Is this true, if so is it obvious?

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The discussion centers on the concept of whether there exists a proper class X such that every totally ordered set is isomorphic to a subclass of X. It is suggested that the class of all sets could satisfy this requirement, as every set, including totally ordered sets, is a subclass of it. However, the class of all totally ordered sets is proposed as a stricter alternative, leading to the question of whether it qualifies as a proper class. The participants agree that the initial assertion holds true, confirming the complexity of the topic. Ultimately, the conversation highlights the nuances of class theory within ZFC.
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.
 

Thread 'Deductive proof in logic formal systems'

I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
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