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I have learned (in naive set theory) that intersection, union, difference, cartesian product, etc. are defined as operations on sets. But now when I'm studying abstract algebra, I came to know that an algebraic structure is actually, sets with operations defined on it. But then I thought that any arbitrary set has the operations like the ones mentioned above, defined on them. So is it possible that there exists an algebraic structure where the underlying set is the set of all sets while the operations defined on this structure are intersection, union, difference, cartesian product, etc? If this is true can you suggest me any source from where I can find further details about this concept?

Also when we define the operation (also functions and relations) we use the concept of the cartesian product, so this raises a question in my mind that how can we define operation using an operation itself? This process appears a bit cyclic to me.

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  • $\begingroup$ The set of all sets does not exist. For further reading, I would recommend Enderton's book on set theory (from personal experience). I'm sure that other set theorists on this website would be able to guide you better. $\endgroup$ – Karthik Kannan May 23 '20 at 15:40
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    $\begingroup$ You can consider a set $U$ whatever and you consider its power-set $\mathcal P(U)$. On it, consider the Algebra of sets mathematical structure. $\endgroup$ – Mauro ALLEGRANZA May 23 '20 at 15:45
  • $\begingroup$ There are some ways to bypass the Russel's paradox here(set of all set issue). One way is to extend the definition of algebraic structures to allow for Classes, or we can just play in a Universe. Intersection, union and negation is defined for a Boolean lattice. There is a nice bijective correspondence, up to isomorphism, between Boolean algebras and Boolean rings(commutative rings with $x^2=x$ for all $x$) $\endgroup$ – Ariana May 23 '20 at 16:47
  • $\begingroup$ how can we define operation using an operation itself --- For similar seemingly circular constructions, see Is there a such thing as an operator of operators in mathematics? AND Topologies on topologies. For an elementary example, one consider consider a matrix (or an array) as a list of lists, namely several horizontal lists of objects are arranged into a vertical list. $\endgroup$ – Dave L. Renfro May 23 '20 at 18:47
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There exists nothing like a set of sets. This is so because it violates the definition of a set. If A is the collection of all sets and we assume that A is a set itself, then A must belong the set as per the definition of the collection. But this is not possible as any set cannot be a member of itself. You can take Power set of any set U , given by P(U) which is the set of all subsets of U into consideration. Here you can define union, intersection and difference as binary operations from P(U)×P(U) to P(U). A binary operation, (binary means involving two operands) of this kind will pick two elements and perform an operation on them. You can refer Naive Set Theory by Paul Halmos

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As has been pointed out, there is no such thing as a set of all sets. In order to avoid Russel's paradox, it's usually best to start with something $S$ that is known to be a set and take this as the ambient space for your work, dealing only with subsets of $S$, or with very simple extensions of $S$ that are known to be non-problematic. For example, $S$ could be such that $x \in S$ iff $x$ is a sequence of real numbers.

Next, defining an operation in terms of an operation is not so cyclic. Addition of complex numbers is defined in terms of addition of real numbers. It would only be cyclic is we also tried to go the other direction and define addition of real numbers in terms of addition of complex numbers (which we don't do).

Third, you have to be careful about calling the Cartesian product an operation, an operation is a map $S \times S \to S$, so... what would $S$ be? Making an operation out of Cartesian product is tricky, though you could probably do it.

Finally, there are plenty of meaningful ways in which one can define a commutative ring of sets! I refer you to the bottom of page 335 of these notes: http://www.cip.ifi.lmu.de/~grinberg/t/19s/notes.pdf.

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