Timeline for Is it consistent with ZF that $V \to V^{\ast \ast}$ is always an isomorphism?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 12, 2018 at 18:49 | comment | added | David Lampert | @JeremyRickard Le Bourgeois Gentilhomme! Yet I wonder if there's some meaning in this coincidence of choices. | |
| Jan 12, 2018 at 18:04 | comment | added | Jeremy Rickard | @DavidLampert No. Turns out I can make jokes without even trying! | |
| Jan 12, 2018 at 15:45 | comment | added | David Lampert | @JeremyRickard I wonder if your choice of "choose" was an intentional coincidence. | |
| Jan 12, 2018 at 8:25 | comment | added | Jeremy Rickard | @AaronBergman Hmmm... If I’m forced to choose, then can I cheat and pick $V\oplus W$? :) | |
| Jan 12, 2018 at 3:06 | comment | added | Aaron Bergman | So, which is it? | |
| Jan 11, 2018 at 16:18 | comment | added | Emil Jeřábek | Ah! You are right, I didn't read the proof carefully enough. | |
| Jan 11, 2018 at 16:04 | comment | added | David E Speyer | @EmilJeřábek He doesn't. The argument breaks down into two cases. Case 1: $W$ has a nonzero map to $k$. Then the composition $V \to W \to k$ is an element of $V{\ast \ast}$ not in $V$. Case 2: $W^{\ast}=0$. Then $W^{\ast \ast}=0$, so $W \to W^{\ast \ast}$ has a kernel. | |
| Jan 11, 2018 at 14:45 | vote | accept | David E Speyer | ||
| Jan 11, 2018 at 14:45 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |