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A set is collection of distinct objects: https://en.wikipedia.org/wiki/Set_(mathematics).

The word distinct implies the identity relation on each set: an element of a set is equal to itself, or $a = a$
(Does the word 'distinct' in the definition of Set implies an equivalence relation between the objects of the collection?).

Thus, every set by the definition is a relational structure with the identity relation.

The identity relation is a binary relation $(a, a)$ that implies the unary identity operation $f(a) = a$
(https://math.stackexchange.com/a/3719648/427611).

Thus, every set by the definition is an algebraic structure with the identity operation.

But Wikipedia states that a set is "a degenerate algebraic structure S having no operations"
(https://en.wikipedia.org/wiki/Algebraic_structure#One_set_with_operations).

Which statement is correct?
If a set is an algebraic structure, can we apply algebraic terms in set theory and vice versa?
E.g., can we call equinumerous sets isomorphic?

Update

Based on the discussion:

  1. The concept of an identity https://en.wikipedia.org/wiki/Identity_(philosophy) is more fundamental, then the concept of a set.

  2. Without the concept of identity it is not possible to introduce a set, since if an element $a$ of a set $A$ is not equal to itself than the statements "$a$ is an element of $A$" and "$a$ is not an element of $A$" are true at the same time.

  3. Therefore, no matter how exactly we define a set, it has the identity relation.

  4. The binary identity relation "an element is equal to itself" is the unary identity operation "an element is paired with itself".

  5. Therefore, no matter how exactly we define a set, it is an algebraic structure with the identity operation.

Would it be correct? And many thanks to all who responded.

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  • $\begingroup$ "no binary operations". The identity relation is not an operation, and $f(a) = a$ is not a binary operation. $\endgroup$
    – xxxxxxxxx
    Jun 26, 2020 at 20:18
  • $\begingroup$ @MorganRodgers No. Two rows below they say: "Unary system: S and a single unary operation over S." Thus, they do not consider a set as a unary system. $\endgroup$
    – Alex C
    Jun 26, 2020 at 20:20
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    $\begingroup$ It is true that "isomorphisms in the category of sets" are exactly bijections. But this is very dangerous to use this term. My experience with students is that they don't understand yet that there is an additional context in the word "isomorphism", where the structure is usually implicit. So "bijection" is better since it is clearer what structure is required to be preserved: none. $\endgroup$
    – Asaf Karagila
    Jun 26, 2020 at 20:34
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    $\begingroup$ @AlexC They are probably just distinguishing a "Unary system" as having a nontrivial unary operation. This is why they say that if a set is considered as an algebraic object, it is "degenerate" in a sense. This is not a critical distinction. Also notice that "unary system" does not have a wikipedia link; that is because this is also not especially interesting as an algebraic structure. $\endgroup$
    – xxxxxxxxx
    Jun 27, 2020 at 6:11

2 Answers 2

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I'm going to answer in terms of ZF set theory since that is what most of us need.

In ZF, there is no definition for a set. It is a primitive idea. All you have is the notion set and membership and that gives you equality of sets, and in turn equality of members.

Whether or not two elements of a set are equal is again a matter of set equality. You don't need anything extra to tell you when two elements are equal.

It certainly can't be part of the definition of a set, because an equivalence relation is essentially a special subset of $X\times X$, and if you haven't accepted what a set is yet, you shouldn't be discussing thing like subsets of $X\times X$. You will just be going around in circles.

You can grant $X$ an equivalence relation given by the partition of the set into singletons, so that you get the "identity relation", but it doesn't tell you anything new.

E.g., can we call equinumerous sets isomorphic?

Sure you can. In the category of sets, the isomorphisms are precisely the bijections. There are "isomorphic in the category of sets."

You don't need operations to do this... a category can be made up of non-algebraic objects. That is isomorphism and homomorphism are not algebraic-only concepts.

Thus, every set by the definition is an algebraic structure with the identity operation.

It would be more plausible to say that a set is an algebraic structure with no operations. I don't know if universal algebra accepts this empty case, but they may.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Jun 28, 2020 at 17:19
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The Wikipedia article only gives the degenerate algebraic structure as an example of an algebraic structure with no binary operation, which can be defined on an arbitrary set. Your identity unary operation is another example of an algebraic structure without a binary operation that can be put on any set, that is technically not degenerate, but honestly almost is.

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  • $\begingroup$ Two rows below they say: "Unary system: S and a single unary operation over S." Thus, they do not consider a set as a unary system. $\endgroup$
    – Alex C
    Jun 26, 2020 at 20:22
  • $\begingroup$ @AlexC A set is a set. Without defining a unary operation, it has no unary operation. But you defined one, and if you pair the set with that unary operation you have a unary algebraic structure. While it can be defined on any set, one couldn't say that it's "part of" the set. It's an addition. $\endgroup$ Jun 26, 2020 at 20:23
  • $\begingroup$ The definition of a set defines the unary operation, not me. $\endgroup$
    – Alex C
    Jun 26, 2020 at 20:25
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    $\begingroup$ @AlexC As rschwieb says, there is no "definition" of a set. A set is an undefined thing. You defined the unary operation, which is itself a set. It can't be part of the definition of a set, especially since there's no definition. $\endgroup$ Jun 26, 2020 at 20:27

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