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Let $C^\prime$ be the Cantor set and let $C = C^\prime \times \{0\}$ (viewed as a subset of $\mathbb{R}^2$). For $c \in C$, let $L(c)$ denote the half-closed line segment connecting $(c,0)$ to $(\tfrac{1}{2}, \tfrac{1}{2})$ (including $(c,0)$ but excluding $(\tfrac{1}{2}, \tfrac{1}{2})$).

If $c \in C$ is $(0,0)$, $(1,0)$, or an endpoint of an interval deleted in the Cantor set, let $X_{c} = \{ (x,y) \in L(c) : y \in \mathbb{Q} \}$. For all other $c \in C$, let $X_{c} = \{ (x,y) \in L(c) : y \notin \mathbb{Q} \}$.

Cantor's Teepee is the set $\bigcup_{c \in C} X_{c}$ equipped with the subspace topology inherited from the standard topology on $\mathbb{R}^2$.

I've found several references to the fact that Cantor's Teepee is totally disconnected, but I cannot quite prove it.


It was brought to my attention in the comments that I had taken the wrong definition for totally disconnected when first working on this problem, as demonstrated in the following example. The property I had been attempting to prove was totally separated, which lead to the quandary I was experiencing below.

In particular, I cannot seem to separate two points that belong to the same $X_c$. For concreteness:

What are two separated open sets $A$ and $B$ such that

  • $A$ and $B$ witness the disconnectedness of Cantor's Teepee,
  • $(0,0) \in A$, and
  • $(\tfrac{1}{4}, \tfrac{1}{4}) \in B$?

Notice $(0,0), (\tfrac{1}{4}, \tfrac{1}{4}) \in X_0$. Presumably the method that works for this example can be easily modified to work for any pair of points belonging to the same $X_c$.

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  • $\begingroup$ Do $0$ and $1$ count as endpoints of intervals deleted in the Cantor set? $\endgroup$ – Alex Becker Apr 18 '14 at 0:58
  • $\begingroup$ @AlexBecker I've updated the definition to include $(0,0)$ and $(1,0)$ as endpoints. Whether they are included or not should not affect the total disconnectedness. $\endgroup$ – Austin Mohr Apr 18 '14 at 1:05
  • $\begingroup$ I recall that the purpose of this example is to show that a connected set can become totally disconnected after removing one point, namely $(1/2,1/2)$. Have you removed said point? Also, on MathOverflow. $\endgroup$ – user127096 Apr 18 '14 at 1:22
  • $\begingroup$ @cheapeffectivedietpills Rather than define the "leaky tent" and the remove the point $(1/2,1/2)$ to obtain the teepee, I removed the point from the beginning by explicitly defining $L(c)$ to exclude $(1/2,1/2)$. I saw the MathOverflow question, but it does not contain a proof of total disconnectedness. $\endgroup$ – Austin Mohr Apr 18 '14 at 1:27
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    $\begingroup$ It seems the thesis takes "totally disconnected" to mean "quasicomponents are singletons". I think the more common definition is "components are singletons", which is a weaker property. $\endgroup$ – user127096 Apr 18 '14 at 1:31
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Let $T$ be the Cantor teepee. Every point $t\in T$ lies on unique interval of the form $[c, (1/2, 1/2)]$, where $c\in C'$. The map $f: t\mapsto c$ is continuous on $T$. Suppose $A\subset T$ is a connected subset. Then $f(A)$ is also connected. Since $C'$ is totally disconnected, $f(A)$ is a single point. Thus, $A$ is contained in subset $L_c$. But each $L_c$ is clearly totally disconnected. Hence, $A$ is a single point. Thus, $T$ is totally disconnected. qed

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The following proof is a clunkier version of that given by studiosus, but I provide some details that I think will be helpful for my students who are involved with this problem.


Let $Y$ be any subset of Cantor's Teepee containing more than one point. Our goal is to produce two sets $A$ and $B$ such that

  • $A$ and $B$ are separated open sets in the subspace topology induced by $Y$ and
  • $Y = A \cup B$.

Since Cantor's Teepee is itself a subspace of the usual topology on $\mathbb{R}^2$, we may revise these conditions to read

  • $A$ and $B$ are separated open sets in the usual topology on $\mathbb{R}$ and
  • $Y \subseteq A \cup B$.

Now, suppose $Y$ contains points $p \in X_a$ and $q \in X_c$ with $a \neq c$ (that is, $Y$ witnesses more than one "strand" of the teepee). Choose any real number $b$ such that $a < b < c$ and $b$ lies in a deleted interval of the Cantor set. To obtain the separated open sets $A$ and $B$, take any open set containing $Y$ (or even the entire teepee) and "split" it along the line connecting $(b,0)$ and $(1/2,1/2)$ (see figure).

enter image description here

Suppose instead that all points of $Y$ belong to a single strand. The $y$-coordinates of points belonging to this strand are either all rational or all irrational. In either case, we can easily separate the strand by passing our open sets through a point $r$ with $y$-coordinate of the opposite type (see figure).

enter image description here

Therefore every subset of Cantor's teepee with more than one point is disconnected, which is to say Cantor's teepee is totally disconnected.

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Will this work? Choose $d\notin C$, and set $A=\{(x,y): \left((x,y) \in X_{c} \mbox{ for some } c<d\right) \mbox{ and } y<\frac{1}{8}\}$, and let $B$ be the complement of this set in the teepee?

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  • $\begingroup$ What would be the open set in the usual topology on $\mathbb{R}^2$ whose intersection with the teepee is $B$? That is to say, I don't think $B$ is open in the subspace topology. $\endgroup$ – Austin Mohr Apr 18 '14 at 1:38
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    $\begingroup$ Hm. I guess the closure of $A$ would contain $(c,\frac{1}{8})$ for the endpoints of deleted intervals. I have to say, though, the thesis appendix you found is odd - seems like lots (all?) references describe the teepee as totally disconnected. $\endgroup$ – Steve Kass Apr 18 '14 at 1:52
  • $\begingroup$ Another commenter noted that the author of that thesis is using a nonstandard definition of "totally disconnected". $\endgroup$ – Austin Mohr Apr 18 '14 at 1:53
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    $\begingroup$ @studiosus I left off "whose union is the entire space", but I take it even this revision would be incorrect. I think you have helped me identify my confusion, however, so please tell me if the following is correct: If I want to show a subset S of the teepee is disconnected, I need to demonstrate separated sets A and B whose union is S, not necessarily the entire teepee. The sets A and B should be open in the subspace topology induced by S, not necessarily open in the teepee's topology. (The bold, if correct, is the key new realization for me.) $\endgroup$ – Austin Mohr Apr 18 '14 at 19:14
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    $\begingroup$ @AustinMohr The stronger property you had in mind when asking this question is called totally separated. $\endgroup$ – user127096 Apr 18 '14 at 22:12

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