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Fix $n\in \mathbb N$. Let $X$ be a separable metric space of (inductive) dimension $n$. Let \begin{align} \Lambda(X)&=\{x\in X:X\text{ is $n$-dimensional at }x\}\\ \\ \Lambda^2(X)&=\Lambda(\Lambda(X))\\ &(=\{x\in \Lambda(X):\Lambda(X)\text{ is $n$-dimensional at }x\})\\ \\ \Lambda^3(X)&=\Lambda(\Lambda^2(X)) \end{align}

Note that there are weakly $n$-dimensional spaces which have the property $\dim(X)=n$ and $\dim(\Lambda(X))<n$. In this case, $\Lambda(X)\neq\varnothing$ while $\Lambda^2(X)=\varnothing$. In particular, $\Lambda^2(X)\neq \Lambda(X)$.

Question. Is $\Lambda^3(X)=\Lambda^2(X)$?

I am primarily interested in this question when $n=1$.

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  • $\begingroup$ Isn't $\Lambda^2(X)$ computed relative to the dimension of $\Lambda(X)$? In that case $\Lambda^2(X)$ need not be empty if $\dim\Lambda(X)<\dim X$. $\endgroup$
    – KP Hart
    Commented Apr 22, 2023 at 7:40
  • $\begingroup$ @KPHart I see your point. I suppose I want to fix $n$ at the outset, so that $\Lambda^2(X)$ is the set of points at which $\Lambda(X)$ is $n$-dimensional, etc. I will edit the question. $\endgroup$
    – D.S. Lipham
    Commented Apr 22, 2023 at 11:31

1 Answer 1

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It seems that the answer in general is "no". The Cook-Lelek continuum $X$ described in Figure 1 of the paper below has a left-most segment (arc) $S$ which can be replaced with a copy of $X$. Since the continuum $X$ is chainable, it should be possible to do this in a natural way. Call the new continuum $Y$. There is a countable set $Q$ intersecting each arc in $Y$, and the counterexample is the space $Y\setminus Q$. For this space it can be proved that $\Lambda^2$ is non-empty whereas $\Lambda^3$ is empty.

Lipham, David S., Totally disconnected subsets of chainable continua, Topology Appl. 317, Article ID 108187, 4 p. (2022). ZBL1495.54019.

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