Euclidean Topology is Product Topology/Proof 2

Proof
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.