Talk:Baire Category Theorem

With respect to whether or not we need to use AoC or ADC, this is presented in 1973: Thomas J. Jech: The Axiom of Choice: $1.$ Introduction: $1.4$ Problems: $5$:

Without using the Axiom of Choice, prove that the set $R$ of all real numbers is not meager (Baire Category Theorem).

Hence it is reasonable to assume that you don't actually need it after all.

At this stage in the book, the Axiom of Dependent Choice has not been introduced, so it's unreasonable to expect the person solving this problem should be using it. --prime mover (talk) 20:00, 24 August 2023 (UTC)

BCT definitely needs something Choicey. The exercise just shows that ZF proves that $\R$ is non-meager, not anything about other complete metric spaces. The use of Choice here is just in the construction of $\sequence {x_n}_{n \mathop \in \N}$. (we reason that for each $n \in \N$ there exists elements satisfying a certain property in terms of $n$, and we pick one for each $n$ using Choice) It's a use of Choice that we don't think about much in analysis, but since the relation between BCT and the various types of Choice is of some interest we should probably make this explicit. I personally haven't looked into the weaker forms of Choice so wouldn't know the details. Caliburn (talk) 20:49, 24 August 2023 (UTC) Caliburn (talk) 20:49, 24 August 2023 (UTC)

Quick google says that BCT is actually equivalent to Dependent Choice. E.g. https://mathoverflow.net/questions/373197/bct-equivalent-to-dc Caliburn (talk) 20:51, 24 August 2023 (UTC)

Okay, I think I've got it: while you need ADC for the general BCT, you don't need ADC to prove that $\R$ is not meager. I think this book is a bit too advanced for me. --prime mover (talk) 21:11, 24 August 2023 (UTC)

Your understanding is correct yes, there are specific features of $\R$ that allow for a proof without ADC. Caliburn (talk) 21:52, 24 August 2023 (UTC)

See what I did. --prime mover (talk) 06:03, 25 August 2023 (UTC)