Euclidean Topology is Product Topology/Proof 1

Proof
Denote the Euclidean topology on $\R^n$ as $\tau$, and denote the product topology on $\R^n$ as $\tau'$.

Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$.

Then there exists $\epsilon \in \R_{>0}$ such that the open ball $\map {B_\epsilon} x \subseteq U$.

We show that:
 * $\ds B' = \prod_{i \mathop = 1}^n \openint {x_i - \dfrac \epsilon n} {x_i + \dfrac \epsilon n} \subseteq \map {B_\epsilon} x$

For if $y = \tuple {y_1, \ldots, y_n} \in B'$, then:

From Natural Basis of Product Topology of Finite Product, $B' \in \tau'$.

As $B' \subseteq \map {B_\epsilon} x \subseteq U$, Neighborhood Condition for Coarser Topology shows that $\tau' \subseteq \tau$.

Let $U' \in \tau'$.

Let $x = \tuple {x_1, \ldots, x_n} \in U'$.

From Natural Basis of Product Topology of Finite Product, sets of the type $\ds \prod_{i \mathop = 1}^n U'_i$ with $U'_i \in \tau_1$ form an analytic basis for $\tau'$.

From Equivalence of Definitions of Analytic Basis, it follows that we can select $U'_1, \ldots, U'_n \in \tau_1$ such that $\ds x \in \prod_{i \mathop = 1}^n U'_i \subseteq U'$.

By definition of open set, it follows that for all $i \in \set {1, \ldots, n}$, we can find $\epsilon_i \in \R_{>0}$ such that $\openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq U'_i$.

Put $\epsilon = \min \set {\epsilon_i : i = 1, \ldots, n}$.

We show that the open ball $\map {B_\epsilon} x \subseteq U'$.

For if $y = \tuple {y_1, \ldots, y_n} \in \map {B_\epsilon} x$, then $y_i \in \openint {x_i - \epsilon_i} {x_i + \epsilon_i }$, as:


 * $\size {x_i - y_i} < \epsilon \le \epsilon_i$

It follows that:
 * $\ds \map {B_\epsilon} x \subseteq \prod_{i \mathop = 1}^n \openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq \prod_{i \mathop = 1}^n U'_i \subseteq U'$

Then Neighborhood Condition for Coarser Topology shows that $\tau \subseteq \tau'$.

It follows that $\tau = \tau'$.