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I'm doing research in generalised inverse limits, and I'm trying to prove a result about circle-like plane continua.

Definitions

A continuum is a compact, connected metric space.

A plane continuum is a continuum that is homeomorphic to a subcontinuum of $\mathbb{R}^2$. This definition can just be interpreted as "a continuum which is a subset of $\mathbb{R}^2$", as the natural equivalence class on continua is homeomorphism classes.

A cover $\mathscr{C}$ is a $chain\ cover$ if there is an enumeration $\{C_1, C_2, \cdots, C_n\}$ so that $C_p \cap C_q = \emptyset$ if and only if $|p-q|>1$. Intuitively this is any cover that looks like an actual steel chain, but abstract.

A continuum is arc-like if for every $\varepsilon>0$, there is an $\varepsilon$ chain cover of the continuum. (By $\varepsilon$ chain cover I mean every subset in the cover is an $\varepsilon$ ball.)

A continuum is circle-like if it is not arc-like, but for every $\varepsilon>0$, there is an $\varepsilon$ cover $\mathscr{C}$ of the continuum such that $\mathscr{C}$ has an enumeration $\{C_1, C_2, \cdots, C_n\}$ in which $C_p \cap C_q = \emptyset$ if and only if $n-1>|p-q|>1$. Intuitively this looks like a steel chain where the first ring is connected to the last ring. It is the same as taking a chain-cover, but then requiring the first and last subsets in the cover to intersect.

A metric space $X$ is path connected if given any two points $a, b \in X$, there is a path between them. In other words, there is a continuous function $\varphi: [0,1]\rightarrow X$ such that $\varphi(0)=a, \varphi(1)=b$.

A metric space is locally path connected if given any point in the space, there is a neighbourhood around it which is path connected.

Note that these both imply connectedness and locally connectedness respectively.

A continuum is indecomposable if it is not the union of any two of its proper subcontinua. Examples include the Buckethandle set and the Pseudo-arc.

Problem

I'm trying to prove a result about generalised inverse limits, and I've encountered a problem dealing with path connected and locally path connected circle-like continua. I was simply going to try and prove results directly from these facts I know, but this lead to a different question: Are there any "weird" circle-like path connected and locally path connected continua?

Conjecture

Every locally path connected and path connected circle-like continuum is homeomorphic to $S^1$, the circle.

If this is the case, the result I'm trying to prove would get pretty trivial. It's definitely closely to whether or not every locally path connected and path connected arc-like continuum is an arc (a metric space homeomorphic to the unit closed interval.)

The general question I'm asking here on MSE is "Are there any pathological locally path connected and path connected plane continua?" Every strange continuum I know has properties resulting from "infinite things" in such a way that they lose path connectedness and/or locally path connectedness.

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  • $\begingroup$ One well studied pathology is that of the Hawaiian earring, i.e. points at which the continuum is not semilocally simply connected. But it is very far from circle like, and I'm pretty sure one could prove that any space which is not semilocally simple connected is not circle like. $\endgroup$ Commented Mar 30, 2017 at 2:44
  • $\begingroup$ The Hawaiian earring is slc, actually. Slc is equivalent to aposyndetic is equivalent to freely decomposable, which is stronger than decomposable. So any indecomposable circle-like continuum is not slc (e.g. pseudo-circle, circle of pseudo-arcs). $\endgroup$ Commented Sep 26, 2017 at 14:50
  • $\begingroup$ Anyway, if the OP is still curious, could they define what they mean by "pathological"? Maybe some properties they would like to see a counterexample or proof for. Also, the OP may want to know that local connectedness implies local path-connectedness (and thus path-connectedness) for continua, so there is some redundancy. $\endgroup$ Commented Sep 26, 2017 at 14:53

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Yes, indeed, a circle-like Peano continuum $X$ is homeomorphic to the circle. In order to prove this you first show that each 2-point subset $\{x, y\}$ of $X$ separates. Indeed, by taking $\epsilon>0$ small enough, for each "cyclic" cover $C_1,...,C_n$ we have $x\in C_1$, $y\in C_j$, $1\ne j$. Then no two points $z\in C_i$, $w\in C_k$, $1<i<j<k$ can be connected by a path in $X- \{x,y\}$. (If such a path $p$ exists, you refine the cyclic cover to a cyclic cover $C_1',...,C'_{n'}$, where $x\in C_1'$, $y\in C'_{j'}$, such that the path $p$ is disjoint from $C'_1\cup C'_{j'}$, contradicting cyclicity.) Note that we used local path-connectivity to ensure that an open subset of $X$ is connected if and only if it is path-connected.

Once you have the above two-point cut-property, you use Moore's theorem that every nonempty continuum where every two distinct points separate, is homeomorphic to $S^1$. You can find a proof of Moore's theorem e.g. in "Topology of Manifolds" by Wilder.

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