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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.

kaba
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  • It's true, but you should add the assumption that $X$ is nontrivial (i.e. has more than one point). – Tyrone Aug 06 '23 at 03:57
  • @Tyrone Would you have a proof or proof sketch? – kaba Aug 06 '23 at 11:42
  • Macias's book "Topics on Continua" $\S$1.7 is a convenient reference. Theorem 1.7.26 states that a nontrivial continuum $X$ is indecomposable iff each proper subcontinuum has empty interior. On the other hand, if $X$ is locally connected at $x\in X$, then for every open $U\ni x$ there is a subcontinuum $W\subseteq X$ with nonempty interior (note that this property, called almost connected im kleinen at $x$, is much weaker than local connectedness at $x$). – Tyrone Aug 06 '23 at 16:30
  • @Tyrone Thanks. I guess to show local connectedness at $x \in X$ implies decomposability, I take an open connected neighorbood $V \subset X$ of $x$, and then its closure $C$ is compact and connected, so a subcontinuum. The interior of $C$ is $V$, which is non-empty by construction. The only problem I then have is to show that I can choose $V$ such that $C$ is proper. – kaba Aug 06 '23 at 18:26
  • @Tyrone Oh right, I keep forgetting it is a metric space. In that case I can get the closure to stay inside an open ball, and the open ball can be chosen to not cover everything. If you'd make it an answer, then I could accept. – kaba Aug 06 '23 at 18:53

1 Answers1

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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.

Tyrone
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  • Many thanks. I found Macias's book, and that is a good future reference. For some reason I did not get any notice of the answer, and so saw this a bit late. – kaba Aug 11 '23 at 01:38