Comparability of vector spaces over the same field Let $V$ be the collection of linear isomorphism classes of vector spaces over field $F$. Is this class $V$ totally ordered by inclusion?
Argument (possibly erroneously):
(The following assumes that the vector spaces in consideration are all over the same field.)
Does there exist an embedding map between any two vector spaces? My reasoning is that, this holds for finite dimensions and since every vector space has a basis (assuming Zorn's lemma), the difference between any two vector spaces  essentially lies in the cardinality of the bases, so if the cardinality itself is ordered (assuming Continuum Hypothesis?) then we should be able to compare any two vector space and say that one of them is contained in the other.
I was in particular thinking about the normed linear spaces but in that case this seems to fail due to the fact that same vector space can be endowed with different topologies by different norms and we can can construct incomparable topologies using inequivalent norms.
PS: Please excuse the general nature of the question. Since I am not entirely proficient at cardinal arithmetic and topics like Continuum Hypothesis, I am chasing my idea based on the little that I know. 
 A: First of all, as remarked in the comments, the collection of all vector spaces over a fixed field is a proper class and not a set. But we can still ask whether it is totally ordered by some relation or not.
Secondly, and more importantly perhaps, is to understand the difference between inclusion and embedding (via an injective linear homomorphism). Inclusion means that one vector space is literally a subset of another. But since every large enough set can be endowed with a vector space structure over your fixed field (all it has to do is be as large as the field), this means that "most sets" can be made into a vector space. So inclusion certainly does not order the vector space via a linear ordering.
So what about embedding? Well, here you get the same problem you get with infinite sets in general. You can easily write an injective linear map between two different vector spaces (and you can also have that neither of the two maps you wrote is an isomorphism). So you get that injective linear maps do not define a partial order, but rather a quasi-order.
But now what? Well, assuming the axiom of choice, every vector space has a basis with a unique cardinality, also known as the dimension of the space. Now it's easy. Since cardinals are linearly ordered, assuming the axiom of choice, the possible dimensions are also linearly ordered, so if you think about isomorphic vector spaces as "being the same", it follows that injective linear homomorphism do in fact linearly order the class of $F$-vector spaces. Note that this has nothing to do with the continuum hypothesis.
Finally, what happens if we do not assume the axiom of choice? Well, then it's possible that there are vector spaces without a basis; or vector spaces with two bases of different cardinalities (so dimension is not a well-defined notion); or just vector spaces that do not have any injective linear maps between one another. So essentially, wherever you've used choice, you had to use choice.
