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 $I$?
First of all, you need to define $X$ such that $x_i \in D$ for some countable dense subset $D\subset \mathbb C$, or $|X| \geq |\mathbb R|$. More of less the same argument will show that $X$ is dense in $\ell^p(I)$.
To show that $X$ has the same cardinality as $I$, define $F : I \to X$ to be the map sending $i_0\in I$ to $(x_i)\in X$ which has the only nonzero entry 1 when $i=i_0$ (Let's say $1\in D$). This map is injective and thus $|I|\leq |X|$. On the other hand, let $X_n$ be the set of all sequence $(x_i)\in X$ with $n$ nonzero terms and define
$$X_n \to (I\times D)^n,\ \ (x_i)\mapsto \big( (i_1, x_{i_1}), \cdots, (i_n, x_{i_n})\big).$$
Then this map is injective and $|X_n|\leq |(I\times D)^n|= |I \times D| = |I|$ as $I$ is infinite ($|I|\geq |D|$). Thus $|X|\leq |I|$. Summing up we have $|I|=|X|$.