Let $\{(X_\alpha,\mathscr T_{\alpha}):\alpha \in \Lambda\}$ be an indexed family of Hausdorff spaces such that each $X_\alpha$ has atleast two points, and let $X=\Pi_{\alpha \in \Lambda} X_{\alpha}$. If $|\Lambda|\leq |\mathbb R|$ and $(X_\alpha,\mathscr T_\alpha)$ is seperable for each $\alpha \in \Lambda$.Then $(X, \mathscr T)$ is separable.
In the proof, If $|\Lambda|\leq |\mathbb R|$ and $(X_\alpha,\mathscr T_\alpha)$ is seperable for each $\alpha \in \Lambda$. For each $\alpha \in \Lambda$, Let $A_\alpha=\{a_{\alpha i}:i\in \mathbb N\}$ be countable dense subset of $X_\alpha.$ Since $|\Lambda|\leq |I|$ . We may assume without loss of generality that $\Lambda \subset I$. For each finite sequence of $I_1,I_2,..., I_k$ of distinct closed interval with rational end points and each finite sequence $n_1,n_2,...n_k$ of natural numbers, define a member $f$($f$ depends upon $I_1,I_2,..., I_k$ and $n_1,n_2,...n_k$) of $X$ as follows:
Let $\alpha \in \Lambda$. If there is an $i$($i=1,2,3,4...,k)$ such that $\alpha \in I_i$, then the value of $f$ at $\alpha$ is $a_{\alpha n_i}.$ If $\alpha \notin I_i$ for $i=1,2,3,4...,k$, then the value at $\alpha$ is $a_{\alpha 1}.$
The subset $A$ of $X$ consisting of al such $f$ is countable. Then also I have to prove $A$ is dense in $X$
Doubt:-
This is the proof given in Foundation of Topology by C. Wayne Patty. I don't understand what are $I_1,I_2,...I_k$, How it is related to $I$? How do I prove that $A$ is countable? I am not able to find a bijective mapping from $A$ to $\mathbb N$. Can you help me to prove $A$ is countable. If I understand what are $I_1,..I_k$. I can read and understand the proof of $A$ is dense in X.