How weak is the weak form of Zorn's lemma?

The weak form being the form equivalent to the axiom of dependent choice. So what I mean is: is it insufficient for the development of functional analysis and everything else Zorn's lemma implies? Or is it simply the reason we can't prove that not every set is measurable, has the property of Baire, and the perfect set property with just DC?

• Zorn seems a red herring here; I think you're really asking, "How much of analysis can be done in ZF+DC alone?" If that's the case, the answer is "most of the good bits" - that is, the parts of analysis that require more than just DC tend to be the counterexamples ("there is a non-measurable set", e.g.) or the "away-from-$\mathbb{R}$" bits (e.g. "every vector space has a basis" and some functional stuff). Indeed, I think for a while there was hope among some logicians that something like ZF+DC+AD might replace ZFC as the "right" foundation for mathematics, although that didn't pan out . . . – Noah Schweber Aug 11 '16 at 20:44
• @NoahSchweber I'd like to see that as an answer ;-) – Stefan Mesken Aug 11 '16 at 20:52
• @Stefan I left it as a comment because it's not really very precise, and I don't have sources for my last sentence; I was hoping someone who could say more would give an answer. – Noah Schweber Aug 11 '16 at 21:08
• @NoahSchweber Fair enough. However, a 'precise' answer can't be expected, because of how opinion based the matter is. – Stefan Mesken Aug 11 '16 at 21:15
• @NoahSchweber that's a shame because I'm actually quite intrigued by your last sentence. I assume AD is the Axiom of Determinacy (the thing having to do with game theory which I know very little about), I would like to learn more about this model and its shortcomings in particular. Do you know where I should research? – Praise Existence Aug 11 '16 at 21:48