EDIT The question was originally stated in a general setting, but now it is in terms of plane continua as these are the cases I care about.

A $continuum$ is a compact, connected metric space.

By a $plane\ continuum$ I mean a continuum which is homeomorphic to a subset of $\mathbb{R}^2$.

Suppose $X$ is a locally path connected plane continuum. (Note that if a space is locally path connected, it is path connected iff it is connected.) Also, suppose $X$ has covering dimension $1$. I have conjectured that every subcontinuum of $X$ must also be locally path connected.

Each of my proof-attempts have been rather fragmented so it's difficult to write anything coherent, but here's my general approach:

Let $X$ be a locally path connected one dimensional plane continuum. Suppose $S\subset X$ is a non-locally path connected continuum. Choose $s\in S$ so that every open set around $s$ is not path connected. Now, using the local path connectivity of $X$, choose a path connected open set $M$ in $X$ with $s \in M$ so that $M\cap S = N$ is connected. (I haven't been able to prove that choosing an $M$ so that $M\cap S$ is connected is possible.)

Now choose $t \in N$. Note that there is no path from $t$ to $s$ in $N$. However, there is a path from $t$ to $s$ in $M$. I should be able to use this to reach a contradiction along the lines of "$X$ must have covering dimension greater than $1$". However, I'm struggling to formalise anything as I'm not sure how to restrict from the possibility that $X$ has many "loops". Note that I also haven't used the fact that $X$ is a plane continuum. Any help will be appreciated!


Is every subcontinuum of a $1$ dimensional locally path connected plane continuum locally path connected?

  • $\begingroup$ This is probably true in the plane. You may want to start there. $\endgroup$ – Forever Mozart Apr 20 '17 at 23:39
  • $\begingroup$ That is actually the only case I'm interested in so I may as well change my question. I haven't used any plane-specific properties in my proof attempts yet. $\endgroup$ – Harambe Apr 21 '17 at 5:01
  • 2
    $\begingroup$ Isn't the Sierpinski carpet a counterexample? It has a subcontinuum $([0,1] \times C) \cup (C \times [0,1])$ where $C$ is the ternary Cantor set and this is definitely not locally connected. $\endgroup$ – Niels J. Diepeveen Apr 21 '17 at 9:58

The Menger curve $M$ is a counter example. The reason is that $M$ is universal in the sense that for every 1-dimensional metrizable compact space $C$, there is an embedding $C\to M$. This was proven by Menger himself. (I will find a more accessible reference when I have more time. This might be in Engelking's book "General Topology" or/and in Kuratowski's "Topology". You are welcome to check if you have either one.) Menger curve is also connected and locally path-connected. Now, take your favorite non-locally connected continuum (say, the infinite comb) and embed it into $M$.

Edit: Actually, it is easy to see that already the Sierpinski carpet contains an infinite comb and, hence, is a counter example. Namely, suppose we are using the standard ternary Sierpinski carpet $S$ (contained in the unit square $[0,1]\times [0,1]$). Let $C$ be the ternary Cantor set in the x-axis. Then for each $c\in C$, $c\times [0,1]\subset S$. The union (an infinite "comb") $$ C\times [0,1]\cup [0,1]\times \{0\}\subset S $$ is connected, compact, but not locally connected.

On the positive side, every 1-dimensional nonseparating planar Peano continuum satisfies your property, because it is a dendroid.

  • $\begingroup$ Engelking Dimension Theory is the ref for this. $\endgroup$ – John Samples Jun 14 '18 at 23:40
  • $\begingroup$ Also, you mean dendrite, not dendroid. Dendroids are ferocious creatures that we know basically nothing about, still haha $\endgroup$ – John Samples Jun 14 '18 at 23:41

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