proof that all bases of a vector space have equal cardinality

Is this a valid proof that every basis for a (not necessarily finite-dimensional) vector space $V$ has the same cardinality?

Independent sets have cardinality no greater than $dim(V)$ and spanning sets have cardinality at least $dim(V)$.

Let A and B be two bases for $V$. Since A spans and B is independent, $|A|\ge dim(V)\ge|B|$. Since B spans and A is independent, $|B|\ge dim(V)\ge|A|$. Therefore $|A| = |B|$.

But this may be faulty because it uses $dim(V)$ which is a meaningless concept before we have shown that all bases have the same cardinality. So, what is a correct approach... to assume there are two bases, and show there is a bijection between them?

• Is $|V|$ supposed to be $dim(V)$? – Guido A. Sep 30 '17 at 5:09
• What's your definition for the dimension of $V$? – John Griffin Sep 30 '17 at 5:12
• For an infinite-dimensional space this will require the Axiom of Choice. – DanielWainfleet Sep 30 '17 at 6:37
• How would you define dim($V$) without first proving the result? – DanielWainfleet Sep 30 '17 at 6:40
• Your confidence is misplaced even in the finite dimensional case, as alluded to in the comment of @DanielWainfleet. The trouble with this proof is that it is circular: the definition of $\text{dim}(V)$, and the proof of its properties, requires knowing that every basis has the same cardinality; but you're applying that definition and its properties to prove that every basis has the same cardinality. – Lee Mosher Sep 30 '17 at 19:16