Find the cardinality of the set of all norms on R^n (hint: show that every norm || || : R n → R is continuous).

Suppose $|| \cdot ||$ is a norm on $\mathbb{R}^n$ and defined the function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ by $f(x)=||x||.$ We prove that $f$ is continuous. Let $\{x_n\}$ be a sequence in $\mathbb{R}^n$ such that $\lim_{n \rightarrow \infty} x_n=x$, i.e. $||x_n-x|| \rightarrow 0$ as $n\rightarrow \infty$.  To prove that $f$ is continuous it is enough to show that
\[\lim_{n \rightarrow \infty} f(x_n)=f(x).\]
Using the revered triangle inequality we have
\[||f(x_n)-f(x)||=||  || x_n|| -||x||  || \leq || x_n -x || \rightarrow 0.\]
Hence 
\[||f(x_n)-f(x)|| \rightarrow 0,\]
which means 
\[\lim_{n \rightarrow \infty}f(x_n)=f(x).\]
So the cardinality of the set of all norms is at most that of the set of continous functions from $\mathbb{R}^n \rightarrow \mathbb{R}.$

Next we show that the cardinality of continuous functions the same as the cardinality of the continuum $c=|\mathbb{R}|$.

The cardinality is at least that of the continuum because every real number corresponds to a constant function. The cardinality is at most that of the continuum because the set of real continuous functions injects into the set of functions from $Q^n \rightarrow R$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function. The Schroeder-Bernstein theorem (https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem) now implies the cardinality is precisely that of the continuum $c$.