Definition of Cantor's leaky tent:

Let $C$ be the Cantor set in the unit interval and $p$ the point $(1/2,1/2)$ in the Euclidean plane. Let $L(c)$ the line segment in the plane connecting the point c in the Cantor set and p. Let $L_\mathbb{Q}(c)=\{(x,y) \in L(c): y \in \mathbb{Q}\}$ and $L_{\mathbb{R}\setminus\mathbb{Q}}(c)=\{(x,y) \in L(c): y \in \mathbb{R}\setminus\mathbb{Q}\}$. Let $E$ be the set of the endpoints of the intervals that are deleted in the construction of the Cantor set, together with $0$ and $1$. Cantor's leaky tent (or Knaster-Kuratowski fan) is defined as $$T=\bigcup_{c \in E} L_\mathbb{Q}(c) \cup \bigcup_{c \in C\setminus E} L_{\mathbb{R}\setminus\mathbb{Q}}(c), $$ and equipped with the induced Euclidean topology. By the way, it has the peculiar property that it is a connected space but the subspace $T \setminus \{p\}$ is totally disconnected (see Examples 128 and 129 in Counterexamples in topology by Steen & Seebach).

Definition of Cantor's leakier tent:

Cantor's leakier tent is a kind of inverse or complement of the Cantor's leaky tent. It is defined as $$ T'=\bigcup_{c \in E} L_{\mathbb{R}\setminus\mathbb{Q}}(c) \cup \bigcup_{c \in C\setminus E} L_\mathbb{Q}(c), $$ equipped with the induced Euclidean topology as well.


It it said that Cantor's leakier tent is a zero-dimensional space which means it has a basis consisting of open and closed sets, in other words for every point $x \in T'$ and a neighborhood $U$ of $x$ there is an open and closed set inside $U$ containing $x$. How to see this? Given a point $x \in T'$ and a neighborhood $U$ of $x$, perhaps we could enclose the point $x$ with a finite number of line segments which avoid the set $T'$ completely and stay inside $U$. The points of $T'$ inside this enclosure would then be an open and closed set containing $x$. Would this be possible?

  • $\begingroup$ Note that $E$ is homeomorphic to the Baire space. $\endgroup$
    – Asaf Karagila
    Jun 28, 2011 at 14:26
  • 1
    $\begingroup$ @Asaf: How does this help? $\endgroup$
    – t.b.
    Jun 28, 2011 at 16:51
  • $\begingroup$ @Theo: I'm not sure, but if one can "cook" something about this being some $\omega^\omega\times\mathcal C$ then it is a product of two 0-dimensional spaces, which is zero dimensional. $\endgroup$
    – Asaf Karagila
    Jun 28, 2011 at 17:09
  • $\begingroup$ This is exercise 1.4 C (d) in Engelking's Dimension Theory, so it's probably true... $\endgroup$ Jun 28, 2011 at 18:55
  • $\begingroup$ @Asaf, it's a countable set, so how could it be? Don't you mean $C \setminus E$ as that set, as $E$ is countable and dense in itself, and dense in $C$, so more like a copy of $\mathbb{Q}$ inside $C$. $\endgroup$ Jun 28, 2011 at 18:59

1 Answer 1


LostInMath pointed out a hole in my previous version, which I think I've fixed in this edit. The basic idea remains the same.

Let $X = \left(E \times (\mathbb P \cap I)\right) \cup \left((C \setminus E) \times (\mathbb Q \cap I \right)$, where $\mathbb P = \mathbb R \setminus \mathbb Q$ and $I=[0,1]$. Let $X'$ be the quotient obtained by identifying $Z = (C \setminus E) \times \{1\}$ to a point, which I'll call $z$; clearly $X'$ is homeomorphic to $T'$, with $z$ corresponding to $p$.

Let $D$ be the set of dyadic rationals in $I$. Each $x \in C$ can be written uniquely as $\sum \limits_{k \ge 0} x_k 3^{-k}$, where each $x_k \in \{0,2\}$. The function $f:C \to I$ that sends $\sum \limits_{k \ge 0} x_k 3^{-k}$ to $\sum \limits_{k \ge 0} x_k 2^{-k}$ is continuous and onto; it's even $1$-$1$ except on $E \setminus \{0,1\}$, which it maps $2$-$1$ onto $D \setminus \{0,1\}$.

Let $\phi$ be an order-isomorphism of $D$ onto $\mathbb Q \cap I$; $\phi$ extends to an order-preserving homeomorphism $\hat{\phi}:I \to I$. Let $Y = \left((\mathbb Q \times \mathbb P) \cup (\mathbb P \times \mathbb Q) \right) \cap (I \times I)$, topologized as a subspace of the plane. For any $a,b \in \mathbb Q$ such that $0<a<a+b<1$, the line $y=ax+b$ is disjoint from $Y$, and $U(a,b) = \{(x,y) \in Y:y > ax+b\}$ is clopen in $Y$ and contains $\{(x,y) \in Y:y=1\}$. Let $g:C \times I \to I \times I:(x,y) \mapsto (\hat{\phi}(f(x)),y)$, and let $N(a,b) = g^{-1}[U(a,b)]$; $N(a,b)$ is a clopen nbhd of $Z$ in $X$, so its image in $X'$ under the quotient map is a clopen nbhd of $z$. Clearly the set of such nbhds is a local base at $z$.

Now let $u = (x,y) \in X$; any nbhd of $u$ contains a 'box' $B = \left([e_0,e_1] \times [y_0,y_1]\right) \cap X$ such that $e_0 \le x \le e_1$, $e_0 < e_1$, and $y_0 < y < y_1$. It's not hard to see that there are $a_0,b_0,a_1,b_1 \in \mathbb Q$ such that $$u \in \left( N(a_0,b_0) \setminus N(a_1,b_1) \right) \cap \left([e_0,e_1] \times I \right) \subseteq B.$$ $[e_0,e_1]$ is a clopen nbhd of $x$ in $C$, so $\left( N(a_0,b_0) \setminus N(a_1,b_1) \right) \cap \left([e_0,e_1] \times I \right)$ is a clopen nbhd of $u$ in $X'$, and we're done.

  • $\begingroup$ It looks promising but I'm afraid some details get in the way. Namely "if $a$ and $b$ are dyadic rationals, the line $y=ax+b$ is disjoint from $Y$" does not seem to hold. Take for example $a=3/4 \in D$, $b=1/2 \in D$ and $x=1/3 \in I\setminus D$. Then $y=3/4 \in D$ and $(x,y) \in (I \setminus D) \times D \subset Y$. $\endgroup$
    – LostInMath
    Jun 28, 2011 at 22:18
  • $\begingroup$ @LostInMath: Thanks; fixed (barring another mental hiccup). $\endgroup$ Jun 29, 2011 at 0:00
  • $\begingroup$ Thank you very much. I have thought about this problem for quite a long time and always struggled to find these suitable line segments disjoint from $T'$. Never did I realise that one could move the whole situation to $((\mathbb{Q}\times\mathbb{P})\cup(\mathbb{P}\times\mathbb{Q}))\cap(I\times I)$ where these lines could easily be found. $\endgroup$
    – LostInMath
    Jun 29, 2011 at 15:53

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