Context
The following theorem states a condition under which the Continuum Hypothesis fails:
- Theorem $\,$ If there is a measure$\dagger$ on $2^{\aleph_0}$, then the Continuum Hypothesis fails.
Proof Sketch $\,$ Let us assume that $2^{\aleph_0}=\aleph_1$ and that there is a measure $\mu$ on the set $S=\omega_1$. Let $I$ be the ideal of sets of measure $0$:$$I= \{ X \subseteq S \mid \mu (X) =0 \}.$$It can be easily shown that$$\text{there is no uncountable mutually disjoint collection } \mathcal{S} \subseteq \mathcal{P}(S) \\ \text{ such that } X \not \in I \text{ for all } X \in \mathcal{S}. \tag{*} \label{pro}$$Now let us construct an uncountable mutually disjoint collection of subsets of $S$ such that not all of them are in the ideal $I$, contradicting \ref{pro}.
For each $\xi < \omega_1$, there exists a function $f_{\xi}$ on $\omega$ such that $\xi \subseteq \operatorname{ran} f_{\xi}$. Let us choose one $f_{\xi}$ for each $\xi$, and let$$\boxed{{A_{\alpha n} = \{ \xi < \omega_1 \mid f_{\xi}(n)=\alpha \} \quad ( \alpha < \omega_1, \, n < \omega )}}.$$The sets $A_{\alpha n}$ have the following property:$$\text{For every }\alpha , \, S - \bigcup_{n=0}^{\infty}A_{\alpha n} \text{ is at most countable.} \tag{**}\label{pro2}$$Let $\alpha < \omega_1$ be fixed. By using \ref{pro2}, it can be easily concluded that not all the sets $A_{\alpha n}$, $n \in \omega$, are in the ideal $I$. Thus for each $\alpha < \omega_1$ there exists some $n_{\alpha} \in \mathbb{N}$ such that $A_{\alpha n_{\alpha}} \not \in I$, and so there must exist some $n$ such that the set $\{ \alpha \mid n_{\alpha}=n$ is uncountable. Hence $\mathcal{S} =\{ A_{\alpha n} \mid n_{\alpha}=n \}$ is an uncountable mutually disjoint collection of subsets of $S$ such that not all of them are in the ideal $I$.
Question
All steps of the proof are intuitive to me, except the construction of the sets $A_{\alpha n}$. I cannot understand how such a construction can come into one's mind from scratch. What background and idea leads a one to such a construction? Is there any intuition behind it? What were being thought when someone went to construct such sets?
I have the fish; please teach me how to fish.
P.S. Please note that my math knowledge is very low, so please answer this post in layman's terms.
Footnote
$\dagger$ Let $S$ be a nonempty set. A measure on $S$ is a function $\mu \colon \mathcal{P}(S) \to [0,1]$ such that
- $\mu (\varnothing )=0$, $\mu (S)=1$.
- If $X \subseteq Y$, then $\mu (X) \le \mu (Y)$.
- If $X$ and $Y$ are disjoint, then $\mu (X \cup Y)= \mu (X) + \mu (Y)$.
- $\mu (\{ a \})=0$ for every $a \in S$.
- If $\{ X_n \}_{n=0}^{\infty}$ is a collection of mutually disjoint subsets of $S$, then$$\mu \left (\bigcup_{n=0}^{\infty}X_n \right )= \sum_{n=0}^{\infty}\mu (X_n ).$$
A well-order on $\mathbb{N}$ is a relation $R\subseteq \mathbb{N}\times\mathbb{N}$ with some properties. So $$\operatorname{WO}(\mathbb{N})=\{R\in\mathcal{P}(\mathbb{N}\times\mathbb{N}):R\text{ is a well-order on }\mathbb{N}\}$$ is a set by an instance of the axiom of separation (the property "is a well-order on $\mathbb{N}$" is definable by a first-order formula with parameter $\mathbb{N}$, but I'm trying to be terse so I'll skip writing it out.) The following properties of well-orders are 'well known':
Now define a pre-order $\preceq$ on $\operatorname{WO}(\mathbb{N})$ by
This yields an equivalence relation $\alpha\equiv\beta\iff\alpha\preceq\beta\land\beta\preceq\alpha$ which is if and only if $\alpha,\beta$ are order-isomorphic. Denote $\mathscr{A}=\operatorname{WO}(\mathbb{N})\pmod\equiv$, and let $q:\operatorname{WO}(\mathbb{N})\to\mathscr{A}$ be the quotient map.
$\mathscr{A}$ is well-ordered by (the projection of) $\preceq$ and is order-isomorphic to $\omega_1\setminus\omega$ (ordered by $\in$). I'll skip the details of proving this.
The map $q$ is surjective so by the axiom of choice, it has a right inverse $p$ such that $q\circ p = \text{id}_{\mathscr{A}}$. This is just a fancy way of saying that $p(\alpha)$ is a representative of the equivalence class $\alpha\in\mathscr{A}$. I'll denote the representatives $\alpha^\prime=p(\alpha)$.
Now assume there is a measure $\mu$ on $\mathscr{A}$.
If $\alpha,\beta\in\mathscr{A}$ and $\alpha\prec\beta$ then $\alpha^\prime$ is order-isomorphic to a proper initial segment of $\beta^\prime$. There is then a unique $n=\sigma(\alpha,\beta)\in\mathbb{N}$ such that $n$ determines the proper initial segment of $\beta^\prime$ that $\alpha^\prime$ is order-isomorphic to. Another way of saying it is that $n$ is the $\beta^\prime$-least element of $\mathbb{N}$ that is a strict $\beta^\prime$-upper-bound to the proper initial segment of $\beta^\prime$ that $\alpha^\prime$ is order-isomorphic to.
Now, for each $\alpha\in\mathscr{A}$ we can partition $\mathscr{A}$ as follows $$\mathscr{A}=\{\zeta\in\mathscr{A}:\zeta\preceq\alpha\}\cup\bigcup_{n=0}^\infty\{\zeta\in\mathscr{A}:\alpha\prec\zeta\land \sigma(\alpha,\zeta)=n\}$$
Now, the set $\{\zeta\in\mathscr{A}:\zeta\preceq\alpha\}$ is countable (why?). It follows that one of the sets $A_{\alpha,n}=\{\zeta\in\mathscr{A}:\alpha\prec\zeta\land \sigma(\alpha,\zeta)=n\}$ must have $\mu(A_{\alpha,n})>0$ (does this look familiar?)
So now let $$R=\{\langle \alpha,n \rangle:\alpha\in\mathscr{A},n\in\mathbb{N},\mu(A_{\alpha,n})>0\}$$
We showed above that for each $\alpha\in\mathscr{A}$ there is some $n\in\mathbb{N}$ such that $\langle \alpha,n\rangle\in R$. But $\mathscr{A}$ is uncountable ($|\mathscr{A}|=|\omega_1\setminus\omega|=|\omega_1|$) and $\mathbb{N}$ is countable so by an infinite variant of the pigeonhole principle, there is some $n_0\in\mathbb{N}$ such that the set $X=\{\alpha\in\mathscr{A}:\langle\alpha,n_0\rangle\in R\}$ is uncountable. Then the sets $\{A_{\alpha,n_0}:\alpha\in X\}$ are an uncountable family of mutually disjoint subsets of $\mathscr{A}$ of measure $>0$, giving a contradiction.
I apologize for the verbosity, but the background material on well-orders needed some space and you asked for "layman's terms".