i}}$ . Let $\alpha $ and $\beta $ be the left and right halves of the circle S. By the $\theta $ -curve theorem [Reference Munkres10, Lemma 64.1], $\text {int}(S)\setminus \hat \gamma $ is the union of two disjoint domains U and V whose boundaries are $\alpha \cup \hat \gamma $ and $\hat \gamma \cup \beta $ , respectively. Let W be an open ball centered at $-1$ that misses $\hat \gamma $ . Then $W\cap \text {int}(S)\subset U\cup V$ . Since $W\cap \text {int}(S)$ is connected and $-1\in \partial U$ , we have $W\cap \text {int}(S)\subset U$ . Thus, $U\cap A_3\neq \varnothing $ . It follows that $A_3\setminus \{-1\}\subset U$ . Thus, $b_3\in U$ . Likewise, $b_1\in V$ . Now K is a connected subset of $\text {int}(S)$ meeting both U and V. So $K\cap \hat \gamma \neq \varnothing $ . Therefore, $K\cap \gamma \neq \varnothing $ .

Proof For a contradiction suppose that $Y\subset X$ is totally disconnected and Borel, and $Z=\Lambda (Y)$ is uncountable. By Lemma 1 there exists $\varepsilon>0$ such that for uncountably many z’s, $\operatorname {\mathrm {diam}}(F(z))\geq 5\varepsilon $ . The set of all z’s with this property is closed in Y, thus it is Borel and contains a Cantor set. Let us just assume that $Z $ is a Cantor set in Y with the property $\operatorname {\mathrm {diam}}(F(z))\geq 5 \varepsilon $ for each $z\in Z$ . We may further assume that Z lies inside of an open ball of radius $\varepsilon $ centered at one of its points. Let S and $S'$ be the circles of radii $\varepsilon $ and $2\varepsilon $ , respectively, centered at the point. Observe that every $F(z)$ crosses $S'$ .

In the claims below, all boundaries are respective to X (not $\mathbb R ^2$ ).

Proof of Claim 4 For each z in the uncountable set $Z\cap U\setminus \{p\}$ there is a nondegenerate continuum $K(z)\subset F(z)$ containing z and missing $S\cup \{p\}$ . Just take any closed neighborhood N of z missing $S\cup \{p\}$ , and let $K(z)$ be the component of z in $F(z)\cap N$ ; by the boundary bumping theorem [Reference Nadler11, Theorem 5.4] $K(z)$ is nondegenerate. Note that $K(z)\cap Y=\{z\}$ by Lemma 1.

By the Suslinian property of X there exist

$$ \begin{align*}z_1,z_2,z_3,z_4\in Z\cap U\setminus \{p\},\end{align*} $$

such that $K(z_i)\cap K(z_1)\neq \varnothing $ for each $i\in \{2,3,4\}$ . Let $U_i\in \mathcal U_{z_i}$ be pairwise disjoint. By Remark 1, we may assume that each $U_i$ is path-connected. Since $F(z_i)$ extends beyond $S'$ , so does $U_i$ . Thus, there is an arc $A_i\subset U_i$ from $z_i$ to $a_i\in S'$ that intersects $S'$ only at $a_i$ . By a permutation of indices $i\in \{2,3,4\}$ , we may assume that the $a_i$ ’s are in cyclic order around $S'$ . Let $K=K(z_1)\cup K(z_3)$ . Note that $p\notin K$ .

Case 1: $p\in U_1\cup S'\cup U_3$ . Let W be the component of $z_2$ in $U_2\cap U\setminus S$ . Then W is a connected open subset of U intersecting Z, with $p\notin \overline W$ . Note also that W is clopen in $U_2\cap U\setminus S$ , which means that $\partial W\subset \partial (U_2\cap U\setminus S)$ . The boundary of any finite intersection of open sets is contained in the union of the individual boundaries. Therefore, $\partial W\subset \partial U_2 \cup \partial U\cup \partial (X\setminus S)\subset (X\setminus Y)\cup S$ .

Figure 1

Proofs of Claim 4 (left) and Claim 5 (right).

Case 2: $p\notin U_1\cup S'\cup U_3$ . Then certainly $p\notin A_1\cup S'\cup A_3$ and since $p\notin K$ we find that $p\in O=\mathbb R^2\setminus (A_1\cup S'\cup A_3\cup K)$ . By Lemma 2, $z_2$ and $z_4$ are in different components of O. Of these two components, at least one does not contain p. Let’s say that V is the component of $z_2$ in O, and $p\notin V$ . The component of p in O is an open set missing V, so $p\notin \overline V$ . Let W be the component of $z_2$ in $U_2\cap U \cap V\setminus S$ . Then W is a connected open subset of U intersecting Z, and $p\notin \overline {W}$ . It remains to show that $\partial W\subset (X\setminus Y)\cup S$ . Note the following:

• • $\partial V\subset A_1\cup S'\cup A_3\cup K$ ;

• • $\overline W$ misses $A_1\cup A_3$ because $W\subset U_2$ ;

• • $\overline W$ misses $S'$ because it lies in the closed disc bounded by S.

Hence, $\overline {W}\cap \partial V\subset K\setminus \{z_1,z_3\}\subset X\setminus Y.$ We conclude that

$$ \begin{align*} \partial W&\subset \overline W\cap \big[\partial U_2\cup \partial U \cup \partial V\cup \partial (X\setminus S)\big] \\ &\subset \partial U_2\cup \partial U \cup \big(\overline W\cap \partial V\big)\cup \partial (X\setminus S)\\ &\subset (X\setminus Y)\cup S.\end{align*} $$

This completes the proof of Claim 4.

By Claim 5, there are two connected open sets $W_{\langle 0\rangle }$ and $W_{\langle 1\rangle }$ in $X\setminus S$ , intersecting Z, with disjoint closures and boundaries in $(X\setminus Y)\cup S$ . Assuming that $\alpha \in 2^{<\mathbb N}$ is a finite binary sequence and $W_{\alpha }$ has been defined, apply Claim 5 to get connected open subsets $W_{\alpha ^\frown 0}$ and $W_{\alpha ^\frown 1}$ of $W_\alpha $ , each intersecting Z, with disjoint closures and boundaries in $(X\setminus Y)\cup S$ . Their boundaries must meet S because each $F(z)$ meets $S'$ . So for every infinite sequence $\alpha \in 2^{\mathbb N}$ ,

$$ \begin{align*}K_{\alpha}=\bigcap_{n=1}^\infty \overline{W_{\alpha\restriction n}},\end{align*} $$

is a continuum in X stretching from Z to S. Clearly if $\alpha \neq \beta $ then $K_{\alpha }$ and $K_{\beta }$ are disjoint. Therefore, $\{K_\alpha :\alpha \in 2^{\mathbb N}\}$ is an uncountable collection of pairwise disjoint nondegenerate subcontinua of X, a contradiction to the Suslinian property of X. Hence, $\Lambda (Y)$ must have been countable. This completes the proof of Theorem 3.

Proof Let X be a Suslinian plane continuum. Let $U_0,U_1,U_2,\ldots $ be the connected components of $\mathbb R^2\setminus X$ with $U_0$ unbounded. In each domain $U_m$ , there is a sequence $S^m_0,S^m_1\ldots $ of disjoint, concentric simple closed curves limiting to the boundary of $U_m$ . Let $A^0_0=\varnothing $ , and for each $m\geq 1$ let $A^m_0$ be the closed topological disc in $U_m$ that is bounded by $S^m_0$ . For each $n\geq 1$ , let $A^m_n$ be the closed annulus in $U_m$ that is bounded by $S^m_{n-1}$ and $S^m_{n}$ . In $A^m_n$ there is a finite collection of arcs $\mathcal I^m_n$ covering $\partial A^m_n$ such that each component of $A^m_n\setminus \bigcup \mathcal I^m_n$ has diameter less than $1/(m+n)$ . We can easily arrange that the boundaries of these components are simple closed curves, and $X^m_n=\bigcup \mathcal I^m_n$ is connected. For the reader who’d like more details, one can assume that X is a subset of the Riemann sphere $\hat {\mathbb C}=\mathbb C\cup \{\infty \}$ , so that all complementary components of X are simply connected. Let $\mathbb D=\{z\in \mathbb C:|z|<1\}$ . Given a component U of $\hat {\mathbb C}\setminus X$ , apply the Riemann mapping theorem to get a continuous bijection $f:\mathbb D\to U$ and transfer the circles $\{z\in \mathbb C:|z|=1-2^{-n}\}$ to U to get the curves $S_n$ . Apply uniform continuity on each compact annulus $ \{z\in \mathbb C : 1-2^{-n} \leq |z| \leq 1-2^{-n-1}\}$ to divide it further into a grid whose cells have images with a small enough diameter. See Figure 2.

Figure 2

Dividing a domain into cells via a conformal mapping of $\mathbb D$ .

Put

$$ \begin{align*}X'=X\cup \bigcup_{m=0}^\infty\bigcup_{n=0}^\infty X^m_n.\end{align*} $$

The components of $\mathbb R^2\setminus X'$ form a null sequence of domains whose boundaries are simple closed curves, so the continuum $X'$ is locally connected [Reference Whyburn13, VI Theorem 4.4]. Moreover, $X'$ is Suslinian because $X' \setminus X$ is a countable union of arcs.

Proof Let Z be the plane continuum from [Reference van Mill and Tuncali9] which has $\operatorname {\mathrm {bur}}(Z)=K$ . By construction, $\mathbb R^2 \setminus Z $ is a countable union of disjoint connected open sets with simple closed curves as boundaries. One complementary component of Z, say $W_0$ , is unbounded and the others, say $W_1, W_2, \ldots $ , form a sequence of bounded domains. Let $\mathcal I_n$ , for $n \geq 1$ , be a finite set of arcs in $\overline {W_n}$ such that the collection $\mathcal V_n$ of components of $W_n \setminus \bigcup \mathcal I_n$ has mesh less than $1/n$ . We can easily arrange that the boundary of every element of $\mathcal V_n$ is a simple closed curve, and $Z_n=\partial W_n\cup \bigcup \mathcal I_n$ is connected. Here one can even apply Carathéodory’s theorem to get $Z_n$ from a grid in the closed unit disc. As in Theorem 6,

$$ \begin{align*}Z' =Z\cup \bigcup_{n=1}^\infty Z_n,\end{align*} $$

is a locally connected continuum. Clearly, $\operatorname {\mathrm {bur}}(Z) \subset \operatorname {\mathrm {bur}}(Z')$ . For the other inclusion, note that points in $Z' \setminus Z$ have finite graph neighborhoods in $Z'$ , so are not in $\operatorname {\mathrm {bur}}(Z')$ . Now suppose $z\in Z\setminus \operatorname {\mathrm {bur}}(Z)$ . Then $z\in \partial W_n$ for some n, and there exists $V\in \mathcal V_n$ such that $z \in \partial V$ . So $z\notin \operatorname {\mathrm {bur}}(Z')$ . We conclude that $\operatorname {\mathrm {bur}}(Z') = \operatorname {\mathrm {bur}}(Z)=K$ .

Proof

Begin with a Lelek function $\varphi :C\to [0,1]$ whose positive graph

$$ \begin{align*}E=\big\{\langle x,\varphi(x)\rangle:x\in C\text{ and }\varphi(x)>0\big\},\end{align*} $$

is $1$ -dimensional at every point [Reference Dijkstra and van Mill5]. Let

$$ \begin{align*}L=\bigcup_{x \in C} \{x\}\times [0,\varphi(x)]\subset C\times [0,1].\end{align*} $$

The quotient of L that is obtained by shrinking $C\times \{0\}$ to a point is a continuum known as the Lelek fan. We will identify arcs in $L\setminus E$ to produce a Suslinian continuum (a dendroid, in fact) that homeomorphically contains E.

For each $n=1,2,3,\ldots $ let $\mathscr C_n$ be the natural partition of C into $2^n$ pairwise disjoint intervals of length $3^{-n}$ . For each $i=1,\ldots ,2^n-1$ define

$$ \begin{align*}A_{n,i}=\big\{x\in C:\varphi(x)\geq \textstyle\frac{2i+1}{2^{n+1}}\big\}.\end{align*} $$

Put $\langle x,0\rangle \sim \langle x',0\rangle $ for all $x,x'\in C$ . If $y>0$ then define $\langle x,y\rangle \sim \langle x',y\rangle $ if there exists $n,i$ such that $x,x'\in A_{n,i}$ belong to the same member of $\mathscr C_n$ , and $y\leq i/2^n$ . It is possible to see that $\sim $ is an equivalence relation, the equivalence classes under $\sim $ form an upper semicontinuous decomposition of L, and the quotient $D=L/\sim $ is a Suslinian dendroid with endpoint set E.