College Math Teaching

April 10, 2015

Cantor sets and countable products of discrete spaces (0, 1)^Z

Filed under: advanced mathematics, analysis, point set topology, sequences — Tags: , , , — collegemathteaching @ 11:09 am

This might seem like a strange topic but right now our topology class is studying compact metric spaces. One of the more famous of these is the “Cantor set” or “Cantor space”. I discussed the basics of these here.

Now if you know the relationship between a countable product of two point discrete spaces (in the product topology) and Cantor spaces/Cantor Sets, this post is probably too elementary for you.

Construction: start with a two point set $D = \{0, 2 \}$ and give it the discrete topology. The reason for choosing 0 and 2 to represent the elements will become clear later. Of course, $D_2$ is a compact metric space (with the discrete metric: $d(x,y) = 1$ if $x \neq y$.

Now consider the infinite product of such spaces with the product topology: $C = \Pi^{\infty}_{i=1} D_i$ where each $D_i$ is homeomorphic to $D$. It follows from the Tychonoff Theorem that $C$ is compact, though we can prove this directly: Let $\cup_{\alpha \in I} U_{\alpha}$ be any open cover for $C$. Then choose an arbitrary $U$ from this open cover; because we are using the product topology $U = O_1 \times O_2 \times ....O_k \times (\Pi_{i=k+1}^{\infty} D_i )$ where each $O_i$ is a one or two point set. This means that the cardinality of $C - U$ is at most $2^k -1$ which requires at most $2^k -1$ elements of the open cover to cover.

Now let’s examine some properties.

Clearly the space is Hausdorff ($T_2$ ) and uncountable.

1. Every point of $C$ is a limit point of $C$. To see this: denote $x \in C$ by the sequence $\{x_i \}$ where $x_i \in \{0, 2 \}$. Then any open set containing $\{ x_i \}$ is $O_1 \times O_2 \times...O_k \times \Pi^{\infty}_{i=k+1} D_i$ and contains ALL points $y_i$ where $y_i = x_i$ for $i = \{1, 2, ...k \}$. So all points of $C$ are accumulation points of $C$; in fact they are condensation points (or perfect limit points ).

(refresher: accumulation points are those for which every open neighborhood contains an infinite number of points of the set in question; condensation points contain an uncountable number of points, and perfect limit points are those for which every open neighborhood contains as many points as the set in question has (same cardinality).

2. $C$ is totally disconnected (the components are one point sets). Here is how we will show this: given $x, y \in C, x \neq y,$ there exists disjoint open sets $U_x, U_y, x \in U_x, y \in U_y, U_x \cup U_y = C$. Proof of claim: if $x \neq y$ there exists a first coordinate $k$ for which $x_k \neq y_k$ (that is, a first $k$ for which the canonical projection maps disagree ($\pi_k(x) \neq pi_k(y)$ ). Then
$U_x = D_1 \times D_2 \times ....\times D_{k-1} \times x_k \times \Pi^{\infty}_{i=k+1} D_i$,

$U_y = D_1 \times D_2 \times.....\times D_{k-1} \times y_k \times \Pi^{\infty}_{i = k+1} D_i$

are the required disjoint open sets whose union is all of $C$.

3. $C$ is countable, as basis elements for open sets consist of finite sequences of 0’s and 2’s followed by an infinite product of $D_i$.

4. $C$ is metrizable as well; $d(x,y) = \sum^{\infty}_{i=1} \frac{|x_i - y_i|}{3^i}$. Note that is metric is well defined. Suppose $x \neq y$. Then there is a first $k, x_k \neq y_k$. Then note

$d(x,y) = \frac{|x_k - y_k|}{3^k} + \sum^{\infty}_{i = k+1} \frac{|x_i - y_i|}{3^i} \rightarrow |x_k -y_k| =2 = \sum^{\infty}_{i=1} \frac{|x_{i+k} -y_{i+k}|}{3^i} \leq \frac{1}{3} \frac{1}{1 -\frac{2}{3}} =1$

which is impossible.

5. By construction $C$ is uncountable, though this follows from the fact that $C$ is compact, Haudorff and dense in itself.

6. $C \times C$ is homeomorphic to $C$. The homeomorphism is given by $f( \{x_i \}, \{y_i \}) = \{ x_1, y_1, x_2, y_2,... \} \in C$. It follows that $C$ is homeomorphic to a finite product with itself (product topology). Here we use the fact that if $f: X \rightarrow Y$ is a continuous bijection with $X$ compact and $Y$ Hausdorff then $f$ is a homeomorphism.

Now we can say a bit more: if $C_i$ is a copy of $C$ then $\Pi^{\infty}_{i =1 } C_i$ is homeomorphic to $C$. This will follow from subsequent work, but we can prove this right now, provided we review some basic facts about countable products and counting.

First lets show that there is a bijection between $Z \times Z$ and $Z$. A bijection is suggested by this diagram:

which has the following formula (coming up with it is fun; it uses the fact that $\sum^k _{n=1} n = \frac{k(k+1)}{2}$:

$\phi(k,1) = \frac{(k)(k+1)}{2}$ for $k$ even
$\phi(k,1) = \frac{(k-1)(k)}{2} + 1$ for $k$ odd
$\phi(k-j, j+1) =\phi(k,1) + j$ for $k$ odd, $j \in \{1, 2, ...k-1 \}$
$\phi(k-j, j+1) = \phi(k,1) - j$ for $k$ even, $j \in \{1, 2, ...k-1 \}$

Here is a different bijection; it is a fun exercise to come up with the relevant formulas:

Now lets give the map between $\Pi^{\infty}_{i=1} C_i$ and $C$. Let $\{ y_i \} \in C$ and denote the elements of $\Pi^{\infty}_{i=1} C_i$ by $\{ x^i_j \}$ where $\{ x_1^1, x_2^1, x_3^ 1....\} \in C_1, \{x_1^2, x_2 ^2, x_3^3, ....\} \in C_2, ....\{x_1^k, x_2^k, .....\} \in C_k ...$.

We now describe a map $f: C \rightarrow \Pi^{\infty}_{i=1} C_i$ by

$f(\{y_i \}) = \{ x^i_j \} = \{y_{\phi(i,j)} \}$

Example: $x^1_1 = y_1, x^1_2 = y_2, x^2_1 = y_3, x^3_1 =y_4, x^2_2 = y_5, x^1_3 =y_6,...$

That this is a bijection between compact Hausdorff spaces is immediate. If we show that $f^{-1}$ is continuous, we will have shown that $f$ is a homeomorphism.

But that isn’t hard to do. Let $U \subset C$ be open; $U = U_1 \times U_2 \times U_3.... \times U_{m-1} \times \Pi^{\infty}_{k=m} C_k$. Then there is some $k_m$ for which $\phi(k_m, 1) \geq M$. Then if $f^i_j$ denotes the $i,j$ component of $f$ we wee that for all $i+j \geq k_m+1, f^i_j(U) = C$ (these are entries on or below the diagonal containing $(k,1)$ depending on whether $k_m$ is even or odd.

So $f(U)$ is of the form $V_1 \times V_2 \times ....V_{k_m} \times \Pi^{\infty}_{i = k_m +1} C_i$ where each $V_j$ is open in $C_j$. This is an open set in the product topology of $\Pi^{\infty}_{i=1} C_i$ so this shows that $f^{-1}$ is continuous. Therefore $f^{-1}$ is a homeomorphism, therefore so is $f$.

Ok, what does this have to do with Cantor Sets and Cantor Spaces?

If you know what the “middle thirds” Cantor Set is I urge you stop reading now and prove that that Cantor set is indeed homeomorphic to $C$ as we have described it. I’ll give this quote from Willard, page 121 (Hardback edition), section 17.9 in Chapter 6:

The proof is left as an exercise. You should do it if you think you can’t, since it will teach you a lot about product spaces.

What I will do I’ll give a geometric description of a Cantor set and show that this description, which easily includes the “deleted interval” Cantor sets that are used in analysis courses, is homeomorphic to $C$.

Set up
I’ll call this set $F$ and describe it as follows:

$F \subset R^n$ (for those interested in the topology of manifolds this poses no restrictions since any manifold embeds in $R^n$ for sufficiently high $n$).

Reminder: the diameter of a set $F \subset R^n$ will be $lub \{ d(x,y) | x, y \in F \}$
Let $\epsilon_1, \epsilon_2, \epsilon_3 .....\epsilon_k ...$ be a strictly decreasing sequence of positive real numbers such that $\epsilon_k \rightarrow 0$.

Let $F^0$ be some closed n-ball in $R^n$ (that is, $F^)$ is a subset homeomorphic to a closed n-ball; we will use that convention throughout)

Let $F^1_{(0) }, F^1_{(2)}$ be two disjoint closed n-balls in the interior of $F^0$, each of which has diameter less than $\epsilon_1$.

$F^1 = F^1_{(0) } \cup F^1_{(2)}$

Let $F^2_{(0, 0)}, F^2_{(0,2)}$ be disjoint closed n-balls in the interior $F^1_{(0) }$ and $F^2_{(2,0)}, F^2_{(2,2)}$ be disjoint closed n-balls in the interior of $F^1_{(2) }$, each of which (all 4 balls) have diameter less that $\epsilon_2$. Let $F^2 = F^2_{(0, 0)}\cup F^2_{(0,2)} \cup F^2_{(2, 0)} \cup F^2_{(2,2)}$

To describe the construction inductively we will use a bit of notation: $a_i \in \{0, 2 \}$ for all $i \in \{1, 2, ...\}$ and $\{a_i \}$ will represent an infinite sequence of such $a_i$.
Now if $F^k$ has been defined, we let $F^{k+1}_{(a_1, a_2, ...a_{k}, 0)}$ and $F^{k+1}_{(a_1, a_2,....,a_{k}, 2)}$ be disjoint closed n-balls of diameter less than $\epsilon_{k+1}$ which lie in the interior of $F^k_{(a_1, a_2,....a_k) }$. Note that $F^{k+1}$ consists of $2^{k+1}$ disjoint closed n-balls.

Now let $F = \cap^{\infty}_{i=1} F^i$. Since these are compact sets with the finite intersection property ($\cap^{m}_{i=1}F^i =F^i \neq \emptyset$ for all $m$ ), $F$ is non empty and compact. Now for any choice of sequence $\{a_i \}$ we have $F_{ \{a_i \} } =\cap^{\infty}_{i=1} F^i_{(a_1, ...a_i)}$ is nonempty by the finite intersection property. On the other hand, if $x, y \in F, x \neq y$ then $d(x,y) = \delta > 0$ so choose $\epsilon_m$ such that $\epsilon_m < \delta$. Then $x, y$ lie in different components of $F^m$ since the diameters of these components are less than $\epsilon_m$.

Then we can say that the $F_{ \{a_i} \}$ uniquely define the points of $F$. We can call such points $x_{ \{a_i \} }$

Note: in the subspace topology, the $F^k_{(a_1, a_2, ...a_k)}$ are open sets, as well as being closed.

Finding a homeomorphism from $F$ to $C$.
Let $f: F \rightarrow C$ be defined by $f( x_{ \{a_i \} } ) = \{a_i \}$. This is a bijection. To show continuity: consider the open set $U = y_1 \times y_2 ....\times y_m \times \Pi^{\infty}_{i=m} D_i$. Under $f$ this pulls back to the open set (in the subspace topology) $F^{m+1}_{(y1, y2, ...y_m, 0 ) } \cup F^{m+1}_{(y1, y2, ...y_m, 2)}$ hence $f$ is continuous. Because $F$ is compact and $C$ is Hausdorff, $f$ is a homeomorphism.

This ends part I.

We have shown that the Cantor sets defined geometrically and defined via “deleted intervals” are homeomorphic to $C$. What we have not shown is the following:

Let $X$ be a compact Hausdorff space which is dense in itself (every point is a limit point) and totally disconnected (components are one point sets). Then $X$ is homeomorphic to $C$. That will be part II.