Is an indecomposable continuum nowhere locally connected?

Question

Let $(X, \mathcal{T})$ be an indecomposable continuum. A continuum is a compact connected metric space. A continuum is indecomposable if it is not a union of two proper subcontinuums.

Is it true that $(X, \mathcal{T})$ is nowhere locally connected? How to prove it?

I found this claim from a comment in:

Construct a connected compact set but nowhere locally connected.

Answer 1

Following Sergio Macias's Topics on Continua $\S1.7$, start with the following.

Lemma Let $X$ be a continuum and $W\subseteq X$ a subcontinuum such that $X\setminus W=U\cup V$, where $U,V$ are nonempty, disjoint, open subsets of $X$. Then both $W\cup U$ and $W\cup V$ are subcontinua of $X$. $\;\blacksquare$

The proof is easy and is given in Macias's text. The following is also lifted from that book.

Proposition A nontrivial continuum $X$ is indecomposable if and only each of its proper subcontinua has empty interior.

Proof: Suppose $W\subset X$ is a proper subcontinuum with nonempty interior. We consider two cases.

If $X\setminus W$ is connected, then so is $\overline{X\setminus W}=X\setminus W^\circ$. Hence in this case $X=W\cup(X\setminus W^\circ)$, so $X$ cannot be indecomposable.

On the other hand, if $X\setminus W$ is disconnected, then $X\setminus W=U\cup V$ for disjoint, nonempty, open subsets $U,V\subseteq X$. By the lemma, $W\cup U$ and $W\cup V$ are subcontinuua of $X$ satisfying $X=(W\cup U)\cup (W\cup V)$. Since $U,V\not\subseteq W$ and $U,V$ are disjoint, both $W\cup U,W\cup V$ are proper subcontinua. Again we see that $X$ cannot be indecomposable.

Now assume that every proper subcontinuum of $X$ has empty interior. If $X$ were decomposable, then we could write $X=Y\cup Z$ with $Y,Z\subseteq X$ proper subcontinua. This would imply $Y^\circ\neq\emptyset \neq Z^\circ$, a contradiction. $\;\blacksquare$

Now let $X$ be a continuum and observe the following definitions.

1. $X$ is locally connected at $x\in X$ if $x$ has a neighbourhood base of connected open neighbourhoods.
2. $X$ is connected im kleinen at $x\in X$ if whenever $U\ni x$ is open, then there is a subcontinuum $W\subseteq X$ with $x\in W^\circ\subseteq W\subseteq U$.
3. $X$ is almost connected im kleinen at $x\in X$ if whenever $U\ni x$ is open, then there is a subcontinuum $W\subseteq X$ with $\emptyset\neq W^\circ\subseteq W\subseteq U$.

Clearly connected im kleinen at $x$ implies almost connected im kleinen at $x$. We claim that locally connected at $x$ implies that $X$ is connected im kleinen at $x$. Indeed, let $x\in U\subseteq X$ with $U$ open. Since $X$ is regular there is a closed neighbourhood $A\subseteq X$ with $a\in A\subseteq U$. If $V\subseteq X$ is a connected open neighbourhood of $x$ which is contained in $A$, then $W=\overline V$ is a subcontinuum of $X$ with $x\in W^\circ\subseteq W\subseteq U$.

Now the proposition above states that if $X$ is an irreducible continuum, then $X$ cannot be almost connected im kleinen at any of its points. In particular, $X$ is nowhere locally connected.

Note that in all of the above it is sufficient to assume that $X$ is a compact, connected, Hausdorff space.