# Box and Product Topologies

• Sep 18th 2013, 04:05 PM
Aryth
Box and Product Topologies
I actually have two questions. One about topological comparisons between those two topologies and one about the identity map. First, I just wanted to make sure a proof I was using was correct or not... I've battled with it for awhile and I want some criticism.

The exercise was: Let $\displaystyle \{(\mathbb{R}_i,\Omega_i) : i \in \mathbb{N}\}$ be an indexed family of copies of the real line under the usual topology and let $\displaystyle \mathbb{R}^{\infty} = \prod\{\mathbb{R}_i : i \in \mathbb{N}\}$. Show that the product topology is a proper subset of the box topology on $\displaystyle \mathbb{R}^{\infty}$.

For this proof, it will suffice to show that $\displaystyle P \subset B$ where $\displaystyle B = \prod\{ U_i : U_i \in \Omega_i\}$ and $\displaystyle P = \prod \{ V_i : V_i \in \Omega_i \text{ and all but finitely many } V_i \neq \mathbb{R}_i\}$ are the bases of the Box and Product topologies, respectively. We will first consider $\displaystyle U_i \subset \mathbb{R}_i$. We know, then, that $\displaystyle \prod\{U_i : i\in \mathbb{N}\} \in B$ but not in $\displaystyle P$ since every set in $\displaystyle P$ has $\displaystyle U_i = \mathbb{R}_i$ for all but a finite number of $\displaystyle U_i$. Thus, $\displaystyle B \not\subseteq P$. Now we will show that $\displaystyle P \subset B$. We first remember that, in $\displaystyle P$, any, all or no $\displaystyle V_i$ from $\displaystyle \{V_i : i\in \mathbb{N}\}$ will be $\displaystyle \mathbb{R}_i$. $\displaystyle B$ only requires that each individual $\displaystyle U_i \in \Omega_i$, and this is true whether $\displaystyle U_i \subseteq \mathbb{R}_i$ or $\displaystyle U_i = \mathbb{R}_i$. Hence, we can conclude that any element of $\displaystyle P$ is in $\displaystyle B$ and $\displaystyle P \subset B$.
For each $\displaystyle i \in \mathbb{Z}^+$, let $\displaystyle f_i : \mathbb{R} \to \mathbb{R}$ be the identity map $\displaystyle f_i(x) = x$. Show that each $\displaystyle f_i$ is continuous relative to the usual topology on $\displaystyle \mathbb{R}$.