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For example, when I want to prove a proposition like "Let $V$ be a vector space..." and I need a basis of $V$ to prove it, instead of using the theorem "Every vector space has a basis" which needs the axiom of choice, I can replace the assumption with "Let $V$ be a vector space with a basis".

If the axiom of choice is true, this replacement does not lose generality of the theorem. In addition, we can individually prove the existence of basis of concrete vector spaces such as $\mathbb{R}^n, \mathbb{C}^n, ...$

Is it possible to avoid the axiom of choice?

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  • $\begingroup$ This approach may lose generality if the axiom of choice is false. For instance, if you work in models without AC, there can exist vector spaces that have no basis. Then your proof will not say anything about those vector spaces, even if your conclusion might still be valid if proved in some other way. $\endgroup$ – Nate Eldredge Dec 14 '19 at 13:41
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Yes, you can always require that your objects have the properties you'd need choice for.

Just like you can always require that your sets are well ordered, and simply limit the generality.

The problem is that a lot of times naturally occurring objects will not be well behaved without choice. For example, $\ell_2$ might not have a basis without choice.

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