Timeline for How much of mathematical General Relativity depends on the Axiom of Choice?
Current License: CC BY-SA 4.0
13 events
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| Sep 16, 2023 at 7:20 | comment | added | Sam Sanders | @ElliotGlazer I agree, but note that Z$_2^\Omega$, Z$_2^\omega$, and Z$_2$ prove the same second-order sentences. Moreover, Z$_2^\Omega$ has fourth-order Kleene's $\exists^3$ while Z$_2^\omega$ has only got third-order S$_k^2$. Many logicians are "scared" of the third-order stuff already (as their usual second-order techniques do not work). | |
| Sep 15, 2023 at 14:50 | comment | added | Elliot Glazer | Right that's why I think $Z_2^{\Omega}$ is the "right theory" for choiceless analysis. I expect choiceless analysis of the classes BV, semi-continuous, Baire class to go through under any subsystem of $Z_3$ containing the schema extending $\Pi^1_2$-CA to allow third-order parameters. | |
| Sep 15, 2023 at 14:27 | comment | added | Sam Sanders | @ElliotGlazer I see. The closest thing I can think of right now is the following theorem: "a countable covering of a closed set in [0,1], has a finite sub-covering". This is provable in the weak system RCA$_0^\omega +$WKL$ +$ QF-AC$^{0,1}$ and not provable in Z$_2^\omega$ (note the the lower case). | |
| Sep 15, 2023 at 14:21 | comment | added | Elliot Glazer | That’s not a theorem of $Z_3$ or even ZF. The point of my question is whether choiceless theorems of strong systems tend to remain so over $Z_2^{\Omega}.$ | |
| Sep 15, 2023 at 9:18 | comment | added | Sam Sanders | @ElliotGlazer The obvious one (already noted by Kohlenbach in RM2001) is the statement that 'for a function $f:[0,1] \rightarrow \mathbb{R}$ and any $x\in [0,1]$, $f$ is sequentially continuous at $x$ If and only if $f$ is epsilon-delta continuous at $x$'. | |
| Sep 14, 2023 at 17:56 | comment | added | Elliot Glazer | Or a bit more concretely: do you know a natural common consequence of $Z_3$ and of $Z_2^{\Omega} +QF-AC^{0,1}$ not provable in $Z_2^{\Omega}?$ | |
| Sep 14, 2023 at 17:46 | comment | added | Elliot Glazer | The listed facts are provable in ZF. Do you know of any natural ZF theorems of analysis that need countable choice when working in these sorts of subsystems? Btw since I’m not the only commenter on your answer you need to @ me for me to be notified if you respond. | |
| Sep 14, 2023 at 11:38 | comment | added | Sam Sanders | That would be correct mostly, although one may additionally need a fragment of countable choice here and there, called QF-AC$^{0,1}$ by Kohlenbach. It has the form $$(\forall n\in \mathbb{N})(\exists f\in \mathbb{N}^\mathbb{N})(Y(f, n)=0) \rightarrow (\exists \Phi^{1\rightarrow 0})(\forall n\in \mathbb{N})(Y(\Phi(n), n)=0)$$ for any $Y^2$. | |
| Sep 14, 2023 at 7:40 | comment | added | Elliot Glazer | I see. Sifting through your papers, I think the arguments justifying my previous comment are formalizable in $Z_2^{\Omega},$ and that this theory is a conservative extension of $Z_2$ which has third-order objects in its ontology. Is my understanding correct? | |
| Sep 14, 2023 at 6:50 | comment | added | Sam Sanders | @ElliotGlazer the following third-order statement is consistent with RCA$_0^\omega$+ Z$_2$ and stronger systems: "there is a function of bounded variation on [0,1] that is totally discontinuous". Here, $\text{RCA}_0^\omega$ is Kohlenbach's base theory of higher-order RM, which is $L_2$-conservative over RCA$_0$. This implies that the usual hierarchy of function spaces looks VERY different in weak (and some strong) systems. | |
| Sep 13, 2023 at 4:21 | comment | added | Elliot Glazer | I'm confused about your (A) vs (B) dichotomy. Bounded variation functions and semi-continuous functions are in Baire class 1, and Baire class 2 is the same as effective Baire class 2. These are all provable in much less than ZF (say, $Z_2$). | |
| Aug 23, 2023 at 9:50 | comment | added | Igor Khavkine | Where do $L^2(\mathbb{R}^n)$ or $L^2([0,1]^n)$ functions fit in your (A) vs (B) dichotomy? One could also ask about smooth functions $C^\infty(\mathbb{R}^n)$, but these would need to be accompanied by Schwartz distributions on $\mathbb{R}^n$ as well. | |
| Aug 22, 2023 at 18:51 | history | answered | Sam Sanders | CC BY-SA 4.0 |