What are $I_1,I_2,...I_k$? How it is related to $I$?

Question

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.

Answer 1

As stated, $I_1,\ldots, I_k$ are supposed to be pairwise disjoint intervals with rational endpoints. The confusion probably arises from the unmotivated introduction of $I$ (supposedly the standard interval $I=[0,1]$). Either we should work with $\Bbb R$ instead of $I$ (and the $I_j$ are rational-ended intervals $\subset \Bbb R$) or we should additionally specify that the $I_j$ are $\subset I$. Then again, the argument should still work if $\Lambda\subseteq I$ "instead of" $\Lambda\subseteq \Bbb R$, as that is merely a strengthening.

$A$ is countable, because each element is determined by a finite sequence of rational numbers (the $2k$ end points of the $k$ intervals). The set of finite seqeunces of a countable set is countable (it is the countable union of the countable sets of length 1 sequences, length 2 sequences, length 3 sequences, ...)

Answer 2

Because the index set as such does not matter (just its size), and we know it is of size at most continuum, we can assume WLOG that the spaces are in fact indexed by points of $\mathbb{R}$ (or $I =[0,1]$ if you prefer).

Both these spaces (and their subsets) have, as spaces a "countable description" : a base of intervals with rational endpoints. We thus can consider the set of all finite disjoint families of those intervals (so in fact subsets of the index set) and note that this is still a countable set: it's a countable union of sets in bijection with $\mathbb{N}^n$ over different finite values of $n$. We pick a value $x_i= q_j$ for each $i \in I_j$, where $I_1,\ldots, I_n$ is such a disjoint finite sequence from the countable base, and some fixed value for all $i$ not belonging to any $I_j$. So a point in this set (such a point is a function on $I$) is defined as a function with only finitely many values, one for each $I_j$ and one outside. As for each $I_j$ we only have countably many choices (we pick the points in the countable dense subsets in the $X_i$) we only have countably many such function for each of the countably many choices of interval sequences. So we get a countable set of points of $X$ which then has to be shown dense.

I do this proof in a specific case in my post here (powers of discrete spaces), where I also show that the minimal size of a base for the countable dense subspace of $\{0,1\}^\mathbb{R}$ is also $\mathfrak{c} = |\mathbb{R}|$, showing that a countable completely regular space can be very non-metrisable.