# Class of all vector spaces over $\mathbb{R}$ and Cantor's diagonalization

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?

• 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...) Oct 4, 2018 at 13:38
• @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. Oct 4, 2018 at 13:46
• 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. Oct 4, 2018 at 13:47
• @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 $S$. Oct 4, 2018 at 14:03
• D'oh. I misread on the first read, and it just kind of stuck with me. Sorry. :) Oct 5, 2018 at 7:47