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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 $B_r(u) = \lbrace v\in V : d_{||\cdot||}(u,v) = ||u-v|| <r\rbrace$, 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.

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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.

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