$f: \mathcal{P}([0,1])$ to $[0,1]$ such that $f(A)>f(B)$ whenever $B$ is a proper subset of $A$ and $A\setminus B$ is uncountable

I am curious if it is possible for there to be a function $$f: \mathcal{P}([0,1])\rightarrow [0,1]$$ such that $$f(A)>f(B)$$ whenever $$A\supset B$$ and $$|A\setminus B|$$ is uncountable.

*I am aware of Zermelo's theorem that there is no $$f: \mathcal{P}([0,1])\rightarrow [0,1]$$ such that $$f(A)>f(B)$$ whenever $$B$$ is a proper subset of $$A$$. However, I am not seeing immediately how to adapt the proof to the present context. In particular, one natural starting point is to identify two sets whenever their symmetric differences are countable, i.e. $$A\sim B$$ if $$|A\mathbin\Delta B|\leq\aleph_0$$. Let $$X$$ be a set of representatives of equivalence classes under $$\sim$$. Then there is no such desired $$f$$ just in case there is no $$f: X\rightarrow [0,1]$$ such that, for any $$A, B$$ in $$X$$, $$f(A)>f(B)$$ whenever $$A$$ is a proper subset of $$B$$. This is very close to Zermelo's theorem, especially given that $$|X|=|\mathcal{P}([0,1])|$$ (since each equivalence class only has the cardinality of the continuum, as there is only $$\mathfrak{c}$$-many countable subsets of an uncountable set. But I don't know how to proceed from here...Any suggestion would be greatly appreciated.

• Which do you want: "whenever $A$ is a proper subset of $B$ and $B\setminus A$ is uncountable" like in your title, or "whenever $A\supset B$ and $|A\setminus B|$ is uncountable" like in your body? Commented Dec 25, 2023 at 22:53
• Sorry that was a typo on my part. It should be $f(A)<f(B)$ if $A$ is a proper subset of $B$.
– Y.Z.
Commented Dec 25, 2023 at 23:30
• Although consistency helps understanding, there's no ultimate difference between requiring $f ( A ) > f ( B )$ and requiring $f ( B ) > f ( A )$, since if $f$ satisfies one condition, then $1 - f$ satisfies the other. (But in the interests of the reader's understanding, I'm going to edit the question now to swap $A$ and $B$ in the title, in order to match their use in the question.) Commented Dec 25, 2023 at 23:59
• @TobyBartels Obviously, but I just wanted to warn the OP and let them correct their post themself. Commented Dec 26, 2023 at 1:01

Let $$I=[0,1]$$. Suppose there is a function $$f:\mathcal P(I)\to I$$ such that $$f(A)\gt f(B)$$ whenever $$A\subsetneqq B$$ and $$B\setminus A$$ is uncountable. Using a bijection between $$I$$ and $$I\times I$$, we can then define a function $$g:\mathcal P(I\times I)\to I$$ such that $$g(A)\gt g(B)$$ whenever $$A\subsetneqq B$$ and $$B\setminus A$$ is uncountable. Define $$h:\mathcal P(I)\to I$$ by setting $$h(X)=g(X\times I)$$; then $$h(A)\gt h(B)$$ whenever $$A\subsetneqq B$$, which is impossible.