The product topology on $\Bbb{R}^k$ is equivalent to euclidean topology.

Question

Statement

The product topology $\mathcal{P}$ on $\Bbb{R}^n$ is equivalent to euclidean topology $\mathcal{E}$, where $n\in\Bbb{N}$.

Proof. Let be $(a_1,b_1)\times...\times(a_n,b_n)$ a basic open set on $\Bbb{R}^n$ equipped with product topology $\mathcal{P}$. So for any $x\in(a_1,b_1)\times...\times(a_n,b_n)$ we define $$ A=\{b_i-x_i: i=1,...,n\} $$ and $$ B=\{x_i-a_i: i=1,...,n\} $$ ad so let be $$ \rho:=\min(A\cup B) $$ therefore it follows that $\rho\le b_i-x_i$ and $\rho\le x_i-a_i$ for any $i=1,...,n$. So for any $y\in B(x,\rho)$ and for any $i=1,...,n$ it follows that $$ (y_i-x_i)^2\le\sum_{j=1}^n(y_j-x_j)^2<\rho^2\le(b_i-x_i)^2 $$ and $$ (x_i-y_i)^2\le\sum_{j=1}^n(x_j-y_j)^2<\rho^2\le(x_i-a_i)^2 $$ and so clearly $y_i\in(a_i,b_i)$ for any $i=1,...,n$, that is $y\in(a_1,b_1)\times...\times(a_n,b_n)$. So we conclude that $\mathcal{P}\subseteq\mathcal{E}$.

Now let be $B(x,\rho)$ a basic open set on $\Bbb{R}^n$ equipped with euclidean topology $\mathcal{E}$ and so for any $y\in B(x,\rho)$ we consider $$ \rho'=\rho-d(x,y) $$ therefore it follows that $B(y,\rho')\subseteq B(x,\rho)$. So we choose $\epsilon>0$ such that $$ n\epsilon^2<(\rho')^2 $$ and so for any $z\in(y_1-\epsilon,y_1+\epsilon)\times...\times(y_n-\epsilon,y_n+\epsilon)$ and for any $i=1,...,n$ it follows that $$ y_i-\epsilon<z_i<y_i+\epsilon $$ and so we conclude that $$ |(z_i-y_i)|<\epsilon $$ thus if we squaring and if we sum then it follows that $$ \sum_{i=1}^n(z_i-y_i)^2<n\cdot\epsilon^2\le(\rho')^2 $$ and so it follows that $z\in B(x,\rho)$. So we conclude that $\mathcal{E}\subseteq\mathcal{P}$.

Finally $\mathcal{P}=\mathcal{E}$.

So I ask if my proof is correct. Then I ask if the statement is even true for $\Bbb{R}^k$, where $k$ is an infinite cardinal, in the case it is possible to define the euclidean metric if $k>\omega$: so if this is true could you prove it? if not, could you take a counterexample?

Answer 1

For $n=2$ I showed here that $d_\infty$ on $\Bbb R^2$ induces the product topology on $\Bbb R^2$; this easily generalises to finite $n$.

The metric $d_2$ (the Euclidean metric) gives an equivalent topology on $\Bbb R^n$ as $d_\infty$ (the max metric) does, as

$$\forall x,y \in \Bbb R^n: d_\infty(x,y) \le d_2(x,y) \le \sqrt{n}d_\infty(x,y)$$

so the metrics are uniformly equivalent.

Hence $\mathcal{E} = \mathcal{T}_{d_2}= \mathcal{T}_{d_\infty} = \mathcal{T}_{\text{prod}}=\mathcal{P}$.

For the countably infinite case it gets interesting:

$d_p(x,y)=\sum_n \frac{1}{2^n} \min(|x_n - y_n|,1)$ (we need a weighing factor and truncation to ensure convergence) is a metric on $\Bbb R^{\Bbb N}$ that induces the product topology (see here, e.g.), showing that the countable product is still metrisable. Which you also could have deduced indirectly from Urysohn's metrisation theorem: a countable product of second countable $T_3$ spaces is again second countable $T_3$...

If we use $d_{\sup} = \sup_n \min(|x_n-y_n|,1)$, this does define a metric on $\Bbb R^{\Bbb N}$ but its induced topology is strictly stronger (larger) than the product topology and also strictly weaker than the (non-metrisable) box topology on that set. It's one of Munkres' pet examples (the uniform metric topology).

$\Bbb R^I$ for any uncountable index set $I$ (in the product topology) is not first countable, and also not normal, so such spaces are non-metrisable.

Answer 2

There are,of course,several ways to see this.I am mentioning one,first of all,consider $\mathbb{R}^{n}$ as the space of all functions from $\{1,2,.., n\}$ to $\mathbb{R}$.Then,note that the product topology is nothing but the topology of pointwise convergence,and since the domain is a finite set,the pointwise convergence topology and the uniform convergence topology,which is basically the topology induced by the $l_{\infty}$ norm,coincides,and it's well known that all norms on $\mathbb{R}^{n}$ are equivalent.As for your next query,it is true that a countable product of metrizable spaces is metrizable,but you cannot say the same for uncountable products.In your query,you have asked whether or not it is possible to metrize it when $k>\omega$.Well,as long as $k$ is an ordinal with countable cardinality,for example,if it's the successor ordinal $\omega + 1$ , you can still metrize the product topology on $\mathbb{R}^{k}$.However,if the cardinality becomes uncountable,then it's not possible.Say,for example,you consider $\mathbb{R}^{\mathbb{R}}$ with the product topology.Assume it's metrizable.Then,consider it's subspace $[0,1]^{\mathbb{R}}$.This is compact by Tikhonov's theorem,but has cardinality $2^{c}$,which is a contradiction because the cardinality of compact metrizable spaces can be atmost $c$.