User:Dfeuer/Euclidean Topology is Tychonoff Topology

Theorem
Let $T_1 = \struct {\R, \tau_1}$ be the topological space, where $\tau_1$ is the Euclidean topology on $\R$.

Let $T_n = \struct {\R^n, \tau_n}$ be the topological space, where $\tau_n$ is the product topology on the cartesian product $\ds \R_n = \prod_{i \mathop = 1}^n \R$.

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

Proof
Let $d_2$ be the Euclidean metric,


 * $\ds \map {d_2} {x, y} = \sqrt {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}$

Let $\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$, as in Definition:General Euclidean Metric.

We will show that these metrics are equivalent.

$d_\infty$ is finer than $d_2$
For any $\epsilon > 0$, let $\delta = \dfrac \epsilon {\sqrt n}$.

Let $p$ be any point in $\R^n$.

Suppose that $\map {d_\infty} {p, q} < \delta$.

Then $\ds \sum_{i \mathop = 1}^n \paren {p_i - q_i}^2 \le \sum_{i \mathop = 1}^n \paren {\map {d_\infty} {p, q} }^2$, so