Timeline for Which vector spaces are duals ?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2011 at 9:19 | comment | added | Stephen S | @Andres Caicedo - I'm not claiming that $\beth_{\omega_1}^{\aleph_0}>\beth_{\omega_1}$. I only need that $\beth_{\omega_1}\cdot{\aleph_0}=\beth_{\omega_1}$ (to get $|V|=\dim(V)$) and that there is no cardinal $\lambda$ such that $\aleph_0^\lambda=\beth_{\omega_1}$ - and these statements are true, and don't involve $\beth_{\omega_1}^{\aleph_0}$. | |
| Feb 4, 2011 at 7:37 | comment | added | Andrés E. Caicedo | Stephen: Your "the same is true if..." is not quite correct. For example, $\beth_{\omega_1}^{\aleph_0}=\beth_{\omega_1}$. Here, $\omega_1$ is the first uncountable ordinal. (Your remark is true with $\beth_\omega$ and even with $\beth_\alpha$ whenever $\alpha$ has cofinality $\omega$, for example.) | |
| Feb 3, 2011 at 13:04 | history | edited | Stephen S | CC BY-SA 2.5 |
add explanation of formula
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| Feb 3, 2011 at 10:52 | history | edited | Stephen S | CC BY-SA 2.5 |
remove parentheses
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| Feb 3, 2011 at 10:43 | history | edited | Stephen S | CC BY-SA 2.5 |
comment on R^(R)
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| Feb 3, 2011 at 10:36 | comment | added | François Brunault | For more details on $\beth_\omega$ see en.wikipedia.org/wiki/Beth_number | |
| Feb 3, 2011 at 10:25 | history | edited | Stephen S | CC BY-SA 2.5 |
reworded slightly
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| Feb 3, 2011 at 9:01 | history | answered | Stephen S | CC BY-SA 2.5 |