# Axiom of Choice implies the Well-Ordering Principle

I am trying to understand the proof of this implication we were taught in my set theory module. I cannot seem to tie it together with the final line of the argument... We used this lemma:

Given $$F\colon \mathcal{P}(X)\to X$$, there exists a unique well-ordered set $$(W,<)$$ such that

• $$W\subseteq X$$;
• for all $$x\in W$$, we have $$F(\{y\in W:y; and
• $$F(W)\in W$$.

The proof given is:

Let $$\sigma\colon\mathcal{P}(X)\to X$$ be a choice function (technically defined on $$\mathcal{P}(X)-\{\emptyset\}$$ but we may let $$\sigma(\emptyset)$$ be any member of $$X$$). Take $$F\colon\mathcal{P}(X)\to X$$ defined as $$F(A)=\sigma(X-A)$$ and apply the lemma.

How does the lemma yield the well ordering principle? I assume that we want $$W$$ to be $$X$$ since WO states that the well-ordering relation is on the set $$X$$ (for all $$X$$).

Yes, the point is that $W=X$.
The key to that is the last part of the conclusion of the lemma: $F(W)\in W$, which is the same as $\sigma(X\setminus W)\in W$. If $X\setminus W$ is nonempty, then this contradicts $\sigma$ being a choice function, so the only way we can have $\sigma(X\setminus W)\in W$ is if $X\setminus W=\varnothing$, in which case $\sigma(\varnothing)$ is just some member of $X$.
Since $W\subseteq X$ and $X\setminus W=\varnothing$, it must be that $W=X$.