Euclidean Topology is Product Topology/Proof 2

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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

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.

$\blacksquare$


Sources

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