$I$ and $J$ have the same cardinality if $\ell^p(I)$ and $\ell^p(J)$ are isometrically isomorphic I want to prove that if $I,J$ are infinite sets and $\ell^p(I)$ and $\ell^p(J)$ are isometrically isomorphic ($1\leq p<\infty$), then $|I|=|J|$. I could prove that if $|I|=|J|$, then $\ell^p(I)$ and $\ell^p(J)$ are isometrically isomorphic. Also I know that $\ell^p(I)$ and $\ell^p(J)$ have dense subsets $D_1,D_2$ respectively such that $|D_1|=|I|$ and $|D_2|=|J|$. But can I say from here that $|D_1|=|D_2|$? I am not finding any way out. Please suggest.
This is a problem from Conway's Functional Analysis.
 A: If there is a bijection $\theta\colon I\to J$ the formula $Te^I_i = e^J_{\theta(i)}$ defines an isometry from $\ell_p(I)$ onto $\ell_p(J)$, where $(e^I_i)_{i\in I}$ and $(e^J_j)_{j\in J}$ denote the standard unit vector bases of $\ell_p(I)$ and $\ell_p(J)$, respectively. 
When $I$ and $J$ have different cardinalities the spaces are not even homeomorphic. Without loss of generality we may assume that these sets are infinite. The density of a metric space is the smallest cardinality of a dense set in this space. It is enough to show that the density of $\ell_p(I)$ is the same as cardinality of $I$. This will be enough as density is a topological invariant.
Certainly, density of $\ell_p(I)$ is at least $|I|$ because the standard unit vector basis $(e_i)_{i\in I}$ of $\ell_p(I)$ is a closed and discrete subset of cardinality $I$. As every element of $\ell_p(I)$ may be written as $\sum_{k=1}^\infty a_k e_{i_k}$ for some $(i_k)_{k=1}^\infty$ in $I$, the set $V$ of all finite rational linear combinations of $(e_i)_{i\in I}$ is dense in $\ell_p(I)$. On the other hand, this set has cardinality $\aleph_0\cdot |I| = |I|$. We have thus proved that indeed the density of $\ell_p(I)$ is $|I|$.
