Timeline for $\aleph_1$-calibre
Current License: CC BY-SA 3.0
9 events
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| May 16, 2013 at 22:04 | comment | added | Santi Spadaro | Joel, I believe that calibre was introduced by Shanin of $\Delta$-system lemma fame. In fact, I believe that one of the first applications of the $\Delta$-system lemma was proving that (arbitrary) product of spaces of caliber $\omega_1$ has calibre $\omega_1$ (or some variation on this). | |
| Oct 21, 2011 at 5:16 | comment | added | Paul | @Joel David Hamkins.Thanks,Joel. The argument is very beautiful. I'm very appreciate it. | |
| Oct 19, 2011 at 5:37 | vote | accept | Paul | ||
| Oct 18, 2011 at 16:16 | comment | added | Todd Eisworth | I've run across the spelling "caliber" from time-to-time. Negrepontis's "Banach Spaces and Topology" article from the Handbook of Set-Theoretic Topology uses that spelling. No clue on the origin, though, Joel! | |
| Oct 18, 2011 at 13:31 | comment | added | Joel David Hamkins | Yes, Andreas, I agree with everything you say, especially if one imagines the nonempty intersection forming a gun bore through the original family. I also haven't ever seen "caliber". To whom is the concept originally due? | |
| Oct 18, 2011 at 13:17 | comment | added | Andreas Blass |
Of course this doesn't depend on $X=X$; that is, the proof shows that the product of any two spaces of calibre $\aleph_1$ has calibre $\aleph_1$. By the way, does "calibre" have an American spelling "caliber". In the sense of "the bore of a gun barrel", it does, but I don't recall seeing the "er" spelling for the topological sense. (That doesn't mean I haven't seen it or even that I haven't used it, just that I don't recall.) Finally, I can't resist repeating Peter Hinman's suggested definition of "calibre $\aleph_1$": a big bore.
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| Oct 18, 2011 at 13:08 | comment | added | Joel David Hamkins | The argument generalizes to calibre $\kappa$ for any infinite cardinal $\kappa$. It does not generalize to calibre $(\kappa,\lambda)$ (meaning that every $\kappa$ family has a size $\lambda$ subfamily with nonempty intersection), since after the first reduction, you'd have only $\lambda$ many sets left. But the argument does show that if $X$ has calibre $(\kappa,\lambda)$ and $Y$ has calibre $\lambda$, then $X\times Y$ has calibre $(\kappa,\lambda)$. | |
| Oct 18, 2011 at 12:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 119 characters in body
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| Oct 18, 2011 at 12:46 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |