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Understood How Zorn's Lemma Implies the Axiom of Choice

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Many math books that I’ve read referred me to other books for the proof of the equivalence of the Axiom of Choice and Zorn’s Lemma. This afternoon, I spent more than two hours to understand the proof that the later implies and former in Topology written by Davis.

  1. Set up a non-empty partially ordered set $(\mathcal{P},\le)$
  2. Let $\mathcal{T}$ be any non-empty chain in $\mathcal{P}$.
  3. Prove that $\cup \mathcal{T} \in \mathcal{P}$.
  4. Apply Zorn’s Lemma to get a maximal element $g \in \mathcal{P}$.
  5. Use the maximality of $g$ to claim that the domain of $g$ equals the family of non-empty subsets $(S_i)_{i \in I}$ from which elements $(x_i)_{i \in I}$ are chosen.

In the book, $\le$ means function extension, and

\[ \mathcal{P} = \{f \mid f \text{ is a function, } dom\,f \subseteq (S_i)_{i \in I}, f(x) \in x \,\forall x \in dom\,f\}. \]

Step (3) is proved step-by-step according to the definition of $\mathcal{P}$. Usually, suppose that $(x_1,y_1),(x_2,y_2) \in f$, $x_1 = x_2 \implies y_1 = y_2$. The book uses the contrapositive form of this statement. I was stuck at the sentence Now, for a set to be an element of $dom\cup\mathcal{T}$, it must be an element of some member of $\mathcal{T}$. At first, I omitted the phrase "some member of", and stopped for half an hour. Reading the next sentence Hence $dom\cup\mathcal{T} \subseteq (S_i)_{i \in I}$, I knew how to interpret the sentence where I was stuck: if $S \in dom\cup\mathcal{T}$, $\exists f \in \mathcal{T}$ such that $S \in dom\,f$. Since $f \in \mathcal{T} \subseteq \mathcal{P}$, $dom\,f \subseteq (S_i)_{i \in I}$. Then $S \in (S_i)_{i \in I}$, and thus $dom\cup\mathcal{T} \subseteq (S_i)_{i \in I}$.