A Local Connectedness Condition for Compact, Connected Metric Spaces

Question

I am having trouble proving a result from a paper, which of course includes no proof. I wonder if the author had a simple - but flawed - argument in mind, or if I'm just being a dunce. It is Theorem 3 here: https://www.jstor.org/stable/2372339

It involves the following property:

If $X$ is a metric space and $x, y \in X$, then we say that $X$ is aposyndetic at $x$ with respect to $y$ if there is a compact, connected neighborhood of $x$ not containing $y$. Let $A(x)$ be the set of points $y$ such that $X$ is not aposyndetic at $y$ with respect to $x$ - that is to say, the points $y$ all of whose closed, connected neighborhoods also contain $x$.

The following is what I'm trying to prove: If $X$ is a compact, connected metric space and $x \in X$ is a point, then $A(x)$ is connected.

In fact it's also closed, which I can show. Maybe it is something very simple, but I'm just not seeing it. It may be a useful fact that the nested intersection of compact, connected subsets of $X$ is also connected. Does anyone see the proof?

Answer 1

I remember this is a tricky one.

The proof is nontrivial and appears as Theorem $3.1.21$ of Sergio Macias' Topics on Continua.

Answer 2

As in Macias:

$Ap_y^*(x)$ is closed, so it's sufficient to prove that it's connected. Suppose $Ap_y^*(x) = A \cup B$ is a separation and $x \in A$. Each is closed in $X$, so by normality we may pick an open neighborhood $U$ of $A$ such that $\overline{U} \cap B = \varnothing$. Then for every $u \in \partial{U}$, there is a continuum neighborhood $C_u$ of $u$ not containing $x$ since $\partial{U} \cap Ap_y^*(x) = \varnothing$.

Since $\partial{U}$ is compact it's covered by finitely many such sets, say $C_1, \dots, C_n$. If $V = U \setminus (\cup C_k)$, then by the Boundary Bumping Theorem $V^c$ has only finitely many components. If $y \in B$ and if $D$ is the component of $V^c$ containing $y$, then since $y \notin \overline{U}$ we have $y \in \text{int}(D)$. Since $V$ is open $V^c$ is closed and thus $D$ is closed in $X$. Thus it's a continuum neighborhood of $y$ and therefore $X$ is aposyndetic at $y$ with respect to $x$, a contradiction.