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 $\mathcal{B}_X \subset \mathcal{B}_{X_1}\times \mathcal{B}_{X_2} \times \cdots$.