Proving the existence of an element $x$ in a set $X$ so that $\text{F}(X) \geq \text{f}(x)$ becomes an equality given some rules for f and F

Question

Let $X$ be an arbitrary set. I've defined some functions and rules as follows:

• $\text{F}()$ is a function from $\mathcal{P}(X)$ to $\mathbf{Ord}$.

• $\text{f}()$ is a function from $X$ to $\mathbf{Ord}$.

with $\mathbf{Ord}$ being a set of ordinal numbers.

I know this:

1. $\text{f}(a) = \min\{\text{F}(A) \mid A \in \mathcal{P}(X), a \in A\}$ for $a \in X$.

2. $\text{F}(A \cup \{a\}) = \max\{\text{F}(A), \text{f}(a)\}$ for $A \in \mathcal{P}(X)$ and $a \in X$.

3. $\text{F}(A \setminus \{a\}) \leq \text{F}(A)$ for $A \in \mathcal{P}(X)$ and $a \in X$.

4. $\text{F}(A) \geq \text{f}(a)$ for $A \in \mathcal{P}(X)$ and all $a \in A$.

I am trying to prove the following statement:

For all non-empty $A \in \mathcal{P}(X)$, there exists an $a \in A$ such that $\text{F}(A) = \text{f}(a)$.

Basically I want to make rule 4 an equality by finding a 'maximal' element in A.

I think I know how to do this if $A$ is countable with induction. I don't know if it is possible with any A. It might be a possiblity that this is not possible (even for the countable case).

If this is not provable using 1-4, which change to 1-4 would be the 'smallest' change that I can make to the rules (preferably rule 1 or rule 2) to make it provable? I don't think it is needed to add another rule (like a base case for the empty set) but I am unsure.

Answer 1

This is not possible in general if we only use your rules 1–4. For an uncountable set $X$, define $F \colon \mathcal{P}(X) \to \mathbf{Ord}$ by $$ F(A) = \begin{cases} 0 \quad &\text{if } A \text{ is countable}, \\ 1 \quad &\text{if } A \text{ is not countable,} \end{cases} $$ and define $f \colon X \to \mathbf{Ord}$ by $f(a) = 0$ for all $a \in X$.

Then for any uncountable set $A \in \mathcal P(X)$ there does not exist an $a \in X$ such that $F(A) = f(a)$.