Housekeeping with the maximal theorem.

src/fa/norm/normed.tex

@@ -148,3 +148,16 @@
     Thus $d(x_N, \text{span}\bracsn{x_n|1 \le n < N}) = \norm{x_N + F_N}_{E/F} > \alpha_N$.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:finite-dim-norm}
+    Let $K \in \RC$ and $d \in \natp$, then every norm on $K^d$ is equivalent to each other. 
+\end{proposition}
+\begin{proof}
+    Let $[\cdot]: K^d \to [0, \infty)$ be an arbitrary and
+    \[
+    \norm{\cdot}_{1}: K^d \to [0, \infty) \quad x \mapsto \sum_{j = 1}^n |x_j|
+    \]
+
+    then the identity $(K^d, \norm{\cdot}_1) \to (K^d, [\cdot])$ is continuous. Let $S = \bracsn{x \in K^d|\ \norm{x}_1 = 1}$, then $S$ is compact with respect to $\norm{\cdot}_1$ and hence compact with respect to $[\cdot]$. By continuity of $[\cdot]$, $\min_{x \in S}[x]$ exists and is non-zero. 
+\end{proof}
+