# The Cantor set $K$ in $\mathbb{R}$ homeomorphic to $K\times K$ in $\mathbb{R}^{2}$

It is known to me that $K$ is homeomorphic to $2^{\mathbb{N}}$. So it would suffice to show $2^{\mathbb{N}}$ is homeomorphic to $2^{\mathbb{N}}\times2^{\mathbb{N}}$. I was wondering what people's thoughts were about the Cantor set as a binary tree and the implications thereof.

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Are you asking about the proof that $2^{\Bbb N}\times2^{\Bbb N}$ is homeomorphic to $2^{\Bbb N}$, which is very easy, or are you asking just for comments on the Cantor set as a binary tree? – Brian M. Scott Dec 21 '12 at 9:21
Both. I had a general idea of how the proof went but I wanted to spur discussion about the Cantor set because this topic intrigues me. So any comments are appreciated. – Rustyn Dec 21 '12 at 9:29

Jacob Schlather has answered the first question. The second is pretty open-ended, so I’ll take it in a direction that interests me. If you’ve not yet learned about infinite ordinals, some of it may be a bit mysterious.

Definitions. A tree is a partial order $\langle T,\preceq\rangle$ such that for each $x\in T$, $\{y\in T:y\prec x\}$ is well-ordered by $\prec$. $B\subseteq T$ is a branch if (1) $B$ is well-ordered by $\prec$, and (2) if $B\subsetneqq X\subseteq T$, then $X$ is not well-ordered by $\prec$. Let $\operatorname{Br}(T)$ be the set of branches of $T$.

For $x\in T$, the height of $x$, denoted by $\operatorname{ht}(x)$, is the order type of $\{y\in T:y\prec x\}$. For any ordinal $\alpha$, $\operatorname{Lev}_\alpha(T)=\{x\in T:\operatorname{ht}(x)=\alpha\}$ is the $\alpha$-th level of $T$. The height of $T$ is the least $\alpha$ such that $\operatorname{Lev}_\alpha(T)=\varnothing$.

The Cantor tree is the complete binary tree of height $\omega$, the first infinite ordinal. The easiest way to think of it is to let $T={^{<\omega}\{0,1\}}$, the set of finite sequences of zeroes and ones, ordered so that $\sigma\prec\tau$ iff $\sigma$ is a proper initial segment of $\tau$. Then for any $\sigma\in T$ the height of $\sigma$ is simply its length as a sequence: the empty sequence has height $0$, the sequences $0$ and $1$ have height $1$, the sequences $00,01,10$, and $11$ have height $2$, and so on. Thus, $|\operatorname{Lev}_n(T)|=2^n$ for each $n\in\Bbb N$.

Suppose that $\beta$ is a branch of $T$. Then $\beta$ contains exactly one member of each level of $T$; let $\sigma_n$ be the member of $\beta$ in the $n$-th level of $T$. For example, one branch of $T$ contains the following sequences in addition to the empty sequence: $1,11,111,1111,\dots~$. By the definition of branch we know that $\sigma_0\prec\sigma_1\prec\sigma_2\prec\ldots~$: $\sigma_k$ is an initial segment of $\sigma_n$ whenever $k<n$. Thus, there is a unique infinite sequence $\hat\beta$ of zeroes and ones such that $\sigma_n$ is the initial $n$ terms of $\hat\beta$. In the case of the example, $\hat\beta=\langle 1,1,1,\dots\rangle$, the constant $1$ sequence. Thus, the branches of $T$ correspond in a natural way with the points of the product $2^{\Bbb N}$.

For each $\sigma\in T$ let $B(\sigma)=\{\beta\in\operatorname{Br}(T):\sigma\in\beta\}$; this is the set of branches of $T$ that go through the node $\sigma$. Alternatively, it’s the set of branches $\beta$ such that $\sigma$ is an initial segment of the infinite sequence $\hat\beta$. Once you get past all the notation, it’s not hard to see that $\{\hat\beta:\beta\in B(\sigma)\}$ is a basic open set in $2^{\Bbb N}$, and that the collection of all such basic open sets is a base for the product topology on $2^{\Bbb N}$. Thus, $\{B(\sigma):\sigma\in T\}$ is a base for a topology on $\operatorname{Br}(T)$ such that the map $\operatorname{Br}(T)\to 2^{\Bbb N}:\beta\mapsto\hat\beta$ is a homeomorphism. This topology is sometimes called the tree topology on $\operatorname{Br}(T)$; $\operatorname{Br}(T)$ with the tree topology is homeomorphic to the Cantor set.

All of that is really just recasting things that you already know in a slightly different light. The payoff is that one can do the same thing with any tree.

Let $\langle T,\preceq\rangle$ be a tree. For each $t\in T$ let $B(t)=\{\beta\in\operatorname{Br}(T):t\in\beta\}$, and let $$\mathscr{B}=\{B(t):t\in T\}\;;$$ then $\mathscr{B}$ is a base for a topology on $\operatorname{Br}(T)$, the tree topology.

This topology is always Hausdorff. If $\beta,\gamma\in\operatorname{Br}(T)$ and $\beta\ne\gamma$, then there are $s,t\in T$ such that $s\in\beta\setminus\gamma$ and $t\in\gamma\setminus\beta$. It follows that $s\not\preceq t\not\preceq s$ and hence that $B(s)$ and $B(t)$ are disjoint open nbhds of $\beta$ and $\gamma$, respectively. It has other nice properties as well. For instance, for any $\alpha$ less than the height of the tree $\{B(t):t\in\operatorname{Lev}_\alpha(T)\}$ is a partition of $\operatorname{Br}(T)$ into open sets, so each of these sets is also closed. This shows that each $B(t)\in\mathscr{B}$ is a clopen set, and since $\operatorname{Br}(T)$ has a base of clopen sets, it’s a zero-dimensional space. This immediately implies that it’s a Tikhonov (= Hausdorff and completely regular) space.

Notice that for any $B(s),B(t)\in\mathscr{B}$, either $B(s)\subseteq B(t)$, $B(t)\subseteq B(s)$, or $B(s)\cap B(t)=\varnothing$. (Such a base is sometimes called a non-Archimedean base.) It turns out that this implies that if $\operatorname{Br}(T)$ is metrizable, it has a compatible non-Archimedean metric (or ultrametric), meaning a metric $d$ that satisfies the following very strong version of the triangle inequality: for any $\beta,\gamma,\delta\in\operatorname{Br}(T)$,

$$d(\beta,\delta)\le\max\{d(\beta,\gamma),d(\gamma,\delta\}\;.$$

Spaces constructed in this way are of considerable importance in set-theoretic topology. The structure of the tree determines the topology of the space of branches. In particular, certain combinatorial properties of trees are known to correspond to certain topological properties of the branch space, and it’s often easier to construct a tree with a specific combinatorial property than it is to construct from scratch a space with the corresponding topological property.

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With regards to your first question we have $\newcommand{\N}{\mathbb N} 2^\N \times 2^\N \approx 2^{\N + \N}$ and since $|\N + \N|=|\N|$ it follows that $2^\N \approx 2^{\N + \N}\approx 2^{\N} \times 2^{\N}$.

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Cardinality argument, yeah i remember now – Rustyn Dec 21 '12 at 9:22
@Jacob: The correct argument would be $2^\mathbb N\times2^\mathbb N=2^{\mathbb{N+N}}$. But it's all the same because $\mathbb{|N\times N|=|N+N|=|N|}$. – Asaf Karagila Dec 21 '12 at 12:07
@AsafKaragila good point. – JSchlather Dec 21 '12 at 18:09