# Box and Product Topologies

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

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