# Understanding why a finite dimensional vector space is a topological manifold

I wish to understand the details of why a finite dimensional vector space is a topological manifold, particularly following Jonh Lee's Introduction to smooth manifolds. I know that this questions has been asked at least Here and here, and while I convinced myself that it is the case, I would not be able to fill in the details if challenged, hence I decided to open a new post with the relevant questions and approach.

Example 1.24 (Finite-Dimensional Vector Spaces). Let $$V$$ be a finite-dimensional real vector space. Any norm on $$V$$ determines a topology, which is independent of the choice of norm (Exercise B.49). With this topology, $$V$$ is a topological $$n$$-manifold, and has a natural smooth structure defined as follows. Each (ordered) basis $$(E_1,...,E_n)$$ for $$V$$ defines a basis isomorphism $$E:R^n \to V$$ by $$E(x) = \sum_{i=1}^nx^iE_i.$$ This map is a homeomorphism [...].

I was filling the details and wanted to discuss my proofwriting when I was told that it was obvious because any real finite dimensional vector space is isomorphic to $$\mathbb{R}^n$$. However at my stage I don't understand how that translates into showing each of the technical details in the definition of a topological manifold. Here are my questions:

How do I precisely show that $$V$$ is a topological manifold?

My reasoning is this: As stated, all norms are equivalent in a finite-dimensional vector space and generate the same topology, but which one is it? well, since a norm induces a metric, I thought it will be the metric topology (denote it by $$\tau_{||\cdot||}$$), i.e., the one generated by the basis $$\mathcal{B} = \lbrace B_r(v) : u\in V \; \text{and}\; r>0 \rbrace$$ where $$B_r(u) = \lbrace v\in V : d_{||\cdot||}(u,v) = ||u-v|| , which would then be unique. Then $$(V,\tau_{||\cdot||})$$ would be my topological space. I can show that it is Hausdorff but I am not sure how to show it is second countable. any suggestion?

Then I need to show it is locally Eucliean of dimension $$n$$. For this, I need to show that for every point in $$V$$ there is a neighborood which is mapped to an open subset of $$\mathbb{R}^n$$ (or $$\mathbb{R^n}$$ itself). The given suggestion is the isomorphism $$E$$ (in linear algebra, a linear transformation which is also a bijection) which the author claims is a homeomorphism. I am guessing it means precisely a homeomorphism from $$E: (V,\tau_{||\cdot||}) \to (\mathbb{R}^n,\tau_U)$$ where $$\tau_U$$ is the usual topology. So we would be showing that $$V$$ is homeomorphic to $$R^n$$, which implies $$V$$ is locally Euclidean, is this correct? Now I already now that $$E$$ is a biyection and since $$\mathcal{B}$$ is a basis for $$\tau_{||\cdot||}$$, I could show that $$E(\mathcal{B})$$ is a basis for $$\tau_U$$, which should be duable since $$n$$-balls form a basis for $$\tau_U$$ as well, and I think I would be done. Is that right? Is this a good way of proceeding? I will probably also have questions regarding on how to write these ideas, but one step at the time. Unfortuately I am learning all of this at the same time as opposed to a typical math degree curriculum, so I appretiate your patience.

If you show that $$V$$ with the metric topology induced by any norm is homeomorphic to $$\Bbb R^n$$ using the suggested vector space isomorphism, then since $$\Bbb R^n$$ is Hausdorff, second countable, and locally Euclidean, the same things apply to $$V$$, so you will have automatically verified all three properties of what it means to show that $$V$$ is a topological manifold.