Is this true, if so is it obvious?

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SUMMARY

The discussion confirms that there is no proper class X such that every totally ordered set is isomorphic to a subclass of X. The terms "proper class" and "isomorphic" are utilized in the context of Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC). The class of all sets is mentioned as a subclass that includes every totally ordered set, but the participants agree that the class of all totally ordered sets is indeed a proper class.

PREREQUISITES
  • Understanding of Zermelo-Fraenkel set theory (ZFC)
  • Familiarity with the concepts of proper classes and subclasses
  • Knowledge of totally ordered sets and their properties
  • Basic grasp of isomorphism in set theory
NEXT STEPS
  • Research the implications of proper classes in Zermelo-Fraenkel set theory
  • Study the properties and examples of totally ordered sets
  • Explore the concept of isomorphism in mathematical structures
  • Investigate the relationship between classes and sets in set theory
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Mathematicians, logicians, and students of set theory who are exploring the foundational concepts of classes and orders in mathematics.

Dragonfall
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There is no proper class such X such that every totally ordered set is isomorphic to a subclass of X.

I'm using "proper class" and "isomorphic" rather liberally here, but you can assume them to be formulas in ZFC, or something.
 
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? What about the class of all sets? Every set is a subclass of that, so certainly every totally ordered set is a subclass of it... Also, if you want something stricter; since the class of all sets is a class, you can consider the class of all totally ordered sets. Is your question whether or not that is a proper class? I'd think that it would be, but the way you asked your question, it sounds like the class of all sets should satisfy your requirement.

Maybe I'm misunderstanding something here?
 
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No, you're right.
 

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