Downward Lowenheim-Skolem and the Theory of Infinite-Dimensional Vector Spaces

Question

Let $T$ be the theory of infinite dimensional vector spaces. It seems to me that by downward Lowenheim-Skolem we should be able to produce a countable model for this theory. But that means that there is a vector space which is countable with infinite dimension, which ought to be impossible. Could anyone explain where I have gone wrong?

Answer 1

A couple notes on "the theory of infinite-dimensional vector spaces":

• For the usual single-sorted, first-order treatment of vector spaces, scalar multiplication is represented as a family of function symbols, one for each element of the field. So there is not one theory of vector spaces, but rather a theory of vector spaces over $F,$ for each field $F.$
• Even if we fix a field $F$, there is not necessarily a such thing as the theory of infinite-dimensional vector spaces, since this is not necessarily an axiomatizable class. In fact this is the case if and only if $F$ is finite (in which case it coincides with the class of infinite vector spaces, which is axiomatizable).

Note that since there is a function symbol for each scalar, the size of the language is the same as the size of the field. So, e.g. the theory of real vector spaces has an uncountable language, so LS does not imply there are no countable models (and if we preclude the trivial vector space, there aren't).

As mentioned in the comments, we can also use two-sorted first-order logic to axiomatize vector spaces, in which case "the theory of infinite-dimensional vector spaces" makes sense.

All that said, the main issue was already addressed by Mark Saving in his answer: It's just wrong that an infinite-dimensional vector space can't be countable. Consider the direct sum of countably many copies of any finite field.

Answer 2

It depends on which field you are using. Let’s say you’re using a countable field $k$ (since otherwise even a 1-D vector space is uncountable).

In this case, a vector space with countably infinite dimension is countable. In particular, suppose $V$ is a vector space with basis $\{v_i \mid i \in \mathbb{N}\}$. Then we can easily show that $V = \bigcup\limits_{j \in \mathbb{N}} span(v_0, \ldots, v_j)$. Each vector space $span(v_0, \ldots, v_j)$ is finite-dimensional, hence countable. And the countable union of countable sets is countable.