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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)

Andrés E. Caicedo
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Sphere
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2 Answers2

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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.

Jens Schwaiger
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  • 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? – Lee Mosher Jul 30 '20 at 14:42
  • I certainly think so. But writing on a tablet is top cumbersome to write detailed proofs. . – Jens Schwaiger Jul 30 '20 at 15:53
  • You could say a vector space $W$ satisfies $()$ if it is of the form $W=V\times {A}$ where $V$ is a vector space and the operations of W are inherited from $V$. If $W$ satisfies $()$ we define $W_$ as the set such that $W=V \times {W_}$ for some vector space $V$.

    Suppose now that there exists $\mathcal{V}$ the set of all vector spaces and define the property $$P(x,y) \equiv (x \text{ is not a v.s. satisfing () } \wedge y=\emptyset) \vee (x \text{ is a v.s. satisfing () } \wedge y=x_*)$$

     – Eparoh May 14 '21 at 09:02
  • We have then that for every set $x$ there exists a unique set $y$ such that $P(x,y)$ is true, and by the axiom of replacement, there exists a set $B$ such that for every $x \in \mathcal{V}$, there is $y \in B$ for which $P(x, y)$ holds.

    Now the contradiction is that $B$ contains every set as its element, brcause given a set $A$, we define $W=V \times {A}$ a vector space satisfying $(*)$ and by definition of $P$ and $B$, we conclude that $A \in B$.

     – Eparoh May 14 '21 at 09:03
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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.

Daniel Teixeira
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  • But I don't know about module theory or abstract algebra, so I can't understand this proof. – Sphere Jul 30 '20 at 14:17
  • https://planetmath.org/freevectorspaceoveraset – Daniel Teixeira Jul 30 '20 at 14:19
  • The point is that there are more vector spaces than sets, since each set defines at least one vector space. – Daniel Teixeira Jul 30 '20 at 14:20
  • 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? – Lee Mosher Jul 30 '20 at 14:40