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$\textbf{Well Ordering Principle }:$ Every nonempty set $X$ can be well ordered.

$\textbf{The Axiom of Choice}:$ if $\{X_{\alpha}\}_{\alpha \in A}$ is a nonempty collection of nonempty sets, then $\prod_{\alpha \in A} X_{\alpha}$ is nonempty.

In the introduction of Real Analysis by Folland, it is written that: Let $X=\cup_{\alpha \in A}X_{\alpha}$. Pick a well ordering on $X$ and, for $\alpha \in A$, let $f(\alpha)$ be the minimal element of $X_{\alpha}$. Then $f \in \prod_{\alpha \in A} X_{\alpha}$.

My question is that why we need to choose the minimal element of $X_{\alpha}$. Since $X_{\alpha}$ is not empty so why we cannot say it contains an element like $p_{\alpha}$ and we define $f(\alpha)=p_{\alpha}$.

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  • $\begingroup$ Did you notice you didn't say what is this about? You only say what Folland says....but we don't know what for. $\endgroup$ Commented Jun 7, 2019 at 20:21
  • $\begingroup$ For proving the axiom of choice. $\endgroup$ Commented Jun 7, 2019 at 20:39
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    $\begingroup$ “Why we cannot say...?” because that inference uses the axiom of choice, which is the thing we’re trying to prove. $\endgroup$ Commented Jun 7, 2019 at 20:42

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We don't need to choose the minimal element. It's just a convenient way of ensuring a uniform way to choose an element from each set.

The axiom of choice is, essentially, the way you are allowed to move from "for every $\alpha$, $X_\alpha$ is non-empty" to the statement "choose $p_\alpha$ from each $X_\alpha$".

In other words, your argument is just appealing to the axiom of choice. But since we want to prove the axiom of choice, from the well-ordering principle, we need to start with the fact every set can be well-ordered, and prove that there is a choice function.

And the most convenient way of using a well-ordering, is to pick minimal elements from each non-empty subset of a fixed, well-ordered set. Which is how Folland comes to this proof of the axiom of choice, assuming the well-ordering principle.

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