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?
Downward Lowenheim-Skolem and the Theory of Infinite-Dimensional Vector Spaces
logicmodel-theory
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Here's a partial answer showing that your question consistently has a negative answer (and Andreas Lietz showed the consistency of the opposite):
Under $ZFC+V=L$, your theory does not have any countable models.
Basically, this comes down to the fact that $V=L$ gives us a very nicely definable well-ordering of the reals, and we can leverage that to pin down a specific bijection between $\omega$ and any fixed countable set.
(EDIT: This isn't actually true - as Asaf Karagila commented, you've omitted the axiom of infinity and so $V_\omega$ satisfies your theory. I'm assuming you meant to include it, though. I'm also assuming that by "Powerset" you mean first-order powerset, that is "$\forall x\exists y\forall z(z\in y\leftrightarrow z\subseteq x)$" rather than "$\forall x\exists y\forall A(A\subseteq x\leftrightarrow\exists z(z\in y\wedge z=A))$," admitting the appropriate abbreviations, since second-order powerset trivially requires uncountability.)
First, note that any model of your theory is transitive (this is a good exercise). Now suppose $M$ were a countable model of your theory. Note that we can quantify over countable levels of $L$ in second-order logic over $M$ ("There is a binary relation on $\omega$ coding a structure such that ..."). Since $V=L$ we have that $M\in L_\alpha$ for some countable ordinal $\alpha$, and therefore inside $M$ (in second-order logic) we can talk about the least ordinal $\beta$ such that $L_\beta\models ZFC+V=L$ and there is a structure $N\in L_\beta$ with $L_\beta\models\vert N\vert=\aleph_0$ which is isomorphic to $M$.
Now note that there is in $L_\beta$ an "$L$-least" bijection $b$ between $N$ and $\omega$. The key point now is that such an $N$ is uniquely isomorphic to $M$ since $M$ is transitive, so we can use $b$ to second-orderly define in $M$ a well-ordering of $M$ of ordertype $\omega$ (which is a clear contradiction), roughly as follows:
$x\trianglelefteq y$ iff there is a structure $X$ with domain $\omega$ such that
$X\models ZFC+V=L$,
there is some $N\in X$ which is isomorphic to $M$,
$X$ is minimal with the above two properties, and
$x$ precedes $y$ in the $L$-ordering according to $X$.
(In fact since the $L$-ordering is appropriately absolute we don't need $X$ to be minimal here, but it makes the picture nicer.)
In fact, the above approach shows that more generally under $ZFC+V=L$ we have that any model of your theory is correct about cardinality (if $\kappa\in M$ is not a cardinal in reality, think about the $L$-least map witnessing that ...). As a consequence, assuming $ZFC+V=L$ we have that any model of your theory is extremely large: roughly speaking, if we can define an ordinal $\alpha$ in a reasonably simple way then any model of your theory must have cardinality $>\aleph_\alpha$.
Of course, this all breaks horribly if we work in an ambient model of ZFC with no nicely-definable well-ordering of the reals, and Lietz's above-mentioned answer shows that there's no getting around that.
Well, given an infinite set $A$, in order to prove $|A\times A|=|A|$, you only really care about the cardinality of $A$: in other words, it suffice to prove that $|B\times B|=|B|$ for some $B$ such that $|B|=|A|$ (since you can transport a bijection $B\times B\to B$ along a bijection between $B$ and $A$). So it doesn't matter what specific sets we get in our models, as long as we hit every possible cardinality.
Unfortunately, though, your argument doesn't work: starting from $\omega$ and going up and down as your statement of Löwenheim-Skolem allows, you cannot reach all infinite cardinalities in the absence of AC. In particular, your version of Downward Löwenheim-Skolem will never guarantee the existence of a model of any cardinality that is not greater than or equal to $\aleph_0$ (because the conclusion has $|N|\leq |A|+\aleph_0+|L|$ rather than just $|N|\leq |A|$). Without AC, it is not necessarily true that every infinite cardinality is greater than or equal to $\aleph_0$.
Here, then, is a more careful version of the argument you propose in the special case that $|A|\geq \aleph_0$. Starting from the model $\omega$, Upward Löwenheim-Skolem gives a model $M$ of cardinality at least $|A|$. Picking a subset of $M$ which is in bijection with $A$, Downward Löwenheim-Skolem then gives a submodel $N$ of $M$ such that $|A|\leq |N|$ (since $N$ contains our chosen subset of size $|A|$) and $|N|\leq |A|+\aleph_0$. But since $|A|\geq \aleph_0$, $|A|+\aleph_0=|A|$ (since $|A|\geq\aleph_0$, we can write $|A|=\aleph_0+|B|$ for some $B$, and then $|A|+\aleph_0=(|B|+\aleph_0)+\aleph_0=|B|+(\aleph_0+\aleph_0)=|B|+\aleph_0=|A|$). Thus $|N|=|A|$, and since we have $|N\times N|=|N|$ we conclude that $|A\times A|=|A|$.
Of course, this still leaves the issue: what if $|A|\not\geq\aleph_0$? Well, it turns out that if you look at the proof that $|A\times A|=|A|$ for all infinite $A$ implies AC, it actually only ever uses sets $A$ such that $|A|\geq\aleph_0$. (Specifically, it uses $A$ of the form $X\sqcup \aleph(X)$ where $X$ is an infinite set and $\aleph(X)$ is its Hartogs number, and $\aleph(X)$ always contains $\omega$.) So actually, the weaker conclusion obtained above is still enough to deduce AC.
Best Answer
A couple notes on "the theory of infinite-dimensional vector spaces":
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.