Finite Unions of Dendrites

Question

The question is a bit specific, but seems to be the most general question to ask after handling some obvious counterexamples.

Initially, I was wondering the following. Let $X$ be a one-dimensional Peano continuum. If $X$, aside from its points whose neighborhoods are arcs, has only finitely many local cut points, can $X$ be written as a finite union of dendrites $D_1, \dots, D_n$? If so, does $n$ depend on $X$ specifically, or just the number/orders of the local cut points?

A Peano Continuum is a compact, connected, locally connected metric space. A Dendrite is a Peano continuum which contains no copies of the circle. By a local cut point I mean a point $x \in X$ such that $x$ has an open, connected neighborhood $U$ such that $U \setminus x$ is not connected. By the 'order' of a local cut-point $x$ I mean the maximal number of components it cuts a neighborhood $U$ into as $U \rightarrow x$.

Answer 1: No. The Sierpinski Carpet is a one-dimensional Peano continuum which contains all one-dimensional, planar continua: It happens to be locally connected with no local cut-points. In particular, it contains the Cantor Fan, which is just $C \times [0,1] / [\sim]$ where $C$ is the Cantor Set and $\sim$ collapses $C \times \lbrace 0 \rbrace$ to a point. Its vertex has order the continuum, clearly impossible for a finite union of dendrites (or any subset thereof).

A useful concept is hereditarily locally connected (hlc), i.e. every subcontinuum is locally connected. Such a continuum is also regular (a stronger condition): Every point has a neighborhood base whose elements have finite boundary. However, it is known that the union of two dendrites need not be hlc; there is a counterexample constructed in Nadler's Continuum Theory.

It is also known that the union of rational continua is rational: A rational continuum is one whose points have neighborhood bases with countable boundary. Dendrites are rational. Thus "connected, finite unions of dendrites" forms a class of curves nicer than rational continua, but either independent of, or weaker than, hlc. To my knowledge it has not been studied, but such a class sits in a very useful place in the hierarchy of curves.

Main question: What continua can be written as the finite union of dendrites? What hlc (resp. regular) continua cannot be written as the finite union of dendrites?

In particular, can anyone give an example of an hlc continuum which can't be written as the union of a finite collection of dendrites? I tried looking at examples of hlc-but-not regular continua; however, the standard examples I found can be written as the finite union of dendrites. In each case I was able to construct them as the union of only two dendrites.

Examples of continua which can be written as the union of three dendrites, but not two, would be useful. A trivial example would be three copies of the Cantor Tree identified at the endpoints - visually trivial, I should say, but if someone wants to write the details then go for it. In a sentence, too many arcs are produced at the Cantor Set for one dendrite to 'seep' into one of the others and pick up some extra space.

Q2: Suppose that $X$ is locally cyclically connected (every point is contained in arbitrarily small open neighborhoods $U$ each of whose pair of points is connected by a simple closed curve in $U$). Can $X$ be written as the union of two dendrites? Three?

Q2b: Suppose that $X$ is locally cyclically connected and known to be the finite union of dendrites. Can it be two? Three?

Q3: Can the Sierpinski Triangle be written as the finite union of dendrites? How many are required, if so?

I made this one a separate question, since it can potentially be solved without any topological sophistication. Sierpinski Triangle as Finite Union of Dendrites

It's been solved, now. It's not possible.

Q4: (Stronger than Q2) If $X$ is locally $n$-connected (every point is contained in arbitrarily small open neighborhoods $U$ each of whose pair of points is connected by $n$ arcs in $U$ intersecting only in their end points), can we bound the necessary number of dendrites? Is it $n$ (or $n+1$)?

Q5: What continua can be written as the union of two (or finitely many) dendrites each of which intersect at most in a subset of their end points? This problem has some tangential relevance to conformal welding theory and SLE, so has probably been studied in the geometric function theory literature if it hasn't already been studied back in the early 1900's when curve theory was a hot area.

Regarding Q5, I don't think there's anything interesting to ask about dendrites which only intersect in endpoints or vertices. It seems to be an equivalent question to Q1, as any 'vertex' intersection can be obtained as a regular intersection of two dendrites augmented by small triods attached along arcs where we'd want a vertex. But this might be different:

Q5b: Classify finite Dendrite unions $D = D_1 \cup \cdots \cup D_n$ such that if $x \in D_i \cap D_j$ for $i \neq j$ then $x$ is an endpoint of some $D_k$. Here we don't require that $x$ be an endpoint of any other dendrite it belongs to.

Q6: What if we restrict how many dendrites can intersect in a single point to 2? Or $k < n$? What if we require any intersection point to belong to at least three (or $k$) dendrites? This second part has some problems with (local) duplicates to iron out to get an interesting question, probably.

Q7: Anything nice happen if you restrict to continua in the plane?

Have fun!

Answer 1

Ok, here's an hlc continuum which isn't the finite union of dendrites. Not going to give the explicit proof, but it's similar to one of the spaces mentioned in the question.

Let $C_i$ be copies of the Cantor Tree, with end point sets $E_i$. To $C_1$ attach $C_2, C_3$ to the 'left' and 'right' halves of $E_1$ along $E_2, E_3$ respectively. Now iterate: Attach two more copies to the left and right halves of $E_2$, then to $E_3$ etc. so that $C_i \rightarrow E_1$. Continue and take the limit. This limit space $X$ is clearly a continuum, and is hlc, which can be observed due to the following theorem:

A continuum is hlc if and only if it contains no convergence continuum.

A convergence continuum is a non-trivial (i.e. not just a point) subcontinuum $Y$ such that there is a sequence of pairwise-disjoint subcontinua converging to $Y$ (in the Hausdorff topology) no one of which intersects $Y$.

Each point $p$ of $X$ is either in one of the dendrites away from $E_1$ or in $E_1$ itself. Dendrites are hlc, so there are no convergence continua containing any $p$ outside of $E_1$. Thus a convergence continuum would have to be contained in $E_1$. But $E_1$ is the Cantor Set.

This shows that the class "finite union of dendrites" and the class "hlc continua" are independent of each other. So the question is now more interesting. Incidentally, it also shows that "hlc continua which are the countable union of dendrites" is independent from "finite union of dendrites" as well. Actually, the continuum $X$ is regular and even completely regular, i.e. every non-degenerate subcontinuum has non-empty interior in $X$.

The tricky part is showing that it's not the finite union of dendrites. It's 'geometrically obvious' but I'm still trying to think of a slick way to prove it in just a line or two.