Euclidean Topology is Product Topology

Theorem
Let $T_1 = \left({\R, \tau_1}\right)$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$.

Let $T_n = \left({\R^n, \tau_n}\right)$ be the topological space such that $\tau_n$ is the Tychonoff topology on the cartesian product $\displaystyle \R_n = \prod_{i \mathop = 1}^n \R$.

Then the Euclidean topology on $\R^n$ and the Tychonoff topology on $\R^n$ are the same.

Proof
We first show that the Euclidean topology on $\R^n$ is the same as the box topology on $\R^n$.

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

Let $U \in \tau$, and let $x = \left({x_1, \ldots, x_n}\right) \in U$.

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

We show that $\displaystyle B' = \prod_{i \mathop = 1}^n \left({x_i - \dfrac \epsilon n \,.\,.\, x_i + \dfrac \epsilon n }\right) \subseteq B_\epsilon \left({x}\right)$.

For if $y = \left({y_1, \ldots, y_n}\right) \in B'$, then:

By definition of box topology, $B'\in \tau'$.

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

Let $U' \in \tau'$, and let $x = \left({x_1, \ldots, x_n}\right) \in U'$.

From Synthetic Basis and Analytic Basis are Compatible, it follows that sets of the type $\displaystyle \prod_{i \mathop = 1}^n U'_i$ with $U'_i \in \tau_1$ form an analytic basis for $\tau'$.

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

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

Put $\displaystyle \epsilon = \min_{i \in \left\{ {1, \ldots, n}\right\} } \left({\epsilon_i}\right)$.

We show that the open ball $B_\epsilon \left({x}\right) \subseteq U'$.

For if $y = \left({y_1, \ldots, y_n}\right) \in B_\epsilon \left({x}\right)$, then $y_i \in \left({x_i - \epsilon_i \,.\,.\, x_i + \epsilon_i }\right)$, as:


 * $\left\vert{x_i - y_i}\right\vert < \epsilon \leq \epsilon_i$

It follows that $\displaystyle B_\epsilon \left({x}\right) \subseteq \prod_{i \mathop = 1}^n \left({x_i - \epsilon_i \,.\,.\, x_i + \epsilon_i }\right) \subseteq \prod_{i \mathop = 1}^n U'_i \subseteq U'$.

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

It follows that $\tau = \tau'$, so the Euclidean topology is the same as the box topology.

The result now follows from Box Topology on Finite Product Space is Tychonoff Topology.