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How can I prove that the set of all vector spaces doesn't exist? (In other words, if I gather all vector spaces, then it cannot be a set)

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    $\begingroup$ Do you know the proof that the set of all sets doesn't exist? $\endgroup$ – Lee Mosher Jul 30 '20 at 14:07
  • $\begingroup$ Yes I do know briefly. $\endgroup$ – Sphere Jul 30 '20 at 14:17
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Take some fixed vector space $V$. Then for any set $A$ the set $V\times \{A\}$ is also a vector space in an obvious manner. Since the class of all sets is a proper class, and not a set, the collection of all vector space $V\times\{A\}$ is also a proper class. Thus the collection of all vector space which is a super class of the class above is also a proper class.

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  • $\begingroup$ Does one formally need an argument by contradiction to fill out this proof? Something that uses the axiom of replacement together with one or two definable formulas, to conclude that if the set of vector spaces exists then the set of sets exists? $\endgroup$ – Lee Mosher Jul 30 '20 at 14:42
  • $\begingroup$ I certainly think so. But writing on a tablet is top cumbersome to write detailed proofs. . $\endgroup$ – Jens Schwaiger Jul 30 '20 at 15:53
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You can see that the set of all sets don't exist in this answer. On the other hand, every set $S$ defines a vector space, namely the free vector space on $S$. Then if there existed a set of all vector spaces, its elements would be in 1-1 correspondence with sets, i.e. it would define a set of all sets. Contradiction.

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  • $\begingroup$ But I don't know about module theory or abstract algebra, so I can't understand this proof. $\endgroup$ – Sphere Jul 30 '20 at 14:17
  • $\begingroup$ planetmath.org/freevectorspaceoveraset $\endgroup$ – Daniel Plácido Jul 30 '20 at 14:19
  • $\begingroup$ The point is that there are more vector spaces than sets, since each set defines at least one vector space. $\endgroup$ – Daniel Plácido Jul 30 '20 at 14:20
  • $\begingroup$ Does one formally need an argument by contradiction to fill out this proof? Using the formula for this 1--1 correspondence together with the axiom of replacement, to conclude that if the set of vector spaces exists then the set of sets exists? $\endgroup$ – Lee Mosher Jul 30 '20 at 14:40

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