Why is it true that every finite-dimensional inner product space is a Hilbert space? A Hilbert space is, by definition, a complete inner product space. If $(V,|.|)$ is finite dimensional inner product space of dimension $n$ then it is (topologically) isomorphic to $\mathbb{R}^n$ which is of course complete. My instinct here is to say "and therefore, $V$ is also a Hilbert space". I'm not sure about this step however since completeness depends on the norm and the norm depends on the selected inner product (assuming the induced norm $||x|| = (x | x)^{1/2}$ is used). So, must we place some condition on the inner product or is the topological isomorphism between $\mathbb{R}^n$ enough to guarantee that V is Hilbert?
 A: As a matter of fact, the compactness of the unit sphere in a finite dimensional space can tell us that all norms on that space are equivalent. Therefore, in any norm (induced by inner product or otherwise), finite dimensional normed linear spaces are always complete.
A: An $n$-dimensional real inner product space is not just topologically isomorphic to $\mathbb{R}^n$ (this would indeed not imply completeness, since it is not a topological property). It is isometric to it. Indeed, let $V$ be such a space and let $v_1,\dots,v_n$ be an orthonormal basis in $V$. Let $e_1,\dots,e_n$ be the standard basis in $\mathbb{R}^n$. Define a linear transformation $T:V \to \mathbb{R}^n$ by declaring $T(v_i)=e_i$ for $i=1,\dots,n$ and extending linearly. Then $T$ is clearly a bijection and $\left\langle T\left(v_{i}\right),T\left(v_{j}\right)\right\rangle =\left\langle e_{i},e_{j}\right\rangle =\delta_{ij}=\left(v_{i},v_{j}\right)$, so $T$ respects the inner product structures. That is, $T$ is an isomorphism of inner product spaces and therefore any property of $V$ (as an inner product space) is implied by the corresponding property in $\mathbb{R}^n$. In particular it is Hilbert.
A: Well, how does proving that $V$ and $\mathbb{R}^n$ are topologically equivalent work? You conctruct an orthonormal basis and map it to $e_1,e_2,...,e_n$ right?
Then you extend linearly and prove that this map is a continuous linear bijection. But it is even more: It's an Isometry, and thus preserves completeness, as is easily checked.
A: The step is incomplete, but the gap isn't hard to patch. Let $V$ be a finite-dimensional real vector space equipped with a Hausdorff topology such that addition and scalar multiplication are continuous, and let $e_1, ... e_n$ a basis of it. Consider the function
$$f : \mathbb{R}^n \ni (x_1, ... x_n) \mapsto x_1 e_1 + ... + x_n e_n \in V.$$
By assumption, $f$ is continuous. The image of the closed ball $B_r$ of radius $r$ about the origin is Hausdorff by assumption, so $f$ restricted to $B_r$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Hence $f$ is a homeomorphism.
If $V$ is a finite-dimensional inner product space, then run the above argument with $e_1, ... e_n$ an orthonormal basis. Then $f$ is an isometry. 
More generally, it's true that any two norms on a finite-dimensional real or complex vector space are equivalent (so if one is complete, they all are). 
