I am reading the following proof (from page 124 here) which is a part of a larger proof for the statement:
Exercise 151: Show that there exist spaces $X$ and $Y$ such that, $t(C_p(X)) = \omega$ and $t(C_p(Y)) = \omega$ while $t(C_p(X) \times C_p(Y)) > \omega$.
I will state here the Claim ant it's proof. My problem is that I don't manage to understand what is the idea behind trying to prove this claim. More specifically, why, for example, it is so important that the projections on each coordinate will be injections? Is it attempt to somehow approximate the cardinality of powers of the Cantor set? For example like an outer and inner measure?
Here is the proof with the required definitions:
Definitions: $\mathbb K$ is the Cantor set. For any $n \in \mathbb N$, let $p_i^n: \mathbb K^n \rightarrow \mathbb K $ be the natural projection of $\mathbb K^n$ onto the $i$'th factor. Given a set $A \subset \mathbb K^n$, let $|A|_n = sup \{ |B| : B \subset A, p_i^n |_B $ is an injection for each $ i \leq n \}$. The cardinal $|A|_n$ will be called $n$-cardinality of $A$. If $x = (x_1,...,x_n) \in \mathbb K^n$, let $b(x) = \{ x_1,...,x_n \} \subset \mathbb K$.
Proposition (Fact 3 in the proof) Given $A \subset \mathbb K^n$, suppose that $n$-cardinality of $A$ is infinite. Then we have: 1. $|A|_n = min \{ |Y| : Y \subset \mathbb K$ and $A \subset \bigcup (p_i^n)^{-1}(Y) : i \leq n \}$ 2. $|A|_n = sup \{ |B| : B \subset \mathbb K$ and $b(x) \cap b(y) = \emptyset$ for any distinct $x,y \in B \}$
Proof: Denote the $n$ crdinality of $A$ by $k$, and let $k_i$ be the cardinal defined in the condition $(i)$ for $i \in \{ 1,2 \}$. It is clear that if $b(x) \cap b(y) = \emptyset$ for distinct $x,y \in B$, then, $p_i^n |_B$ is an injection for each $i \leq n$. This shows that $k_2 \leq k$. Now, fix a set $Y \subset \mathbb K$ of cardinality $k_1$ with $A \subset \bigcup \{ (p_i^n)^{-1}(Y) : i \leq n \}$. If $B \subset A$ is such that $p_i^n|_B$ is an injection for all $i \leq n$, then, for each $x \in B$, there is $i(x) \leq n$ with $q(x) = p_{i(x)}^n(x) \in Y$. It is easy to see that $q: B \rightarrow Y$ and $|q^{-1}(y)| \leq n$ for any $y \in Y$ and therefore $|B| \leq |Y|$. As a consequence, $k \leq k_1$.\ It is left to show that $k_1 \leq k_2$. So, take $Y \subset \mathbb K$ with $|Y| = k_1$ and $A \subset \bigcup \{ (p_i^n)^{-1}(Y) : i \leq n \}$. Choose a point $z_0 \in A$ arbitrarily. Suppose that $\alpha < k_1$ and we have chosen points $\{ z_\beta : \beta < \alpha \} \subset A$ so that $b(z_\beta \cap b(z_{\beta'}) = \emptyset$ if $\beta \neq \beta'$. The set $Y' = \bigcup \{ b(z_\beta) : \beta < \alpha \}$ has cardinality less then $k_1$ which implies $A \not \subset \bigcup \{ (p_i^n)^{-1}(Y') : i \leq n \}$. Take any $z_{\alpha} \in A \setminus \bigcup \{ (p_i^n)^{-1}(Y') : i \leq n \}$ and observe that $b(z_\alpha) \cap Y' = \emptyset$ and hence the family $\{b(z_\beta) : \beta \leq \alpha \}$ is disjoint. Thus, we can construct a set $\{ z_\alpha : \alpha < k_1 \} \subset A$ such that $\{ b(z_\beta) : \beta \leq k_1 \}$ is a disjoint family. This shows that $k_1 \leq k_2$ which finishes the proof.
Thank you!
The idea of the proof is to apply a Michael line construction on subsets of $\mathbb{K}$ (the Cantor set). The Michael line $\mathbb{M}$ (see this post and others on the same site) is the topology on $\mathbb{R}$ that we get by adding to the usual topology all subsets of the irrationals $\mathbb{P}$, i.e. we make the irrationals an open discrete subspace of the new topology. This construction was used to see that $\mathbb{M} \times \mathbb{P}$ is not normal in the product topology, which is then an example of a hereditarily paracompact space (so normal in a strong way) whose product with a separable (completely) metrisable space (namely $\mathbb{P}$) is not normal, and served as an example to guide theorems on normality of products.
But we can do this for other subsets $A$ of $\mathbb{R}$ as well: isolate $A$ to make a Michael line-like space and consider the product with $A$ (as a normal metric subspace). This was used with $A$ a Bernstein set etc. (special sets constructed by transfinite recursion). This example is a variation on this: the author uses $A \subset \mathbb{K}$, which is convenient as $\mathbb{K}$ is compact (and zero-dimensional) and has the property that is homeomorphic to each of its finite powers. The example he uses is just $X$ as the Michael line-like space for $A$ (but then on $\mathbb{K}$ instead of $\mathbb{R}$) and $Y = A$ as a subspace of $\mathbb{K}$.
In order to get the properties about the tightness of the corresponding function spaces, one needs that $X^n$ is Lindelöf for all $n$ and similarly for $Y$. As $Y$ is just separable metrisable, for $Y$ this is clear, but to get it for $X$ we need $A$ to be "special". He needs the property from fact 5: there are $A$ and $B$ disjoint in $\mathbb{K}$ such that for every $n$ and for every "$n$-uncountable" $F \subset \mathbb{K}^n$, both $A^n$ and $B^n$ intersect $F$. This is a sort of Bernstein set like property, as I see it.
Also he needs $X \times Y$ to be non-Lindelöf, which follows from the Michael line construction idea.
The set $A$ is constructed by transfinite recursion and in order to let everything go through, he also needs all sort of facts about cardinalities of subsets of $\mathbb{K}$ etc. But these (as the notion of $n$-uncountable, and $n$-cardinality) are all auxiliary to the above idea.
I would start reading the construction of $A$ and the proof of the fact that $X^n$ is Lindelöf for all $n$ to see where these special properties of $A$ are used exactly. This will give more motivation about why one would need all the other technical ideas from the beginning.