User:Dfeuer/Euclidean Topology is Tychonoff Topology

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

Let $T_n = \left({\R^n, \tau_n}\right)$ be the topological space, where $\tau_n$ 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
Let $d_2$ be the Euclidean metric,


 * $\displaystyle d_2(x,y) = \sqrt{\sum_{i=1}^n (x_i-y_i)^2}

Let $d_\infty (x,y) = \max_{i=1}^n |x_i - y_i|$, 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 $\delta = \frac \epsilon {\sqrt n}$.

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

Suppose that $d_\infty (p, q) < \delta$.

Then $\sum_{i=1}^n (p_i-q_i)^2 \le \sum_{i=1}^n (d_\infty (p,q))^2$, so