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I was perusing through some advanced linear algebra texts and could only find proofs for the dimension theorem for infinite basis that require the use of some advanced set-theoretic properties. I understand the need to use Zorn's lemma but I wonder if we could make do without the Cantor–Bernstein theorem and rely solely on linear algebra theorems. I managed to find a paper that seems to show that it is possible. The paper can be found here.

What I don't get is why would he use linear transformations instead of normal functions for the poset he created and secondly for each element in the poset he defined, after restriction of the domain, there is a bijection between the bases. Isn't that the property we are trying to prove?

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    May help: http://vigo.ime.unicamp.br/MT401-2016/HamelCard.pdf – 2017-02-04
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    To me, the proof presented in that paper doesn't actually rely heavily on countability. If you define the $A_n,B_n$ sequences like they do and add that $A_{\alpha} = \bigcup_{\beta < \alpha}A_{\beta}$ for limit ordinals $\beta$ (and likewise define $B_{\alpha}$), and suppose that the theorem is true for all smaller cardinalities, the proof should go through. – 2017-02-04
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    Cantor–Schröder–Bernstein is hardly an "advanced set theoretic property". It's one of the most basic facts about cardinality, and it has an extremely concrete proof - there's no reason to be scared of it. The dimension theorem for infinite-dimensional vector spaces is a statement about the cardinalities of infinite sets, so it's not surprising that you have to use *some* basic set theory. – 2017-02-04
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    Also, what leads you to believe that the proof in Moore's paper doesn't cover the case of uncountable bases? I read the proof and couldn't find anywhere where a countability assumption stated or used. At one point, he points out that $A_\infty$ is countable, but $A_\infty$ is just a set that's being added to the basis for $W$ at one stage in the proof, it's never claimed to be a basis for the whole space. – 2017-02-04
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    Yes. Sorry that was a mistake on my part I misread the proof thanks – 2017-02-04
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    I am not sure what you are asking, specifically. – 2017-02-04
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    @CarlMummert Could you explain why is he using linear transformations when ordinary functions would do? What's the purpose in the proof of including the properties of a linear transform thanks. – 2017-02-05

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The proof does cover the case for uncountable dimension. What the proof does is to use Zorn's lemma to assert the existence of some maximal automorphism on a subspace, and assuming that the domain of it is not the whole space, it is always possible to come up with a countable extension, contradicting its maximality. If we have a vector space of uncountable dimension, this extension is likely to fall short of the entire space, but this is not an issue because we only need a candidate subspace to contradict maximality, rather than explicitly constructing the automorphism for the whole space.

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    That makes more sense, thanks. But if he wants to show that the maximality is contradicted why can't he just use two vectors not in the domain. Why go through the process of creating all those sets? – 2017-02-04
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    There is a very stringent condition on the poset that is being considered such that Zorn's lemma can be applied, that is, the subspaces created by the new vectors must be the same. This is not going to be the case most of the time by picking any two old random vectors. This may seem weird if you just look at the bases as sets, but this is not the case. A certain structure has to be preserved in a vector space, and if you take any two random vectors, what's stopping one of them, together with W, from spanning the whole space, while the other one doesn't? – 2017-02-04
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    Hi could you explain why he would use linear transformations and not just functions with from one vector space to another with the same properties – 2017-02-05
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    Secondly in the poset he created isn't he already using the property we want to proof i.e. Bijection between bases? – 2017-02-05