On Wilder, Strongly Wilder, Continuumwise Wilder, D, $D^*$, and $D^{**}$ Continua

Abstract

We study properties of Wilder, strongly Wilder, continuumwise Wilder, D, $D^*$, and $D^{**}$ Hausdorff continua. We present an example of a colocally connected continuum that is not a $D^*$-continuum, answering a question by Espinoza and Matsuhashi. We give several positive answers to this question for unicoherent continua. We also present some equivalences for the class of homogeneous Hausdorff continua with the property of Kelley.

1 Introduction

We study properties of Wilder, strongly Wilder, continuumwise Wilder, D, $D^*$, and $D^{**}$ Hausdorff continua. We always consider Hausdorff continua unless it is stated otherwise. We present an example of a colocally connected metric continuum that is not a $D^*$-continuum, answering a question by Espinoza and Matsuhashi [5, Question 1] (Example 3.1). We give several positive answers to this question for unicoherent continua. For instance, if the continuum X is colocally connected (Corollary 3.4), or aposyndetic (Theorem 3.6), then X is a $D^*$-continuum. If the continuum X is hereditarily unicoherent, the following are equivalent: X is a Wilder continuum, X is a hereditarily Wilder continuum, X is an arcwise connected continuum (a dendron), and X is a continuumwise Wilder continuum (Theorem 3.8, we note that this result is mentioned in [11, Proposition 1.4] for metric continua). We observe that the colocally connected continua are strongly Wilder continua (Proposition 3.9). We also prove that freely decomposable continua are Wilder continua (Theorem 3.10). As a consequence of this result, we obtain that aposyndetic continua are Wilder continua (Theorem 3.11, a result originally proved by Wilder in [28, Theorem 1] for metric continua). It is known that all these classes of continua are decomposable, if we have a metric D-continuum for which its n-fold hyperspace is finite dimensional, then that continuum is hereditarily decomposable (Theorem 3.13). We note that if a continuum X contains at least two nondegenerate proper terminal subcontinua, then X cannot belong to any of the classes mentioned above (Lemma 3.15). We show that if X is a homogeneous continuum with the property of Kelley (not all homogeneous Hausdorff continua have the property of Kelley, see next paragraph), then the following properties are equivalent: X is freely decomposable, X is aposyndetic, X is semi-aposyndetic, X is Wilder, and X is a D-continuum (Theorem 3.20). If we consider a metric homogeneous continuum, we can add to the previous list: X is colocally connected and X is a strongly Wilder continuum (Theorem 3.22).

Charatonik constructed a homogeneous continuum without the property of Kelley [3]. In [1] Bellamy and Porter constructed an example of a homogeneous continuum which is not Effros. However, the continuum is locally connected. Hence, by [3, Proposition 2.6] continuum has the property of Kelley. Observe that, by [19, Theorem 9.6], the continuum defined by Bellamy and Porter does not have the uniform property of Effros. A global definition of the property of Kelley may be found in [18, Definition 1.6.17].

2 Definitions

A continuous function between topological spaces is a map. Given a topological space Z and a nonempty subset A of Z, Int(A) and Cl(A) denote the interior and the closure of A in Z, respectively. We add subscripts when necessary. A decomposition of Z is a family $\mathcal {G}$ of nonempty and mutually disjoint subsets of Z such that $\bigcup \mathcal {G}=Z$. A decomposition $\mathcal {G}$ of a Hausdorff topological space Z is said to be upper semicontinuous if the quotient map $q:Z\rightarrow \!\!\!\!\!\rightarrow Z/\mathcal {G}$ is closed. The decomposition $\mathcal {G}$ is continuous provided that q is open and closed. A decomposition $\mathcal {G}$ is monotone provided that each element of $\mathcal {G}$ is connected.

Let Z be a Hausdorff space. If V and U are subsets of $Z\times Z$, then

$$\begin{aligned} -V=\{(z',z)\ |\ (z,z')\in V\} \end{aligned}$$

and

$$\begin{aligned} V+U=\{(z,z'')\ |\ \hbox {there exists}\, z'\in Z \,\hbox {such that}\ (z,z')\in V\ \hbox {and}\ (z',z'')\in U\}. \end{aligned}$$

We write $1V=V$ and for each positive integer n, $(n+1)V=nV+1V$.

The diagonal of Z is the set $\Delta _Z=\{(z,z)\ |\ z\in Z\}$. An entourage of the diagonal of Z is a subset V of $Z\times Z$ such that $\Delta _Z\subset V$ and $V=-V$. The family of entourages of the diagonal of Z is denoted by $\mathfrak {D}_Z$. If $V\in \mathfrak {D}_Z$ and $(z,z')\in V$, then we write $\rho _Z(z,z')< V$.

Let Z be a nonempty set. A uniformity on Z is a subfamily $\mathfrak {U}$ of $\mathfrak {D}_Z\setminus \{\Delta _Z\}$ such that:

• (1) If $V\in \mathfrak {U}$, $U\in \mathfrak {D}_Z$ and $V\subset U$, then $U\in \mathfrak {U}$.

• (2) If V and U belong to $\mathfrak {U}$, then $V\cap U\in \mathfrak {U}$.

• (3) For every $V\in \mathfrak {U}$, there exists $U\in \mathfrak {U}$ such that $2U\subset V$.

• (4) $\bigcap \{V\ |\ V\in \mathfrak {U}\}=\Delta _Z$.

A uniform space is a pair $(Z,\mathfrak {U})$ consisting of a nonempty set Z and a uniformity on the set Z. For any uniformity $\mathfrak {U}$ on a set Z, the family

$$\begin{aligned} \mathfrak {G}=\{G\subset Z\ |\ \hbox {for every}\ z\in G,\ \hbox {there exists}\ V\in \mathfrak {U}\ \hbox {such that}\ B(z,V)\subset G\} \end{aligned}$$

is a topology on the set Z [4, 8.1.1]. The topology $\mathfrak {G}$ is called the topology induced by the uniformity $\mathfrak {U}$. It is well known that a topology is induced by a uniformity if and only if it is Tychonoff [4, 8.1.20].

Remark 2.1

Let Z be a Tychonoff space and let $\mathfrak {U}$ be a uniformity of Z that induces its topology. If $V\in \mathfrak {U}$, then we define the cover of Z, $\mathfrak {C}(V)=\{B(z, V)\ |\ z\in Z\}$.

Remark 2.2

Note that, by [4, 8.3.13], for every compact Hausdorff space Z, there exists a unique uniformity $\mathfrak {U}_Z$ on Z that induces the original topology of Z.

A compactum is a compact, Hausdorff space. A continuum is a connected compactum. A continuum X is decomposable provided that there exist two proper subcontinua K and L of X such that $X=K\cup L$. The continuum X is indecomposable if it is not decomposable. A subcontinuum K of a continuum X is terminal provided that for each subcontinuum L of X such that $L\cap K\not =\emptyset$, we have that either $L\subset K$ or $K\subset L$. A decomposition $\mathcal {G}$ of X is terminal if each element of $\mathcal {G}$ is a terminal subcontinuum of X. An arc is a continuum having exactly two points whose complement is connected. A metric arc is a continuum homeomorphic to [0, 1].

Given a continuum X and a positive integer n, we define the n-fold symmetric product of X, denoted $\mathcal {F}_n(X)$, by

$$\begin{aligned} \{A\subset X\ |\ A\ \hbox {is nonempty and has at most}\, n \,\hbox {points}\} \end{aligned}$$

and the n-fold hyperspaces of X, denoted $\mathcal {C}_n(X)$, by

$$\begin{aligned} \{A\subset X\ |\ A\ \hbox {is nonempty, closed and has at most}\, n \,\hbox {components}\}. \end{aligned}$$

Both are topologized by the Vietoris topology [17, Theorem 1.8.14]. More information about these hyperspaces may be found in [17].

A continuum X is a finite graph provided that it can be written as the union of finitely many metric arcs any two of which are either disjoint or intersect only in one or both of their endpoints.

A continuum X is a $\theta$-continuum ($\theta _n$-continuum, for some positive integer n) if for each subcontinuum K of X, we have that $X\setminus K$ only has a finite number of components ($X\setminus K$ has at most n components) [7]. Following [24], we say that a metric $\theta$-continuum ($\theta _n$-continuum) is of type A provided that it admits a monotone upper semicontinuous decomposition $\mathcal {D}$ whose quotient space is a finite graph, and it is of type $A'$ if, in addition, the elements of the decomposition have empty interior. A metric $\theta$-continuum ($\theta _n$-continuum) of type $A'$ for which the decomposition $\mathcal {D}$ is continuous is a continuously type $A'$ $\theta$-continuum $(\theta _n$-continuum).

Notation 2.3

Let Z be a continuum and let $\mathcal {H}(Z)$ be its group of homeomorphisms with compact-open topology. If z is an element of Z, then define $\gamma _z:\mathcal {H}(Z)\rightarrow Z$ by $\gamma _z(h)=h(z)$.

A homogeneous continuum Z is said to be an Effros continuum if $\gamma _z$ is an open map for all points z of Z (Notation 2.3).

Let Z be a Tychonoff space and let $\mathfrak {U}$ be a uniformity that induces the topology of Z. Then Z has the uniform property of Effros with respect to $\mathfrak {U}$, provided that for each $U\in \mathfrak {U}$, there exists $V\in \mathfrak {U}$ such that if $z_1$ and $z_2$ are two points of Z with $\rho _Z(z_1,z_2)<V$, there exists a homeomorphism $h:Z\rightarrow \!\!\!\!\!\rightarrow Z$ such that $h(z_1)=z_2$ and $\rho _Z(z,h(z))<U$, for all $z\in Z$. The entourage V is called an Effros entourage for U. A homeomorphism $h:Z\rightarrow \!\!\!\!\!\rightarrow Z$ satisfying $\rho _Z(z,h(z))<U$, for all $z\in Z$, is called a U-homeomorphism.

A continuum X is semi-aposyndetic if for each pair of distinct points x and y of X, there exists a subcontinuum W of X such that $\{x,y\}\cap Int(W)\not =\emptyset$ and $\{x,y\}\cap (X{\setminus } W)\not =\emptyset$. The continuum X is aposyndetic provided that for each pair of distinct points x and y of X, there exists a subcontinuum W of X such that $x\in Int(W)\subset W\subset X{\setminus }\{y\}$. Let n be a positive integer. Then the continuum X is n-aposyndetic if for each subset A of X having n points and every element x of $X\setminus A$, there exists a subcontinuum W of X such that $x\in Int(W)\subset W\subset X\setminus A$. The continuum X is finite set aposyndetic provided that it is n-aposyndetic for each positive integer n. The continuum X is colocally connected if each point of X has a local base of open subsets whose complements are connected. The continuum X is said to be weakly continuumwise aposyndetic provided that for every subcontinuum K of X that is irreducible about a finite set and each point $x\in X\setminus K$, there exists a subcontinuum L of X such that $x\in Int(L)\subset L\subset X\setminus K$ [23].

Charatonik [3] and Makuchowski [21] defined the pointwise version of the property of Kelley as follows:

A continuum X has the property of Kelley at a point $x_0$ if for every subcontinuum L of X containing $x_0$ and each open subset $\mathcal {A}$ of $\mathcal {C}_1(X)$ containing L, there exists an open subset $K(x_0,L,\mathcal {A})$ of X such that $x_0\in K(x_0, L,\mathcal {A})$ and if $x\in K(x_0, L,\mathcal {A})$, then there exists a subcontinuum M of X such that $x\in M$ and $M\in \mathcal {A}$. The open set $K(x_0,L,\mathcal {A})$ is called a Kelley set for $\mathcal {A}$, L and $x_0$. The continuum X has the property of Kelley provided that it has the property of Kelley at each of its points.

Wilder defined C and $C'$ spaces in [26, 27], respectively, for topological spaces. We change the name of $C'$-space to Strongly Wilder space. For the case of metric continua, the authors of [11] change the name C-continuum to Wilder continuum. The rest of following concepts are defined for metric continua in [6, 11, 22]; we extend their definitions to topological spaces.

Let Z be a topological space, then Z is a:

• Wilder space if it has at least three points and for each three distinct points x, y and z of Z, there exists a subcontinuum K of Z such that $x\in K$, $\{y,z\}\cap K\not =\emptyset$ and $\{y,z\}{\setminus } K\not =\emptyset$.

• Strongly Wilder-space provided that it has at least three points, and if x, y, and z are three distinct points of Z, there exists a subcontinuum K of Z such that $\{x,y\}\subset K$ and $z\in Z{\setminus } K$.

• Continuumwise Wilder space provided that it contains at least three points and for each three mutually disjoint subcontinua A, B and C of Z, there exists a subcontinuum L of Z such that $A\subset L$, either $B\subset L$ and $L\cap C=\emptyset$ or $C\subset L$ and $B\cap L=\emptyset$.

• D-space if for each two nondegenerate disjoint subcontinua K and L of Z, there exists a subcontinuum M of Z such that $K\cap M\not =\emptyset$, $L\cap M\not =\emptyset$, $K\setminus M\not =\emptyset$ or $L\setminus M\not =\emptyset$.

• $D^*$-space if for each two nondegenerate disjoint subcontinua K and L of Z, there exists a subcontinuum M of Z such that $K\cap M\not =\emptyset$, $L\cap M\not =\emptyset$, $K{\setminus } M\not =\emptyset$ and $L{\setminus } M\not =\emptyset$.

• $D^{**}$-space if for each two disjoint subcontinua K and L of Z, there exists a subcontinuum M of Z such that $K\cap M\not =\emptyset$, $L\cap M\not =\emptyset$ and $L{\setminus } M\not =\emptyset$.

Remark 2.4

Note that, by [13, Theorem 5.18], every Wilder continuum is a D-continuum. Observe that, in fact, the compactness of the original space not used in the proof of [13, Theorem 5.18]. Hence, every Wilder space is a D-space. Also, we have that every $D^*$-space is a $D^{**}$-space and every $D^{**}$-space is a D-space.

Remark 2.5

Observe that there exists a metric Wilder continuum that is not a continuumwise Wilder continuum [11, Example 1.3]. Also, there exists an aposyndetic metric continuum that is not a continuumwise Wilder continuum [11, Example 1.5] (compare with Theorem 3.11).

3 Results

Espinoza and Matsuhashi ask: If X is a semi-aposyndetic (or aposyndetic, or colocally connected) continuum, then is X a $D^*$-continuum? [5, Question 1]. The following example gives a negative answer to this question. The example is taken from [8, Example, p. 225]; we present the details of the appropriate modifications for the convenience of the reader.

Fig. 1

Example 3.1

Example 3.1

Let K be the Knaster indecomposable continuum [12, Example 1, p. 204] and let L be the set of all points p of K such that p is an endpoint of a semicircle of K. Define $X=(K\times \{0,1\})\cup (L\times [0,1])$ (see Fig. 1, we use convex arcs instead of semicircles).

We see that X is colocally connected. To this end, note that if q is a point of $K\times \{0\}$, there exists arbitrarily small open subsets U of X containing q such that each point of $X\setminus U$ can be joined with an arc to an arc of the form $\{c\}\times [0,1]$, where $c\in L$. Hence, there exists an arc from q to $K\times \{1\}$ and $X{\setminus } U$ is connected. Similarly if $q\in K\times \{1\}$. Assume that $q\in L\times [0,1]$. Then there exists arbitrarily small open subsets W of X such that either $W\cap (K\times \{0\}\cup K\times \{1\})=\emptyset$ or $W\cap (K\times \{0\}\cup K\times \{1\})\not =\emptyset$. If $W\cap (K\times \{0\}\cup K\times \{1\})=\emptyset$, then we may assume that W is of the form $C\times (a,b)$, where C is an open subset of L and (a, b) is an open subset of [0, 1]. If $W\cap (K\times \{0\}\cup K\times \{1\})\not =\emptyset$, then we may assume that W is of the form $C\times [0,b)$, where C is an open subset of L and [0, b) is an open subset of [0, 1], when $W\cap (K\times \{0\})\not =\emptyset$, or W is of the form $C\times (a,1]$, where C is an open subset of L and (a, 1] is an open subset of [0, 1], when $W\cap (K\times \{1\})\not =\emptyset$. In either case, it is clear that $X\setminus W$ is connected. Therefore, X is colocally connected.

Let A and B be two subcontinua of K in distinct composants of K. Let C be a subcontinuum of X such that $(A\times \{1\})\cap C\not =\emptyset$ and $(B\times \{0\})\cap C\not =\emptyset$. Let $\pi :K\times [0,1]\rightarrow \!\!\!\!\!\rightarrow K$ be given by $\pi ((k,t))=k$. Note that $\pi (C)=K$ ($\pi (C)$ intersects two different composants of K). Let W be a subcontinuum of C irreducible with respect to being mapped onto K. By [12, Theorem 4, p. 208], W is an indecomposable continuum. Let U be an open subset of K such that $Cl_K(U)\cap L=\emptyset$. Let

$$E=W\cap \pi ^{-1}(Cl_K(U))\cap (K\times \{1\})\ \hbox {and}\ G=W\cap \pi ^{-1}(Cl_K(U))\cap (K\times \{0\}).$$

Since $\pi (E)$ and $\pi (G)$ are closed subsets of K and $Cl_K(U)=\pi (E)\cup \pi (G)$, either $\pi (E)$ or $\pi (G)$ contains a nonempty open subset of K. Without loss of generality, assume that $\pi (E)$ contains a nonempty subset of V of K. Since $L\cap Cl_K(U)=\emptyset$, $E\cap \pi ^{-1}(V)$ is a nonempty subset of X. Hence, the indecomposable subcontinuum W of C contains a nonempty open subset of X. Since $[(K\times \{0\})\cup (K\times \{1\})]\cap W\not =\emptyset$, we have that either $K\times \{0\}\subset W\subset C$ or $K\times \{1\}\subset W\subset C$. Hence, either $B\times \{0\}\subset C$ or $A\times \{1\}\subset C$. Therefore, X is not a $D^*$-continuum.

Theorem 3.2

If X is a unicoherent colocally connected continuum, then X is n-aposyndetic for each $n\ge 1$.

Proof

Let $a_1,\ldots ,a_n$ and x be distinct points of X. Since X is colocally connected there exist open subsets $U_1,\ldots , U_n$ of X such that, for each $i,j\in \{1,\ldots ,n\}$, $a_i\in U_i$, $U_i\cap U_j=\emptyset$ whenever $i\ne j$, $x\notin Cl_X(U_i)$, and $X\setminus U_i$ is connected. Note that since X is unicoherent, $L=\bigcap _{i=1}^n(X\setminus U_i)$ is connected. Also, $x\in X\setminus \bigcap _{i=1}^n Cl_X(U_i)\subseteq L$. Thus, $x\in Int_X(L)$ and $\{a_1,\ldots ,a_n\}\cap L=\emptyset$. Therefore, X is n-aposyndetic. $\square$

Remark 3.3

In [6, Proposition 3.8], the authors prove that each 2-aposyndetic metric continuum is a $D^*$-continuum. Note that proof can be given without assuming the continuum is metric.

Since each 2-aposyndetic continuum is $D^*$-continuum [6, Proposition 3.8] (Remark 3.3), we have the following result. In particular, it gives a positive answer to [5, Question 1] for the class of unicoherent colocally connected continua.

Corollary 3.4

If X is a unicoherent colocally connected continuum, then X is a $D^*$-continuum.

Remark 3.5

Note that an aposydetic $\theta$-continuum or an aposyndetic hereditarily unicoherent continuum X is locally connected [18, Theorem 2.1.39 and Corollary 2.2.16, respectively]. Hence, in either case, X is a $D^*$-continuum.

In the following result, we prove that each unicoherent aposyndetic continuum is a $D^*$-continuum.

Theorem 3.6

Let X be a unicoherent continuum. If X is aposyndetic, then X is a $D^*$-continuum.

Proof

Let A and B be nondegenerate subcontinua of X such that $A\cap B=\emptyset$. Let $a\in A$. Since X is aposyndetic, there exists an open set U such that $a\in U$, $A{\setminus } Cl(U)\ne \emptyset$, $Cl(U)\cap B=\emptyset$ and $X\setminus U=L_1\cup \cdots \cup L_n$ where $L_1,\ldots ,L_n$ are the components of $X\setminus U$ [18, Theorem 1.4.26]. Since $A\setminus Cl(U)\ne \emptyset$, there exists $k\in \{1,\ldots ,n\}$ such that $Int(L_k)\cap A\ne \emptyset$. Note that $A\setminus L_k\ne \emptyset$. We consider two cases:

Case (1). $L_k\cap B=\emptyset$.

Since $B\subseteq X\setminus U$, there exists $l\in \{1,\ldots ,n\}$, $l\ne k$ such that $B\subseteq L_l$. Observe that $B\subseteq Int(L_l)$. Let $b\in B$ and let V be an open subset of $Int(L_l)$ such that $b\in V$, $B\setminus Cl(V)\ne \emptyset$ and $X{\setminus } V=\bigcup _{j=1}^m S_j$, where $S_1,\ldots ,S_m$ are the components of $X\setminus V$. Since $S_j\cap Cl(V)\ne \emptyset$, for each $j\in \{1,\ldots ,m\}$, and $Cl(V)\subseteq L_l$, we have that $L_l\cap S_j\ne \emptyset$, for every $j\in \{1,\ldots ,m\}$. Let $r\in \{1,\ldots ,m\}$ be such that $Int(S_r)\cap B\ne \emptyset$. Since $V\subseteq L_l$ and $X\setminus V=\bigcup _{j=1}^m S_j$, we have that

$$X=S_r\cup \left( L_l\cup \left( \bigcup _{j\ne r}S_j\right) \right) .$$

Since X is unicoherent, $S_r\cap (L_l\cup (\bigcup _{j\ne r}S_j))=S_r\cap L_l$ is connected. Let $K=S_r\cap L_l$. We have that $Int(K)\cap B\ne \emptyset$, $B{\setminus } K\ne \emptyset$ and $A{\setminus } K\ne \emptyset$. If $K\cap A\ne \emptyset$, then K is the continuum that we want. Hence, suppose that $K\cap A=\emptyset$. We consider two subcases:

Subcase (1.1). $L_k\cap K\ne \emptyset$.

In this case, $L_k\cup K$ is the required subcontinuum of X.

Subcase (1.2). $L_k\cap K=\emptyset$.

The proof of this subcase is similar to the proof of subcase (2.2), below

Case (2). $B\subseteq L_k$.

Since $B\cap Cl(U)=\emptyset$, $B\subseteq Int(L_k)$. Let $b\in B$ and let W be an open subset of X such that $b\in W\subseteq Int(L_k)$, $B{\setminus } Cl(W)\ne \emptyset$, $A\cap Cl(W)=\emptyset$, and $X{\setminus } W=R_1\cup \cdots \cup R_d$, where $R_1,\ldots ,R_d$ are the components of $X\setminus W$. Since $A\cap Cl(W)=\emptyset$, there exists $s\in \{1,\ldots ,d\}$ such that $A\subseteq R_s$. Hence, $A\subseteq Int(R_s)$. We have that $X=R_s\cup (L_k\cup (\bigcup _{j\ne s}R_j))$ and, since X is unicoherent, $R_s\cap (L_k\cup (\bigcup _{j\ne s}R_j))=R_s\cap L_k$ is a continuum. Let $G=R_s\cap L_k$. We consider two subcases:

Subcase (2.1). $G\cap B\ne \emptyset$.

Since $A\subseteq R_s$ and $B\subseteq L_k$, G intersects both A and B. Also, $A{\setminus } L_k\ne \emptyset$ and $B{\setminus } R_s\ne \emptyset$. Thus, G is the continuum that we want.

Subcase (2.2). $G\cap B=\emptyset$.

There exists $r\ne s$ such that $Int(R_r)\cap B\ne \emptyset$. Observe that $A{\setminus } G\ne \emptyset$, $Int(G)\cap A\ne \emptyset$ and $G\cap R_r=\emptyset$. Thus, by the Cut Wire Theorem [18, Theorem 1.4.8], there exists a component U of $X\setminus (G\cup R_r)$ such that $Cl(U)\cap G\not =\emptyset$ and $Cl(U)\cap R_r\not =\emptyset$. Let $W=Cl(U)$. Then W is a subcontinuum of X. We need to consider three subcases.

Subcase (2.2.1). $W\cap (A\cup B)=\emptyset$.

In this case, $W\cup G\cup R_r$ is the required subcontinuum of X.

Subcase (2.2.2). $W\cap (A\setminus G)\not =\emptyset$ and $W\cap (B\setminus R_r)\not =\emptyset$.

Here, W is the subcontinuum of X that we want.

Subcase (2.2.3). $W\cap A=\emptyset$ and $B\cap W\not =\emptyset$ (or $W\cap A\not =\emptyset$ and $B\cap W=\emptyset$).

In this case, $W\cup G$ (or $W\cup R_r$) is the required subcontinuum of X.

Therefore, X is a $D^*$-continuum. $\square$

Remark 3.7

Let X be a metric continuum, by [15, Theorem 8], $\mathcal {F}_n(X)$ is 2-aposyndetic for each $n\ge 2$. Thus, $\mathcal {F}_n(X)$ is a $D^*$-continuum [6, Proposition 3.8]. Also, for $n\ge 3$, by [15, Lemma 3], $\mathcal {F}_n(X)$ is colocally connected, note that it is also unicoherent [14, Theorem 8]. Hence, $\mathcal {F}_n(X)$ is a unicoherent that is not necessarily arcwise connected [2, Proposition 2.7].

Note that [11, Proposition 1.4] is true for continua.

Theorem 3.8

Let X be a hereditarily unicoherent continuum. Then the following are equivalent:

• (1) X is a Wilder continuum.

• (2) X is a hereditarily Wilder continuum.

• (3) X is an arcwise connected continuum (a dendron).

• (4) X is a continuumwise Wilder continuum.

Proof

Since X hereditarily unicoherent, we have that (1) implies (2). Suppose X is a hereditarily Wilder continuum and let x and y be two distinct point of X. Let K be an irreducible continuum between x and y [9, Theorem 2–10]. Now, by [26, Theorem 1], K is an arc. Thus, (2) implies (3). It is clear that (3) implies (4) and that (4) implies (1). $\square$

Proposition 3.9

If X is a colocally connected continuum, then X is a strongly Wilder continuum.

Proof

Let x, y and z be three distinct points of X. Since X is colocally connected, we have that there exists an open subset U of X such that $z\in U$, $\{x,y\}\cap U=\emptyset$ and $X{\setminus } U$ is connected. Let $K=X\setminus U$. Then K is a subcontinuum of X, $\{x,y\}\subset K$ and $z\in X{\setminus } K$. Therefore, X is a strongly Wilder continuum. $\square$

A continuum X is freely decomposable if for each pair of distinct points p and q of X, there exist two subcontinua P and Q of X such that $X=P\cup Q$, $p\in P{\setminus } Q$ and $q\in Q{\setminus } P$.

Theorem 3.10

If X is a freely decomposable continuum, then X is a Wilder continuum.

Proof

Let x, y and z be three points of X. Since X is freely decomposable, there exist two subcontinua K and L of X such that $X=K\cup L$, $y\in K{\setminus } L$ and $z\in L{\setminus } K$. Since x belongs to X, either $x\in K$ or $x\in L$. Assume that $x\in K$. Then $\{y,z\}\cap K=\{y\}$ and $\{y,z\}{\setminus } K=\{z\}$. Therefore, X is a Wilder continuum. $\square$

As a corollary, we obtain a different proof of the original result by Wilder [28, Theorem 1]. Note that, in this case, the proof is done for the case of Hausdorff continua.

Theorem 3.11

If X is an aposyndetic continuum, then X is a Wilder continuum.

Proof

If X is an aposyndetic continuum, then X is a freely decomposable continuum [18, Theorem 1.4.28]. The theorem now follows from Theorem 3.10. $\square$

Remark 3.12

Regarding Theorem 3.11, more can be said since Espinoza and Matsuhashi proved that every semi-aposyndetic continuum is a Wilder continuum [5, Theorem 3.6] (their proof works for the nonmetric case, we only need to use the Hausdorff version of the Boundary Bumping Theorem [18, Theorem 1.4.36]) and Lončar showed that each semi-aposyndetic continuum is a D-continuum [13, Lemma 5.11].

It is mentioned on [5, p. 2] that metric D-continua are decomposable; with an extra hypothesis, we obtain that they are hereditarily decomposable.

Theorem 3.13

If X is a metric D-continuum and $\dim (\mathcal {C}_n(X))<\infty$, then X is hereditarily decomposable.

Proof

Suppose that X is not hereditarily decomposable and Z is an indecomposable subcontinuum of X. Since $\dim (\mathcal {C}_n(X))<\infty$, we have that $\dim (\mathcal {C}_1(X))<\infty$ and, by $({}^*)$ of [20, Lemma 3.4], only finitely of many composants of Z are accessible from $X\setminus Z$. Let $\kappa _1$ and $\kappa _2$ be two composants not accessible form $X\setminus Z$. Let K be a subcontinuum of Z contained in $\kappa _1$, and let L be a subcontinuum of Z contained in $\kappa _2$. Then for any subcontinuum M of X such that $M\cap K\not =\emptyset$ and $M\cap L\not =\emptyset$, we have that $K\cup L\subset M$. Therefore, X is not a D-continuum. $\square$

The following result extends [5, Theorem 3.2] to strongly Wilder continua, continuumwise Wilder continua, $D^{**}$-continua, since $D^*$-continua are $D^{**}$-continua.

Theorem 3.14

If X is a continuum-chainable continuum, then X is a Wilder continuum, a D-continuum, a $D^*$-continuum and a $D^{**}$-continuum.

Lemma 3.15

Let X be a continuum. If K and L are two disjoint nondegenerate, proper, terminal subcontinua of X, then X is neither a Wilder continuum, a strongly Wilder continuum, a continuumwise Wilder continuum, a D-continuum, a $D^*$-continuum, nor a $D^{**}$-continuum.

Proof

Suppose X is a Wilder continuum. Let $x\in K$ and let y and z be points of L. Since X is a Wilder continuum, there exists a subcontinuum M of X such that $x\in M$, $\{y,z\}\cap M\not =\emptyset$ are $\{y,z\}\setminus M\not =\emptyset$. Note that $M\setminus K\not =\emptyset$ and $M\setminus L\not =\emptyset$. Since K and L are disjoint, nondegenerate, proper, terminal subcontinua of X, we have that $K\cup L\subset M$, a contradiction. Therefore, X is not a Wilder continuum. Hence, it is not a strongly Wilder continuum or a continuumwise Wilder continuum.

Now, we show that X is not a D-continuum; the proofs of the other two cases are similar. Let K and L be two disjoint, nondegenerate, proper, terminal subcontinua of X. If M is a subcontinuum of X such that $M\cap K\not =\emptyset$ and $M\cap L\not =\emptyset$, then, since K and L are terminal subcontinua of X, $M{\setminus } K\not =\emptyset$ and $M{\setminus } L\not =\emptyset$, we obtain that $K\cup L\subset M$. Therefore, X is not D-continuum. $\square$

Theorem 3.16

Let X be a a continuously type $A'$ metric $\theta$-continuum that is not a finite graph. If the quotient map has at least two nondegenerate fibers, then X is neither a Wilder continuum, a strongly Wilder continuum, a continuumwise Wilder continuum, a D-continuum, a $D^*$-continuum, nor a $D^{**}$-continuum.

Proof

Note that, by [16, Theorem 3.6], the quotient map q from X to a finite graph, T, is an atomic map. Hence, by [17, Theorem 8.1.25], for each $t\in T$, $q^{-1}(t)$ is a terminal subcontinuum of X. Since q has at least two nondegenerate fibers, the theorem follows from Theorem 3.15. $\square$

Theorem 3.17

Let X be a decomposable homogeneous continuum with the property of Kelley. If X is either a Wilder continuum, a strongly Wilder continuum, a continuumwise Wilder continuum, a D-continuum, a $D^*$-continuum, or a $D^{**}$-continuum, then X is aposyndetic.

Proof

We do proof for D-continua; the other cases are shown in a similar way. Suppose X is not aposyndetic. Then there exists a point $x_0$ of X such that $\mathcal {T}(\{x_0\})\not =\{x_0\}$ [18, Theorem 2.1.34]. Since X is homogeneous, by [18, Lemma 3.3.1], $\mathcal {T}(\{x\})\not =\{x\}$, for any $x\in X$. Hence, by [18, Theorem 3.3.2], $\mathcal {G}=\{\mathcal {T}(\{x\})\ |\ x\in X\}$ is an upper semicontinuous terminal decomposition of X. Therefore, by Lemma 3.15, X is not a D-continuum. $\square$

As a consequence, we obtain a generalization of [28, Theorem 4] to homogeneous plane strongly Wilder continuum, continuumwise Wilder continuum, D-continua, $D^*$-continua and $D^{**}$-continua.

Theorem 3.18

Let X be a homogeneous plane continuum. If X is either a Wilder continuum, a strongly Wilder continuum, a continuumwise Wilder continuum, a D-continuum, a $D^*$-continuum, or a $D^{**}$-continuum, then X is a simple closed curve.

Proof

We prove the result for D-continua. The other cases are done similarly. Suppose X is a homogeneous plane D-continuum. Since metric continua have the property of Kelley [17, Theorem 4.2.36], by Theorem 3.17, X is an aposyndetic continuum. Hence, by [10, Theorem 1], X is a simple closed curve. $\square$

Corollary 3.19

Let X be a decomposable continuum with the uniform property of Effros. If X is either a Wilder continuum, a strongly Wilder continuum, a continuumwise Wilder continuum, a D-continuum, a $D^*$-continuum, or a $D^{**}$-continuum, then X is aposyndetic.

Proof

Note that, by [18, Theorem 1.4.59], X is a homogeneous continuum. Since continua with the uniform property of Effros have the property of Kelley [18, Theorem 1.6.22], the corollary now follows from Theorem 3.17.$\square$

For the class of homogeneous continua with the property of Kelley, we have that several properties are equivalent.

Theorem 3.20

Let X be a decomposable homogeneous continuum with the property of Kelley. Then the following are equivalent:

• (1) X is a freely decomposable continuum.

• (2) X is an aposyndetic continuum.

• (3) X is a semi-aposyndetic continuum.

• (4) X is a Wilder continuum.

• (5) X is a D-continuum.

Proof

By [18, Theorem 1.4.28], (1) and (2) are equivalent. It is clear that (2) implies (3). Now, by [5, Theorem 3.6] (see Remark 3.12), we have that (3) implies (4). By [13, Theorem 5.18], we obtain that (4) implies (5). Since X is a decomposable homogeneous continuum with the property of Kelley, by Theorem 3.17, we obtain that (4) implies (2).$\square$

Corollary 3.21

Let X be a decomposable continuum with the uniform property of Effros. Then the following are equivalent:

• (1) X is a freely decomposable continuum.

• (2) X is an aposyndetic continuum.

• (3) X is a semi-aposyndetic continuum.

• (4) X is a Wilder continuum.

• (5) X is a D-continuum.

Proof

Note that, by [18, Theorem 1.4.59], X is a homogeneous continuum. Since continua with the uniform property of Effros have the property of Kelley [18, Theorem 1.6.22], the corollary now follows from Theorem 3.20. $\square$

For the class of metric homogeneous continua, we can say more:

Theorem 3.22

Let X be a decomposable metric homogeneous continuum. Then the following are equivalent:

• (1) X is a freely decomposable continuum.

• (2) X is an aposyndetic continuum.

• (3) X is a semi-aposyndetic continuum.

• (4) X is a Wilder continuum.

• (5) X is a D-continuum.

• (6) X is a colocally connected continuum.

• (7) X is a strongly Wilder continuum.

Proof

By [17, Theorem 4.2.36], X has the property of Kelley. Hence, by Theorem 3.20, we have that (1), (2), (3), (4), and (5) are equivalent. Suppose X is an aposyndetic metric continuum. Since every continuum has at least two non-cut points [9, Theorem 2–18], a homogeneous continuum does not have cut points. Now, since X is an aposyndetic metric continuum, we have that X is semilocally connected [18, Theorem 1.4.26]. Thus, if X is a semilocally connected homogenous metric continuum, by [25, (6.22), p. 737], X is colocally connected. Hence, (2) implies (6). By Proposition 3.9, (6) implies (7). It is clear that (7) implies (4). $\square$

Theorem 3.23

If X is a weakly continuumwise aposyndetic continuum, then X is a $D^*$-continuum.

Proof

Note that, by [23, Theorem 7, p. 22], X is a finite set aposyndetic continuum. In particular, X is a 2-aposyndetic continuum. Hence, by [5, Proposition 3.8], X is a $D^*$-continuum $\square$