Let $\Omega\neq \emptyset$ and let $\{X_\lambda\}_{\lambda \in \Omega}$ be a collection of sets. Define the Cartesian product of the sets $X_\lambda$ to be $\prod_{\lambda \in \Omega}X_\lambda := \{f : \Omega \to \cup_{\lambda \in \Omega}X_\lambda : f(\lambda ) \in X_\lambda \,\forall \lambda \in \Omega\}.$ Prove that the following $3$ statements are equivalent:
If $\Omega\neq \emptyset$ and for each $\lambda \in \Omega, X_{\lambda}\neq \emptyset,$ then $\prod_{\lambda \in \Omega}X_\lambda \neq \emptyset$.
Suppose $\Omega \neq \emptyset$ and that
- for all $\lambda \in \Omega, X_\lambda \neq \emptyset$, and
- $X_\lambda \cap X_\alpha = \emptyset$ if $\lambda \neq \alpha \in \Omega.$
Then $\prod_{\lambda \in \Omega}X_\lambda \neq \emptyset$.
Given a nonempty set $X,$ there exists a function $f : \mathcal{P}(X)\backslash \{\emptyset\}\to X$ so that $f(A)\in A$ for all $A \in \mathcal{P}(X)\backslash \{\emptyset\},$ where $\mathcal{P}(X)$ is the power set of $X$.
The above are all versions of the axiom of choice. I'm thinking of showing that $(1)\Leftrightarrow (3)$ and $(1)\Leftrightarrow (2).$ We first show $(1)\Rightarrow (3).$ Suppose $X$ is a nonempty set. Consider $\Omega := \mathcal{P}(X)\backslash \{\emptyset\}\neq \emptyset$ as $X\neq \emptyset.$ Then for each $\lambda \in \Omega,$ define $X_\lambda := \lambda.$ Then $\prod_{\lambda \in \Omega}X_\lambda \neq \emptyset$ by $(1)$. Let $f \in \prod_{\lambda \in \Omega}X_\lambda.$ Then observe that $f : \Omega \to X.$ Also, $f(A)\in A$ for each $A\in \Omega.$ But how can one show $(3)\Rightarrow (1)$? I'm thinking of constructing some sort of power set, but all I have is a nonempty set $\Omega$ and that for each $\lambda \in \Omega, X_{\lambda}\neq \emptyset.$
$(1)\Rightarrow(2)$ is obvious. But $(2)\Rightarrow (1)$ requires me to explicitly construct disjoint $X_\lambda$'s. If $\Omega$ is uncountable, would this result in problems? How could I work around them?