0

I am currently writing a paper on the Axiom of Choice and Zorn's lemma. I'm finding a lot of information on the topic, but unfortunately I find a lot of it uses syntax and jargon that I've forgotten or simply never learned. I have a section of my paper on how Zorn's Lemma implies the Axiom of Choice. I'll include what I have below. I was hoping someone could fact check what I have in case I misunderstood some of my readings, as frankly I've used some informaion without fully understanding it. I understand the 1st paragraph, but my second paragraph is very confusing, specifically chosing the choice function and what it means. If someone could explain in plain english I would really appreciate it. Thank you!

As we are working with sets, an intuitive starting point is to let $X$ be a nonempty set. Then, we can consider the set $S=\{f:f\text{ is a choice function on } Y\subset\mathcal{P}(X)\}$. Remember that a choice function is simply a function that chooses one element out of each non-empty subset of a set. Also, note that the power set of $X$, $\mathcal{P}(X)$, is the set of all possible subsets of $X$. Therefore, this power set creates a perfect family of nonempty sets needed for the AC. Then, we can introduce a partial order on the set by defining $f_1\leq f_2$ if and only if $f_1\subset f_2$. Note this is a partial ordering as ``$\subset$'' is always a partial ordering in $\mathcal{P}(X)$.

Then we can see that $S$ is nonempty because for every $x\in X$ there is an obvious partial choice function on $\{x\}$. Specifically, let $f=\{(\{y\},y)\}$ be the choice function $f:\{\{y\}\}\to \{y\}$, with $f(\{y\})=y$. With this choice function, $f\in S$ for all $y\in Y$, thus explicitly showing $S$ is nonempty. If $C$ is a chain in this partially ordered subset, then we can define $\bar{Y}=\cup _{(Y,f)\in C}Y$ to be the union of the functions in $C$. Then $(\bar{Y},\bar{f})$ is an upper bound for C.

Suppose Zorn's Lemma is true. That is to assume, ``If $X$ is a partially ordered set such that every chain in $X$ has upper bound, then $X$ contains a maximal element.'' Then, by Zorn's Lemma there is some maximal element we will call $(f,Y)$. If $x\in X \backslash Y$, then we can extend from $Y$ to $Y\cup {x}$ by defining $f(S)$ to be equal to $S$ for any $S$ containing $x$. This contradicts maximality, and so $X \backslash Y = \null$, so $f$ is a choice function for $X$.

Robin McManus
  • 59
  • The domain of a choice function must not contain the empty set, i.e. it should be $P(X) \setminus \varnothing$ not $P(X)$. – Kenny Lau Nov 15 '21 at 16:09
  • 1
    This is not an expeditious way to fill in the gaps in your understanding. You ask Readers to review your write-up, and guess from that what it is you don't know. Does the paragraph you wrote beginning with "Then we can see that $S$ is nonempty..." require attention? It would be best to set up the problem (how is $S$ defined) and set as the goal to show $S$ is nonempty. Or something similar if this is already clear to you. – hardmath Nov 15 '21 at 16:14
  • "Remember that a choice function is simply a function that chooses one element out of each non-empty subset of a set." — this statement is already wrong, I'm sorry to say. (It is, rather, a function that chooses one element out of each member of a set of sets.) I can understand the interest in the topic — it's a great one! — but choosing to write a paper on something you don't understand well enough to write up seems like a poor idea. Is there a reason you chose this topic particularly? – Steven Stadnicki Nov 15 '21 at 16:18

0 Answers0