Borel Sets of Infinite Binary Sequences

Note: The content of this blog is an expansion of some results in Chapter 2 of Parthasarathy's Book on Probability measures on Metric Spaces.

$M=\{0,1\}^\infty$ is a compact metric space as shown previously. Let $X,Y$ be two metric spaces. Borel sets $B_1 \subset X, B_2 \subset Y$ are said to be isomorphic if there exists a bijective map, $\phi : B_1 \to B_2$ such that both $\phi, \phi^{-1}$ are measurable. This is denoted as $B_1 \sim B_2$.

Here the sigma algebra on $B_1$ is $B_1 \cap \mathcal{B}_X$ and similarly on $B_2$. Measurability of $\phi,\phi^{-1}$ should be thought of in this sense. (Note that $B_1 \cap \mathcal{B}_X = \mathcal{B}_{B_1}$). Throughout, $\mathcal{B}_X$ represents the $\sigma$-algebra generated by the open sets in $X$.

There exists a Borel set $E \subset M$ such that $E \sim [0,1]$

Define $\tau : M \to [0,1]$ by $\tau(\{x_1,x_2,\cdots\}) = \sum_{i=1}^\infty \frac{x_i}{2^i}$. Then $\tau$ is continuous from the uniform limit theorem. $\tau$ is also onto since every number in $[0,1]$ has a binary expansion. Let $E$ be set of all binary sequences which contain infinitely many zeros. Then, $M - E$ is countable. As each singleton in a metric space is a closed set, $M-E$ is a Borel set. Hence, $E$ is also Borel.

The restriction of $\tau$ to $E$ is a bijective map as every number in $[0,1]$ has exactly one binary expansion which contains infinitely many zeros. As $\tau$ is continous, it is measurable map from $M \to [0,1]$ and hence $\tau : E \to [0,1]$ is also measurable.

It needs to be shown that $\tau^{-1}$ is also measurable. $\tau^{-1} : [0,1] \to E$ and it needs to be shown that $\tau(A)$ is Borel in $[0,1]$ for all $A$ Borel in $E$ i.e. $A \in \mathcal{B}_E$. As $\mathcal{B}_E = E \cap \mathcal{B}_M$ and the fact that $\mathcal{B}_M$ is generated by the sets $F_j = \{ \{x_1,x_2\cdots\} | x_j = 1\}$ it is enough to show that $\tau(A)$ is Borel for all $A$ of the form $E\cap F_j$.

To understand what $\tau(E\cap F_j)$ is, it helps to fix the first $j-1$ entries and let the entries from $j+1$ to $\infty$ be arbitrary. Each such set gives rise to finite union of intervals whose end points are dyadic rationals. Hence $\tau(E\cap F_j)$ is Borel.

$M$ is homeomorphic to $M^\infty$

Let $x$ be an infinite binary sequence. From this we construct a sequence of infinite binary sequences as follows. From the given sequence $x$ choose elements alternatingly to obtain the first sequence. After removing this subsequence from the original sequence, the procedure is repeated ad infinitum. This can be written mathematically as $\{x_1,x_2,x_3,\cdots,\} \to \{x^1,x^2,x^3,\cdots\}$ where each $x^j \in M$ and $i$th element of the $j$th sequence $x^j_i = x_{2^{j-1}2(i-1)}$. Clearly, this map is a bijection. \begin{array} dd(\{x^1,x^2,x^3,\cdots\}, \{y^1,y^2,y^3,\cdots\}) &= \sum_{j=1}^\infty \frac{1}{2^j} \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_M(x^j_i,y^j_i)}{1+d_M(x^j_i,y^j_i)} \\ &= \sum_{j=1}^\infty \frac{1}{2^j} \sum_{i=1}^\infty \frac{2^{2^{j-1} + (2i-1)}}{2^i} \frac{1}{2^{2^{j-1} + (2i-1)}} \frac{d_M(x_{2^{j-1} + (2i-1)},y_{2^{j-1} + (2i-1)})}{1+d_M(x_{2^{j-1} + (2i-1)},y_{2^{j-1} + (2i-1)})} \\ &\geq d_M(x,y) \end{array} The other direction is not too hard to show.

There exists a Borel set $E_1 \subset M$ such that $E_1 \sim I^\infty$

Let $\tau,E$ be the same as above. Define $\tau^\prime : M^\infty \to I^\infty$ by $\tau^\prime(x^1,x^2,\cdots) = (\tau(x^1),\tau(x^2),\cdots)$ where $x^i = \{x^i_1,x^i_2,\cdots\} \in M$. We show that $\tau^\prime$ is continuous.

\begin{array} dd_{I^\infty}(\tau^\prime(x),\tau^\prime(y)) &= \sum_{i=1}^\infty \frac{1}{2^i}\frac{|\tau(x^i) - \tau(y^i)|}{1+|\tau(x^i) - \tau(y^i)|} \\ &= \sum_{i=1}^\infty \frac{1}{2^i}\frac{|\sum_{j=1}^\infty \frac{1}{2^j}(x^i_j - y^i_j)|}{1+|\sum_{j=1}^\infty \frac{1}{2^j}(x^i_j - y^i_j)|} \\ &\leq \sum_{i=1}^\infty \frac{1}{2^i} \sum_{j=1}^\infty \frac{1}{2^j} {|x^i_j - y^i_j|} \\ &\leq d_{M^\infty}^0(x,y) \end{array} where $d_{M^\infty}^0$ is equivalent to $d_{M^\infty}(x,y) = \sum_{i=1}^\infty \frac{1}{2^i}\frac{d_M(x^i,y^i)}{1+d_M(x^i,y^i)}$. Hence $\tau^\prime$ is continuous and therefore measurable. Now, it will be shown that $\tau^\prime : E^\infty \to I^\infty$ is an isomorphism.

• $\tau^\prime$ is bijective as $\tau : E \to I$ is bijective.
• $E^\infty$ is a measurable set in $\mathcal{B}_{M^\infty}$
• $(\tau^\prime)^{-1} : I^\infty \to E^\infty$ is measurable. $\mathcal{B}_{E^\infty} = E^\infty \bigcap \mathcal{B}_{M^\infty} = \sigma(\{E^\infty \cap \rho_n^{-1}(U_{ni}), U_n \in \mathcal{D}\})$ where $\mathcal{D}_i$ is a denumerable base for $M$. However, each $U_n$ can in-turn be written as a countable union of finite intersection of sets of the form $V_{m,j}=\{\gamma_m^{-1}(\{j\}), j \in \{0,1\}\}$ (where $\gamma_m : M \to \{0,1\}$ is the co-ordinate projection). Hence, $\mathcal{B}_{E^\infty} = \sigma(\{E^\infty \cap \rho_n^{-1}(V_{mj})\})$. $\tau^\prime(E^\infty \cap \rho_n^{-1}(V_{mj})) = \tau^\prime(E \times E \times \cdots \times E\cap V_{mj} \times E \times E \cdots) =\tau(E) \times \tau(E) \cdots \times \tau(E \cap V_{mj}) \times \tau(E) \times \tau(E) \cdots$. Therefore, $(\tau^\prime)^{-1}$ is also measurable.

As $M$ and $M^\infty$ are homeomorphic, they are also isomorphic (in the sense defined initially) and it follows that $E^\infty$ is isomorphic to some $E_1 \subset M$. Therefore, we get that $E_1 \sim I^\infty$.

Compactness of infinite binary sequences

Let $X=\{0,1\}^\infty$ (countable product). Then $X$ is a compact metric space. We show this via the Cantor set . The Cantor set is denoted by $C$

$C=\{\sum_{k=1}^\infty a_k 3^{-k} : a_k \in \{0,2\} \forall k\}$

The Cantor set is generated in the following way. $C_0 = [0,1], C_1 =[0,1/3] \cup [2/3,1], C_2=[0,1/9] \cup [2/9,1/3] \cup [2/3,7/9] \cup [8/9,1]$ etc. First it is shown easily (by induction) that the left end points (of the disjoint intervals) of $C_k$ are of the form $\{\sum_{n=1}^\infty a_n 3^{-n} : a_n \in \{0,2\}~\forall 1 \leq n \leq k, a_n = 0, k < n\}$ and conversely.

For $k=0$, this is indeed true. Assume this to be true for $k$ and let $[c,d]$ be one of the disjoint intervals in $C_{k}$. Now, this interval gives rise to the intervals $[c,e]$, $[f,d]$ in $C_{k+1}$. In the first case, $c$ already has the required form by induction hypothesis. In the second case, $f=c+2(e-c)/3=c+2/3^{k+1}$ as $e-c = 1/3^k$ by construction. Therefore, $f$ has the required form as well.

Conversely, suppose a sequence $\{a_1,a_2,\cdots,\}$ where $a_n \in \{0,2\}, 1 \leq n \leq k, a_n = 0, k < n$ is given. This sequence represents the left end point chosen in the following way:- If $a_1 = 0$ then choose the left end point of the left partition of $[0,1]$ and if $a_1=2$ then choose the left end point of the right partition of $[0,1]$. If $a_2=0$ then choose the left end point of the left partition of the partition chosen in the previous step and so on. For example, $\{0,2,2,0,0,\cdots\}$ represents choosing the left end point of the left partition of $[0,1]$ (which is $[0,1/3]$) followed by the choosing the left end point of right partition of $[0,1/3]$ (which is $[2/3,1]$) followed by choosing the left end point of the right partition of $[2/3,1]$ (which is $[8/27,1/3]$). Therefore $\{0,2,2,0,0,\cdots\}$ represents the end point $8/27$ (in $C_3$).

The Cantor set contains other points (which are uncountably many) too. Infact all the end points are countable, so the other points are what make up the majority of the cantor set. Take $x \in C$ then $x \in C_k~\forall k$ and hence there exists $[s_k,t_k]$ such that $x \in [s_k,t_k] \subset C_k$. By construction of the Cantor set, $|x-s_k| < \frac{1}{3^k}$ and hence $s_k \to x$. But $s_k = \sum_{n=1}^k a_n3 ^{-n}$ and $\lim_{k \to \infty} s_k = \lim_{k \to \infty} \sum_{n=1}^k a_n3 ^{-n}$ where $a_n \in \{0,2\}~\forall n$.

$\{0,1\}^\infty$ is a compact metric space

Note that the metric on $\{0,1\}^\infty$ is given by $d(x,y) = \sum_{i=1}^\infty \frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$ where $d_i(x_i,y_i) = |x_i-y_i|$. Let $f : \{0,1\}^\infty \to C$ be given by $f(\{x_1,x_2,\cdots,\} = \sum_{k=1}^\infty 2x_k 3^{-k}$. Now it is shown that $f$ is a homeomorphism ($C$ is given the subspace topology induced from $\mathbb{R}$)

• $f$ is injective. Assume that $a = (a_0,a_1,\cdots) \neq b = (b_0,b_1,\cdots)$. Let $k$ be the smallest integer for which $a_k \neq b_k $. WLOG assume that $a_k=0,b_k=1$. Then, $\sum_{n=k+1}^\infty 2a_n3^{-n} \leq \sum_{n=k+1}^\infty 2*3^{-n} \leq 1/3^k < 2*3^k + \sum_{n=k+1}^\infty 2b_n3^{-n}$. Therefore, $f(a) < f(b)$.
• $f$ is onto. If $y \in C$, then $y=\sum_{k=1}^\infty a_k 3^{-k}$ for some $a_k \in \{0,2\} \forall k$. Choose $x_k = a_k/2$. Then $f(x) = y$.
• $f$ is continuous. $|f(x)-f(y)| \leq |\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| \leq \sum_{k=1}^\infty \frac{2}{3^k} |x_k-y_k| \leq 2d^\prime(x,y)$ where $d^\prime(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} d_i(x_i,y_i)$. As $d$ and $d^\prime$ are topologically equivalent , it follows that $f$ is uniformly continuous.
• $f^{-1}$ is continuous: Suppose $d(f(x),f(y)) = |\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| < 3^{-N}$. Let $M$ be the first mismatch between $x_k,y_k$ i.e. $x_M \neq y_M$ and $x_k = y_k, 1 \leq k \leq M-1$. Then $|\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| = |\sum_{k=M}^\infty \frac{2}{3^k} (x_k-y_k)|$. Assume now that $1 \leq M \leq N$. This gives \begin{array} $|\sum_{k=M}^\infty \frac{2}{3^k} (x_k-y_k)| &= |\pm \frac{2}{3^M} + \sum_{k=M+1}^\infty \frac{2}{3^k} (x_k-y_k)| \\ &\geq |\frac{2}{3^M} | - \sum_{k=M+1}^\infty \frac{2}{3^k} (x_k-y_k)| | \\ & \geq \frac{2}{3^M} - \sum_{k=M+1}^\infty |\frac{2}{3^k}| \\ &\geq \frac{1}{3^M} \\ &\geq \frac{1}{3^N}. \end{array} Hence the first mismatch can only occur at $ M > N$. This gives that $d^\prime(x,y) \leq 2^{-N}$.

Hence $\{0,1\}^\infty$ is compact as $C$ is compact.

Product Topology and Metric Spaces

Topology: Let $X$ be a set and $\tau$ be a collection of subsets of $X$. $(X,\tau)$ is called a Topological space if the following are satisfied :

• $\tau$ is closed under arbitrary union
• $\tau$ is closed under finite intersections
• $\phi \in \tau$, $X \in \tau$

$\tau$ is the collection of open sets.

Base for a Topology . A collection of open sets $B \subset \tau$ is called a base for the topology $(X,\tau)$ if any open set can be written as a union of sets in $B$.

Properties of a Base

Let $B \subset \tau$ be a base for the topology $\tau$. Then, $B$ satisfies the following.

• $B$ covers $X$
• If $U_1,U_2 \in B$ and $x \in U_1 \cap U_2$ then there exists $U_3 \in B$ such that $x \in U_3 \subset U_1\cap U_2$.

Let $x \in U_1 \cap U_2$. As $U_1,U_2$ are both open sets (note that the base is a subset of $\tau$), their finite intersection is also contained in $\tau$. Hence $U_1\cap U_2 = \cup_\alpha U_\alpha$ for some sets $U_\alpha \in B$. This implies that $x \in U_\alpha \subset U_1 \cap U_2$ for some $\alpha$. As $X\in \tau$, we obviously have that $B$ covers $X$.

Conversely, suppose the topology $\tau$ is not given and instead we have a collection of sets $B \subset X$. We construct the set of all arbitrary unions of sets in $B$ and call it $\tau$. The question is when is $\tau$ a topology as defined above. The answer is that the collection has to satisfy exactly the same two conditions given above. This is summarized below.

Properties of a Base (Converse )

Let $B \subset X$ be a collection of subsets of $X$ and $\tau$ is the collection of arbitrary unions of elements of $B$. $\tau$ is a topology if (and only if)

• $B$ covers $X$
• If $U_1,U_2 \in B$ and $x \in U_1 \cap U_2$ then there exists $U_3 \in B$ such that $x \in U_3 \subset U_1\cap U_2$.

(Note that the "only if" part has been proved above)

This is easy to check.

With abuse of notation we call a collection $B$ of subsets of $X$ even if we haven't defined a topology on $X$. In such situations, we only mean that $B$ satisfies the properties mentioned previously. If $\{\tau_\alpha\}$ is a collection of topologies on $X$ then $\cap_\alpha \tau_\alpha$ is also a topology on $X$. Therefore, we can define the smallest topology containing given subsets of $X$ as the intersection of all topologies containing those subsets. We also call the smallest topology containing given subsets of $X$ as the topology generated by those sets.

Subbase for a Topology . A collection of open sets $SB \subset \tau$ is called a subbase for the topology $(X,\tau)$ if $\tau$ is the smallest Topology containing $SB$.

(Property of a Sub-Base) Let $\tau$ be a topology and $SB$ be a sub-base for $\tau$. The class of sets consisting of all finite intersections of elements of $SB$ along with $X,\phi$ forms a base for the topology $\tau$.

If $SB$ is a sub-base and $B$ is a base, denote by $\tau(SB)$ be the topology generated by $SB$ and $\tau^\prime(B)$ be the collection of all arbitrary unions of elements of $B$ (which is a topology as shown above). Define $B$ to the class of sets consisting of all finite intersections of elements of $SB$ along with $X,\phi$. Then $B$ is a base. Therefore, $\tau^\prime(B)$ is a topology. It is easy to see that $\tau^\prime (B) \subset \tau$ and as $\tau^\prime(B) \supset SB$ we have $\tau^\prime(B) \supset \tau(SB) = \tau$. Hence $\tau^\prime(B) = \tau$

Product Topology Let $(X_1,\tau_1),(X_2,\tau_2), \cdots $ be a collection of Topological spaces and let $X = X_1 \times X_2 \times \cdots$. The product topology is defined as the topology generated by the co-ordinate projections $\rho_i : X \to X_i, \rho_i (x_1,x_2,\cdots) = x_i$

From the above we identify the product topology as the topology generated by sets of the form $\rho_i^{-1}(U_i)$ where $U_i \in \tau_i$. In other words, $\{\rho_i^{-1}(U_i) : U_i \in \tau_i, i \in \mathbb{N}\}$ forms a subbase for the product topology.

Metric Spaces and Metric Topology : Let $(X,d)$ be a metric space. The metric induces a topology on the set in the following way. Any subset $U\subset X$ is said to be open if for every point $u \in U$, there exists an $r(u) > 0$ such that $B(u,r(u)) := \{x : d(x,u) < r(u)\}$ is contained in $U$. It is clear that the collection of all such sets is a topology.

Separable Metric Space : Let $(X,d)$ be a metric space. If $S \subset X$ such that $S$ is countable and any point in $X$ is a limit point of the set $S$ then $X$ is called a separable metric space.

Any Separable metric space has a countable base

Basically the idea is to take a countable set of points and take countable number of open balls for each of the points and then claim that they form a basis. For the countable set of points, we of-course take $S$. Let $S=\{x_1,x_2,\cdots\}$. Now, for each of the points $x_i$ we consider the balls $B_{ij} := B(x_i,\frac{1}{2^j}), j = 1,2,\cdots$. The class of all $B_{ij}$ is countable as countable union of countable sets is countable . It is easy to see that the $B_{ij}$ form a base for the metric topology.

Topological Equivalence of Metrics : Two metrics are called equivalent if the corresponding metric topologies are same.

Two topologies $\tau_1$ and $\tau_2$ are equal if and only if for every $x \in V_1$ where $V_1 \in \tau_1$ then there exists $V_2 \in \tau_2$ so that $x \in V_2 \subset V_1$ and vice versa.

Let $V_1 \in \tau_1$. For every $x \in V_1$, there exists $V_2(x)$ such that $x \in V_2(x) \subset V_1$. As $\cup_{x \in V_1} V_2(x) = V_1$, $V_1 \in \tau_2$. Similarly, if $V_2 \in \tau_2$ then $V_2 \in \tau_1$. The other direction is trivial.

The metrics $d$ and $d/(1+d)$ are topologically equivalent

A Metric on Countable Product of Metric Spaces : If $(X_i,d_i)$ are metric spaces then define $d(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_i(x_i,y_i)}{1 + d_i(x_i,y_i)}$. Then it can be shown that $d$ is a metric on $X=X_1\times X_2 \cdots$

$d(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_i(x_i,y_i)}{1 + d_i(x_i,y_i)}$ and $d^\prime(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} d_i(x_i,y_i)$ are topologically equivalent if $d_i(x,y) \leq C$ for all $i,x,y$.

As $B_d^\prime(x,a) \subset B_d(x,a)$, we get $\tau \subset \tau^\prime$. If $d(x,y) < a$ then $\sum_{i=1}^\infty \frac{1}{2^i}\frac{d(x_i,y_i)}{1+d(x_i,y_i)} < a$. As $d(x_i,y_i) \leq C$, we have $\sum_{i=1}^\infty \frac{1}{2^i}\frac{d(x_i,y_i)}{1+d(x_i,y_i)} \geq \sum_{i=1}^\infty \frac{1}{2^i}\frac{d(x_i,y_i)}{1+C}$. Therefore $\sum_{i=1}^\infty \frac{1}{2^i}d(x_i,y_i) < (1+C)a$

The Metric topology of $(X, d)$ where $X=X_1\times X_2 \cdots$ and $d(x,y) := \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_i(x_i,y_i)}{1 + d_i(x_i,y_i)}$ is the same as the product topology on $(X_1\times X_2 \cdots)$

We will denote by $\tau_1$ the metric topology and by $\tau_2$ the product topology. Let $x \in V_1 \in \tau_1$, then $x \in B(x,r)$ for some $r > 0$ by definition of open sets in $\tau_1$. Choose $N$ such that $2^{-N} \leq r/2$ and define $U = B_1(x_1,a/2) \times B_1(x_2,a/2) \times \cdots B_N(x_N,a/2) \times X_{N+1} \times X_{N+2} \cdots$. Note that the balls here are w.r.t. to the metric $d_i/(1+d_i)$. As shown earlier such balls are also open in the metric $d_i$. Hence, $U$ is open in the product topology. We now need to show that $x \in U \subset B(x,r)$.

Let $y\in U$ then $d(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_i(x_i,y_i)}{1 + d_i(x_i,y_i)} \leq \sum_{i=1}^N \frac{1}{2^i} \frac{r}{2} + \sum_{i=N+1}^\infty \frac{1}{2^i} \frac{d_i(x_i,y_i)}{1 + d_i(x_i,y_i)} < r$.

Let $x \in V_2 \in \tau_2$. The aim is to find $B(x,r), r > 0$ such that $x \in B(x,r) \subset V_2$. $V_2 = \bigcup_\alpha \left(\cap_{k=1}^{m(\alpha)} \rho_{n(\alpha,k)}^{-1} (U_{n(\alpha,k)})\right)$ by definition. Therefore, for some $\alpha$ and $\forall k = 1,2,\cdots,m(\alpha)$ \begin{align} x &\in \rho_{n(\alpha,k)}^{-1} (U_{n(\alpha,k)}) \\ \rho_{n(\alpha,k)}(x) &\in U_{n(\alpha,k)}\\ \rho_{n(\alpha,k)}(x) &\in B(\rho_{n(\alpha,k)}(x), r_{(\alpha,k)}(x))\\ x &\in \rho_{n(\alpha,k)}^{-1}(B(\rho_{n(\alpha,k)}, r_{(\alpha,k)}(x)))\\ \end{align} Take $r(x) = \min_{k=1,2,\cdots, m(\alpha)} \frac{r(\alpha,k)(x)}{2^{n(\alpha,k)}}$ and define $B = \{y | d(x,y) < r(x)\}$. Then, $x \in B \subset \bigcap_{k=1}^{m(\alpha)}\rho_{n(\alpha,k)}^{-1}(B(\rho_{n(\alpha,k)}, r_{(\alpha,k)}(x))) $

Countable Cartesian Product of Separable Metric Spaces is Separable

Fix points $x_n \in X_n$ and consider the sets $E_m = \left(\Pi_{1 \leq n \leq m}S_n \right)\times\left(\Pi_{n > m} \{x_n\}\right)$ where $S_n$ is a countable dense subset of $X_n$. $E_m$ is a countable set and $E=\bigcup_{m \geq 1}E_m$ is also countable. $E$ is the desired dense set in $X$. To see this let $z=\{z_1,z_2,z_3,\cdots\}\in X$ and fix $\epsilon > 0$. Choose $N$ such that $2^{-N} < \epsilon$ and choose $y_i \in S_i$ such that $d_i(z_i,y_i)/(1+d_i(z_i,y_i)) < \epsilon/2$ for all $1 \leq i \leq N$ and $y_i = x_i$ for all $N+1 \leq i$. Then $d(x,y) < \epsilon$.

We would like to know the countable basis for the cartesian product. Let $\mathcal{D}_i = \{U_{i,1},U_{i,2},\cdots,\}$ be a countable base for $X_i$. Any open set $O$ in $X=X_1\times X_2\times\cdots$ can be written as \begin{array} OO&= \bigcup_\alpha \bigcap_{k=1}^{m(\alpha)} \rho_{n(\alpha,k)}^{-1}(U^\prime_{n(\alpha,k)}) \\ &= \bigcup_\alpha \bigcap_{k=1}^{m(\alpha)} \rho_{n(\alpha,k)}^{-1}\left(\bigcup_{l=1}^\infty U_{n(\alpha,k),j(l)}\right) \end{array} where $j:\mathbb{N}\to \mathbb{N}$. It can be shown that $\bigcap_{k=1}^m\bigcup_{l=1}^\infty A_{ik} = \bigcup_{l_1,l_2,\cdots,l_m =1}^\infty \bigcap_{k=1}^m A_{l_k,k}$. Therefore, $\bigcap_{k=1}^{m(\alpha)} \bigcup_{l=1}^\infty \rho_{n(\alpha,k)}^{-1} U_{n(\alpha,k),j(l)}$ is a countable union of finite intersections of the sets $\rho_i^{-1}(U_{i,j}), i \geq 1, j \geq 1$. Therefore, $O$ is also a countable union of such countable number of sets.

As $X^\infty$ is a topological space the Borel $\sigma$-algebra on $X^\infty$ (the $\sigma$-algebra generated by the open sets in $X^\infty$) denoted by $\mathcal{B}_{X^\infty}$ can be defined. We can also define another $\sigma$-algebra on $X^\infty$ caleed the the product $\sigma$-algebra. This the smallest $\sigma$-algebra such that all the co-ordinate maps are measurable (Note the similar definition used to define product topology).

If $X_1,X_2,\cdots,$ are separable metric spaces and $X=X_1\times X_2\times\cdots$ then $\mathcal{B}_X= \mathcal{B}_{X_1}\times \mathcal{B}_{X_2} \times \cdots$.

$\mathcal{B}_{X_1}\times \mathcal{B}_{X_2} \times \cdots = \sigma(\{\rho_n^{-1}(B_n), B_n \in \mathcal{B}_{X_n}\}) = \sigma(\{\rho_n^{-1}(U_{ni}), U_{n,i} \in \mathcal{D}_i\})$. Clearly, $\mathcal{B}_X \supset \mathcal{B}_{X_1}\times \mathcal{B}_{X_2} \times \cdots$. As every open set in $X$ is a countable union of finite intersections of the sets $\rho_i^{-1}(U_{i,j}), i \geq 1, j \geq 1$, it follows that $\mathcal{B}_X \subset \mathcal{B}_{X_1}\times \mathcal{B}_{X_2} \times \cdots$.