Euclidean Topology is Product Topology

Theorem

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

Let $T_n = \struct {\R^n, \tau_n}$ be the topological space such that $\tau_n$ 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 1

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:

 $\ds \map d {x, y}$ $=$ $\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^2}^{1/2}$ where $d$ denotes the Euclidean metric $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \size{x_i - y_i}$ by Minkowski's Inequality for Sums $\ds$ $<$ $\ds \sum_{i \mathop = 1}^n \dfrac \epsilon n$ $\ds$ $=$ $\ds \epsilon$

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 $\tau' \subseteq \tau$.

$\Box$

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 $\tau \subseteq \tau'$.

$\Box$

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

$\blacksquare$

Proof 2

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

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

Basis for the Induction

$\map P 2$ is the case:

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

This is demonstrated in Euclidean Topology on Cartesian Plane is Product Topology.

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

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

from which it is to be shown that:

the Euclidean topology on $\R^{k + 1}$ and the product topology on $\R^{k + 1}$ are the same.

Induction Step

This is the induction step:

We have that:

$\R^{k + 1} = \R^k \times \R$

By the induction hypothesis:

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

By the basis for the induction:

the Euclidean topology on $\R^k \times \R$ and the product topology on $\R^k \times \R$ are the same.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 0}:$
the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.

$\blacksquare$