Euclidean topology on $\mathbb{R}^{m+n}$ is equivalent to the product topology on $\mathbb{R}^m \times \mathbb{R}^n$

Question

I'm attempting to teach myself topology for graduate school this summer, but I'm having a tough time. I'm trying to prove that the Euclidean topology on $\mathbb{R}^{m+n}$ is equivalent to the product topology on $\mathbb{R}^m \times \mathbb{R}^n$. I realize to do this that I should make a homeomorphism between them, and the identity function would work for this, but I'm unsure on what to do from there.

Here's what I have so far: Let $x \in U \subseteq \mathbb{R}^{m+n}$ be some open set in $\mathbb{R}^{m+n}$, then there exists an open ball $B_{\epsilon}(x) \subseteq U$. But I'm not sure where to go from there.

I have read this Product topology and standard euclidean topology over $\mathbb{R}^n$ are equivalent but I do not understand why you are allowed to assume that each of the subsets in $B^1_{\epsilon}(x_i)$ are open in $\mathbb{R}$. thank you

Answer 1

Hint Suppose $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are the Euclidean norms in $ \Bbb R^n,\Bbb R^m$ respectively. Then $\lVert (x,y)\rVert = (\lVert x\rVert_1^2+\lVert y\rVert_2^2)^{1/2}$ is the Euclidean norm in $\Bbb R^n\times\Bbb R^m$.

Answer 2

Just use $f:\Bbb R^m \times \Bbb R^n \to \Bbb R^{m+n}$. $((x_1,...,x_m),(x_{m+1},...,x_{m+n}))\to (x_1,...,x_{m+n})$ Now see that it is bijective and continuous. Actually it some kind of identity function. [As you are having problem, some things you need to clarify, 1) Eucledean topology is same as the metric topology, here euclidean norm is the metric,$d(x,y)=||x-y||$ (check!!) 2) For finite products, box topology is same as product topology. I don't know whether you come across the term box topology or not, but advise you to make your conception clear in these cases]