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A set which transforms via converse functions is called antiset. Antisets usually arise in the context of Chu spaces. I couldn't understand the notion of antiset and its examples.

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    $\begingroup$ Your links are almost absolutely useless. The Chu-Space Wikipedia article contains a definition of Chu Spaces, but does not mention antisets. Where dd you find examples for antisets? $\endgroup$
    – miracle173
    Apr 17, 2021 at 6:49
  • $\begingroup$ @miracle173 You are right, that link really doesn't provide any idea of antiset and neither I could find a concrete example of antiset from elsewhere. It seems that the notion of antiset is a mystery. One can also read a post on the rise of antiset here google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$
    – gete
    Apr 17, 2021 at 7:29
  • $\begingroup$ @gete The meaning of "antiset" in the blog post you linked to in your last comment is totally unrelated to the meaning of "antiset" described in my answer. It's worth noting that the blog post is fictional (the "little-known logicians named Narl Cowman and Bishi Ranger" do not exist, and the "radical new form of set-theory" described in the blog has not, in fact, "come to be known across math departments across the world as Anti-Set Theory"). $\endgroup$ Apr 17, 2021 at 16:04

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The definition of "antiset" given on the Wolfram MathWorld page you linked to (and also on Wiktionary!) is:

A set which transforms via converse functions.

This "definition" seems meaningless to me without more context. What is a "converse function"? What does it mean for a set to "transform via converse functions"? Is transforming via converse functions a property that a set may or may not have?

Digging a bit deeper, it seems that Vaughan Pratt (who is a leading proponent of Chu spaces, see here) uses the word "antiset" to mean an object of the category $\mathsf{Set}^{\text{op}}$ (the opposite category of the category of sets). See the second full paragraph on p. 3 of his paper Chu Spaces: Automata with quantum aspects.

Since the categories $\mathsf{Set}$ and $\mathsf{Set}^{\text{op}}$ have exactly the same objects, an antiset is just a set! The distinction between sets and antisets comes down to the morphisms between them. A morphism of antisets $X\to Y$ is just an ordinary function $Y\to X$. Now it's somewhat more clear what is meant by "transforming via converse functions"...

The category $\mathsf{Set}^{\text{op}}$ is well-known to be equivalent to the category $\mathsf{CABA}$ of complete atomic Boolean algebras (the equivalence is given by mapping a set to its powerset algebra and mapping a complete atomic Boolean algebra to its set of atoms). So you can also think of an antiset as a complete atomic Boolean algebra $B$ and a morphism of antisets $B\to C$ as a complete Boolean algebra homomorphism $B\to C$.

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    $\begingroup$ Is it sensible to view "antiset" morphisms $X\rightarrow Y$ as maps from subsets of $X$ to their pre-image in $Y$ under the functions $Y\rightarrow X$? Or is it best to just see them as the original, ordinary function? $\endgroup$
    – Sam Jaques
    Apr 17, 2021 at 15:41
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    $\begingroup$ @SamJaques I suppose there are (at least) four equivalent ways to think about a morphism $X\to Y$ in $\mathsf{Set}^{\text{op}}$. (1) An ordinary function $Y\to X$. (2) A relation $R\subseteq X\times Y$ which satisfies the dual property to the definition of a function: For all $y\in Y$, there is a unique $x\in X$ such that $(x,y)\in R$. (3) A complete Boolean algebra homomorphism $\mathcal{P}(X)\to \mathcal{P}(Y)$. Note that since every subset of $X$ is a union of singletons (atoms in $\mathcal{P}(X)$) and a complete Boolean algebra homomorphism preserves unions, ... $\endgroup$ Apr 17, 2021 at 15:53
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    $\begingroup$ ... (3) can be equivalently described by: (4) A function $X\to \mathcal{P}(Y)$ such that distinct points in $X$ map to disjoint subsets of $Y$ and these subsets cover $Y$ (i.e., a partition of $Y$, with empty pieces allowed, indexed by the elements of $x$). Since these perspectives are all equivalent, which to prefer depends on which one seems psychologically and notationally most convenient for your goals. $\endgroup$ Apr 17, 2021 at 15:56

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