Are normed spaces isodyne?

Question

In general, do all non-empty open subsets of a normed space necessarily have the same cardinality?

Answer 1

Let $\phi:B(0,1) \to X$ be given by $\phi(x) = \begin{cases} \tan (\|x\| { \pi \over 2 }){x \over \|x\|}, & x \neq 0 \\ 0, & x=0 \end{cases}$. We see that $\phi^{-1} (x) = \begin{cases} {2 \over \pi}\arctan (\|x\| ){x \over \|x\|}, & x \neq 0 \\ 0, & x=0 \end{cases}$, hence $\phi$ is a bijection.

Hence $\operatorname{card} B(0,1) = \operatorname{card} X$.

If $r>0$, the map $x \mapsto x_0+rx$ is invertible, hence $\operatorname{card} B(x_0,r) = \operatorname{card} B(0,1)$.

If $U$ is open, we have $B(x_0,r) \subset U$ for some $x_0,r$, and so $\operatorname{card} X = \operatorname{card} B(x_0,r) \le \operatorname{card} U \le \operatorname{card} X$.

It follows that all open sets have the same cardinality as $X$.