I am currently studying the proof of the following theorem:
Every finite dimensional normed vector space is complete, and every finite dimensional subspace of a normed vector space is closed.
The proof begins with an isometric isomorphism between the finite dimensional vector space $X$ with $K^n$ (as always $K=\mathbb{R},\mathbb{C}$).
It is clear that we have an isomorphism $T: X\to K^n$. But I do not see why we also get an isometry.
I think we can get a norm $\|\cdot\|_1$ on $K^n$ by constructing it with regards to the norm $\|\cdot\|$ on $X$ like this:
When $x_1,\dotso, x_n$ is a basis of $X$, and $e_1,\dotso, e_n$ is a basis of $K^n$, we get an isomorphism by mapping basis elements to basis elements.
If we then construct the norm $\|\cdot\|_1: K^n\to [0,\infty)$, by letting $\|e_i\|_1=\|x_i\|$ if I see it correctly this actually yields a norm, which is in some sense "induced" by the norm of $X$.
Can you confirm this?
Thanks in advance.