Naming a property of continua

Question

Let me define a continuum $X$ to be nice if for each pair $a,b\in X$ of distinct points, there are subcontinua $M,N\subseteq X$ such that: $a\in M\setminus N$; $b\in N\setminus M$; and $M\cup N=X$. (This is a very strong form of aposyndesis.) If a continuum is both locally connected (base consisting of connected open sets) and colocally connected (base consisting of open sets with connected complements), then it is clearly nice. My question is: does this property have an established name in the continuum-theoretic literature?

Answer 1

Yes, this is called freely decomposable and actually it's equivalent to aposyndesis as a global property of metric continua. Your "colocally connected" is a stronger form of a property called semi-locally connected (slc), where each point has a neighborhood base whose elements have only finitely many components in their complements. The slc property is also equivalent to freely decomposable as a global property.

Let $x, y \in X$ be distinct, where $X$ is a continuum.

Assume $X$ is freely decomposable, i.e. there are subcontinua $A \cup B = X$ with $x \in A \setminus B$ and $y \in B \setminus A$. If $x \notin \text{int}(A)$ then every neighborhood of $x$ intersects $X \setminus A \subset B \implies x \in \overline{B} = B$, a contradiction. Thus $A$ is a neighborhood of $x$ and $A \subset X \setminus \lbrace y \rbrace$, so that $X$ is aposyndetic at $x$ wrt $y$. But $x, y$ were arbitrary, so $X$ is aposyndetic.

Now assume $X$ is aposyndetic and $U$ is a neighborhood of $x$. Since $X$ is aposyndetic at $y$ wrt $x$ for every $y \notin U$ there is a neighborhood continuum $K_y$ of $y$ not containing $x$, and $X \setminus U$ is contained in $\cup K_y$. Since $X \setminus U$ is compact it's contained in finitely many such continua, say $K_1, K_2, \dots, K_n$. Then $V = X \setminus (\cup K_j)$ is a neighborhood of $x$ with $V \subset U$ and $X \setminus V = K_1 \cup \cdots \cup K_n$ has at most $n$ components. Therefore $X$ is slc at $x$ since $U$ was an arbitrary neighborhood of $x$. But $x$ was arbitrary, so $X$ is slc.

(note that the above implies: If $X$ is aposyndetic at $y$ wrt $x$ for every $y \in X \setminus \lbrace x \rbrace$ then $X$ is slc at $x$. As local properties slc is not equivalent to aposyndesis, even for planar curves)

Now assume that $X$ is slc and we show that $X$ is freely decomposable (this is the hardest part). Let $U$ be a neighborhood of $x$ with $y \notin \overline{U}$ such that $X \setminus U$ has only finitely many components $C_1, C_2, \dots, C_k$ with $y \in C_1$. Since $y \notin \overline{U}$ we have $y \in \text{int}(C_1)$. Note that since $U$ was open each $C_j$ is closed.

Since $X$ is slc at $y$ pick an open neighborhood $V \subset C_1$ of $y$ such that $X \setminus V$ has only finitely many components $D_1, D_2, \dots, D_n$ with $x \in D_1$ (similarly, $x \in \text{int}(D_1)$ and the $D_j$'s are closed). Then $\mathcal{A} = \lbrace C_1, \dots, C_k, D_1, \dots, D_n \rbrace$ is a finite cover of $X$ by closed, connected sets.

Then $C_1 \cup D_2, \cup \cdots \cup D_n$ is a neighborhood continuum of $y$ since $C_1$ is a neighborhood and $C_1 \cup D_i$ is connected for any $i$ by the boundary bumping theorem. Similarly, $D_1 \cup C_2 \cup \cdots \cup C_n = D_1$ since $U$ (and thus $U \cup C_2 \cup \cdots \cup C_n$ by bbp) is contained in $D_1$ since $D_1$ is a component of $X \setminus V$. Then these are a decomposition of $X$ between $x$ and $y$ as desired.

The above proofs work fine for Hausdorff continua. Below is the triangle basin $T$. $T$ is locally connected, and thus aposyndetic, at $v$, but is not slc at $v$. It's slc at $x$ but not aposyndetic at $x$ with respect to $v$.