I wish to prove that the class$$\mathcal{V} = \big\{(V, +, \cdot) : (V, +, \cdot) \text{ is a vector space over } \mathbb{R}\big\}$$ is not a set by using Cantor's diagonal argument directly.

Assume that $\mathcal{V}$ is a set. Then the collection of all possible vectors $\bigcup \mathcal{V}$ is also a set.

Let $f : \mathcal{V} \to \bigcup \mathcal{V}$ be an injection, it exists because the free vector space $F(\mathcal{V})$ with the basis $\mathcal{V}$ contains all vector spaces in $\mathcal{V}$ as vectors.

Now consider the following table where the rows $V_1, V_2, \ldots$ are elements of $\mathcal{V}$ and the element at the position $(V, f(W))$ is $1$ if and only if $f(W) \in V$, otherwise it is $0$. For example:

$$\begin{array}{c|c|c} \in & f(V_1) & f(V_2) & f(V_3) & \cdots \\ \hline V_1 & 1 & 0 & 1 & \cdots\\ \hline V_2 & 0 & 1 & 1 & \cdots\\ \hline V_3 & 1 & 0 & 0 & \cdots\\ \hline \vdots & \vdots & \vdots & \vdots & \ddots\\ \end{array}$$

This is only an intuitive concept as $\mathcal{V}$ is uncountable, of course.

We wish to construct a vector space $X \in \mathcal{V}$ such that for all $V \in \mathcal{V}$ we have $f(V) \in X$ if and only if $f(V) \notin V$, therefore making $X$ different from all $V \in \mathcal{V}$ on the diagonal of the table.

Consider the set $$S = \{f(V) : V \in \mathcal{V}, f(V) \notin V\}$$

and let $X = F(S)$ be the free vector space with basis $S$.

Then clearly $X \in \mathcal{V}$ and if $f(V) \notin V$ then $f(V) \in S \subseteq X$.

However, if $f(V) \in V$ then we know that $f(V) \notin S$, but is it possible that still $f(V) \in X$?

I also looked into defining a different vector space which contains $S$, namely we define the operations as something like $V + W := V \oplus W$ and $\alpha \cdot V := V$ for $\alpha \ne 0$ and $0\cdot V = \{0\}$, but it doesn't satisfy all the axioms (e.g. we would get $V = (1 + 1) V = V + V = V \oplus V$ which isn't true).

Is there a way to fix the proof?

  • $\begingroup$ If $V_1$ and $V_2$ have both the same underlying set, but are of different dimension (e.g. $\dim V_1=1$ and $\dim V_2=2$), which one is the free vector space? What if they have the same dimension but just a different realization of the addition (e.g. $V_1$ has $v$ as the $0$ vector and $V_2$ has a different one...) $\endgroup$
    – Asaf Karagila
    Oct 4, 2018 at 13:38
  • $\begingroup$ @AsafKaragila Both of them are free vector spaces over their respective bases, which are necessarily of cardinalities $1$ and $2$. I'm not sure I understand where you're getting at. $\endgroup$ Oct 4, 2018 at 13:46
  • $\begingroup$ Consider $(\Bbb R,+_1)$ and $(\Bbb R,+_2)$ such both give you a vector space over the standard (and fixed) field structure on $\Bbb R$. Both are sent to the same free vector space. And you can do that for any other dimension up to $2^{\aleph_0}$ (realize it as a structure on $\Bbb R$ itself as a set). Moreover, any permutation of $\Bbb R$ possibly defines yet another, non-identical (although isomorphic) vector structure on $\Bbb R$. And you're again hitting the same issue. $\endgroup$
    – Asaf Karagila
    Oct 4, 2018 at 13:47
  • $\begingroup$ @AsafKaragila Ok, lots of different vector spaces generate the same free vector space. However, I'm interested in cases where $x$ is not an element of a set $S$ but is an element of the free vector space with basis $S$. $\endgroup$ Oct 4, 2018 at 14:03
  • 1
    $\begingroup$ D'oh. I misread on the first read, and it just kind of stuck with me. Sorry. :) $\endgroup$
    – Asaf Karagila
    Oct 5, 2018 at 7:47

1 Answer 1


{ (R,+,×)^k : k cardinal number } is a collection
of vector spaces that cannot be a set because
{ k : k cardinal number } is not a set.

  • 1
    $\begingroup$ How does that help the OP? They explicitly said they want to use a diagonalization argument. $\endgroup$
    – Asaf Karagila
    Oct 5, 2018 at 0:03
  • $\begingroup$ Precisely, this doesn't answer my question. $\endgroup$ Oct 5, 2018 at 7:26

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