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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Dec 25, 2023 at 22:53
  • $\begingroup$ Sorry that was a typo on my part. It should be $f(A)<f(B)$ if $A$ is a proper subset of $B$. $\endgroup$
    – Y.Z.
    Commented Dec 25, 2023 at 23:30
  • $\begingroup$ 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.) $\endgroup$ Commented Dec 25, 2023 at 23:59
  • $\begingroup$ @TobyBartels Obviously, but I just wanted to warn the OP and let them correct their post themself. $\endgroup$ Commented Dec 26, 2023 at 1:01

1 Answer 1

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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.

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  • $\begingroup$ That's really brilliant! Thanks! $\endgroup$
    – Y.Z.
    Commented Dec 25, 2023 at 23:28

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