Timeline for Is $(\mathbb{R}, +)$ still injective as long as $(\mathbb{Q},+)$ is?
Current License: CC BY-SA 4.0
17 events
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| Oct 15 at 0:04 | comment | added | David Gao | @GarrettFigueroa Yeah, I was just stating the proof I had in mind. I don’t know Blass’s proof. If it is supposed to be in ZFA, then yeah, my proof won’t work. | |
| Oct 14 at 22:02 | comment | added | Garrett Figueroa | @DavidGao Since the Blass paper works within ZFA I think it's worth bearing that, in ZFA, multiple choice does not entail full choice, at least according to the nlab. Doesn't change anything morally as we're still talking about a fragment of AC, but feels worth noting for the record. | |
| Oct 13 at 2:59 | history | edited | David Gao |
Added two relevant tags
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| Oct 13 at 2:57 | comment | added | David Gao | @HanulJeon I don’t think even that’s the case. The injectivity used is that of the kernel of $\mathbb{Q}[J]\to\mathbb{Q}[I]$, not of $\mathbb{Q}[I]$. (As that’s how the proof of injectivity implies s.e.s. must split goes.) So, I agree, the argument likely helps nothing about the OP’s question. | |
| Oct 13 at 2:49 | comment | added | Hanul Jeon | @DavidGao Nice argument. Your proof seems to suggest if $\mathbb{Q}[I]$ is injective then the multiple choice holds for families indexed by $I$. Unfortunately, I do not think your argument is useful for the particular example $\mathbb{R}$ since we do not know without AC that $\mathbb{R}$ has of form $\mathbb{Q}[I]$ for some $I$. | |
| Oct 13 at 2:42 | comment | added | David Gao | … a multiple choice function. As axiom of multiple choice is equivalent to AC under ZF, this proves the implication. (As far as I can tell, I don’t see a way to adapt this proof to show any fragment of AC using just $\mathbb{Q}$ and $\mathbb{R}$ are injective.) | |
| Oct 13 at 2:38 | comment | added | David Gao | @HanulJeon I don’t know if this is the proof you’re referring to, but the proof outline I had in mind is as follows: by injectivity, any s.e.s. of $\mathbb{Q}$-linear spaces splits, so in particular any surjective $\mathbb{Q}$-linear map has a right inverse. Then, given a surjection $\pi:J\to I$, it induces a surjective $\mathbb{Q}$-linear $f:\mathbb{Q}[J]\to\mathbb{Q}[I]$, which then admits a right inverse $g$. For each $i\in I$, $g(e_i)$ has finite support and the support must contain some elements of $\pi^{-1}(i)$, so $i\mapsto\text{supp}(g(e_i))\cap\pi^{-1}(i)$ is… | |
| Oct 13 at 1:31 | comment | added | Hanul Jeon |
I haven't checked Blass' proof for the implication of AC from "every $\mathbb{Q}$-linear space is injective" (the internet in my home isn't working, so I can't check his paper without risking using much mobile data) but it is possible that "$\mathbb{R}$ as a $\mathbb{Q}$-linear space is injective" may imply a fragment of choice (like, choice for a limited size of family of the limited size of sets under some cardinality.) Also, I recommend adding a set-theory tag.
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| Oct 12 at 21:44 | comment | added | David Gao | Also, a tangential remark: if, instead of just $\mathbb{Q}$ and $\mathbb{R}$, we assume all $\mathbb{Q}$-linear spaces are injective (even in just the full subcategory of $\mathbb{Q}$-linear spaces), we will have full AC. | |
| Oct 12 at 21:17 | comment | added | David Gao | Now that I think about this, even assuming $\mathbb{R}$ has a $\mathbb{Q}$-basis does not ensure injectivity, does it? You still need to choose an extension morphism for each basis element, and since there are infinitely many basis element, AC is still needed. | |
| Oct 12 at 20:53 | comment | added | Emil Jeřábek | @DavidGao You’re right, I messed up the arrows. | |
| Oct 12 at 20:10 | comment | added | David Gao | @EmilJeřábek I'm a bit confused. Doesn't DC holds in Solovay model? If DC implies $\mathbb{Q}$ is injective, we will have a nontrivial additive map $f: \mathbb{R} \to \mathbb{Q}$. But in Solovay model, every subset of reals is Lebesgue measurable. So as measurable additive maps from $\mathbb{R}$ to itself must be $\mathbb{R}$-linear, no such $f$ can exist. | |
| Oct 12 at 17:58 | comment | added | Garrett Figueroa | Good point. I edited that bit about bases to make clear that we're not supposing that every vector space has a basis for this Q. | |
| Oct 12 at 17:50 | history | edited | Garrett Figueroa | CC BY-SA 4.0 |
added 15 characters in body
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| Oct 12 at 17:33 | comment | added | LSpice | Just to have it said, if you want to use a $\mathbb Q$-basis for $\mathbb R$ to argue injectivity, then that is itself a use of choice. (Of course, that doesn't mean that there isn't some other way to argue from the injectivity of $\mathbb Q$.) | |
| Oct 12 at 17:30 | history | edited | Garrett Figueroa |
edited tags
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| Oct 12 at 17:24 | history | asked | Garrett Figueroa | CC BY-SA 4.0 |