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Let $(X, \mathcal{T})$ be the pseudoarc, which is a hereditarily indecomposable continuum. Here hereditary means on every subcontinuum. Subcontinuums of a continuum are exactly its closed and connected subsets.

In a previous question it was established that any indecomposable continuum is nowhere locally connected. Hence the pseudoarc is nowhere locally connected on any closed and connected subspace.

Suppose $U \subset X$ is non-empty and $(U, \mathcal{T}|U)$ does not contain isolated points.

Is $(U, \mathcal{T}|U)$ nowhere locally connected?

kaba
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  • Hi, unrelated to your question here, but you will probably find https://math.stackexchange.com/questions/4760309 interesting, as you have edited wikipedia regarding this notion. – PatrickR Sep 01 '23 at 00:46
  • Pick a point $x$, then take a composant $K_0$ of $X$ containing $x$. By Kuratowski, $K_1 = \cup P_{1, n}$ an increasing union of subcontinua of $X$ containing $x$ and converging to $X$ - aka union of pseudo-arcs. Take a composant $K_1$ of $P_{1, 1}$, repeat, taking composants $K_n$ of $P_{n,n}$. We may assume by taking sufficient indices that the Hausdorff distance $d_H(X, \lim P_{n,n}) > 0$ by density. Then $P = \cap P_{n,n}$ is not a point. By a standard result it's a continuum, aka the pseudo-arc again, not lc. This is how one must try to build a counterexample, but it fails. – John Samples Feb 29 '24 at 19:26
  • @JohnSamples I have trouble understanding what you are saying here. Are you saying that a counterexample, if it exists, must be of this form? Or that you think this`kind of approach could give an counterexample but this particular example didn't? – kaba Feb 29 '24 at 21:38
  • @kaba Any such counterexample must be constructed roughly in this manner (rigorous proof will be a bit messier, taking sub-pseudo-arcs in nbhds), it's just a proof sketch. This is a well-known fact but I don't know a reference for it. – John Samples Mar 01 '24 at 06:09
  • Oh sorry, rather I should say $\lim d_H(X, P_{n,n}) < \epsilon$ for arbitrary $\epsilon$ by taking sufficient indices in the second coordinate. Just being fussy in saying you can construct them so that their intersection isn't a point. – John Samples Mar 01 '24 at 13:35

1 Answers1

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This is a partial answer which provides an affirmitive answer for the non-empty locally closed subsets without isolated points.

  • If $U \in \mathcal{T} \setminus \{\emptyset\}$, then $\mathcal{T}|U$ is nowhere locally connected. For otherwise $\mathcal{T}$ would be not be nowhere locally connected.

  • The following shows a that each non-empty closed subset which does not contain a singleton component is nowhere locally connected. For let $V \subset X$ be such. Then the connected components of $V$ are closed, connected, and non-degenerate, hence pseudoarcs. Now let $x \in V$ and $U \in (\mathcal{T}|V)(x)$ be connected. By connectedness, $U$ must be contained in some component of $C$. Hence $V$ is not locally connected at $x$.

  • The following extends the previous to show that any non-empty closed subset without isolated points is nowhere locally connected. For if some component of $V$ is a singleton $\{x\}$, then there is no connected open subset which contains $x$, since that would imply that $\{x\}$ is open and hence isolated.

  • A corollary is that any non-empty locally closed subset without isolated points is nowhere locally connected. Here locally closed means an intersection of an open set and a closed set.

kaba
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