Let $I$ be an infinite set and $1\leq p<\infty$. Show that $\ell^p(I)$ has a dense set of the same cardinality as $I$.
For this I put
$$X=\{(x_i); x_i\in \Bbb C , x_i=0 \text{ for all but for finitely many } i\in I\}.$$
For every $(\alpha_i) \in \ell^p(I)$ we have $\sum_{i\in I}|\alpha_i |^p < \infty$ . So, for every $\epsilon >0$ , $\sum_{i\in I-\{i_1,...,i_n\}}|\alpha_i |^p < \epsilon$ then $x=(x_i)$ where $x_{i_j}=\alpha_{i_j}$ for $j=1,...,n$ belongs to $X$ and $\|\alpha - x\| < \epsilon$, thus $X $ is dense in $\ell^p(I)$ . Am I right? How can I show that the cardinality $X$ is the same $I$?