Timeline for A set for which it is hard to determine whether or not it is countable.
Current License: CC BY-SA 2.5
19 events
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| Nov 28, 2016 at 19:46 | comment | added | Hans-Peter Stricker | @Fedor: Just to let you know: I received your comment. Thanks for it. | |
| Nov 28, 2016 at 14:44 | comment | added | Fedor Petrov | @Hans for any real number $a$ take a sequence of rationals converging to $a$. Of course this is essentially the same construction as Andrés'. | |
| Mar 24, 2011 at 18:38 | comment | added | Asaf Karagila♦ | @David: Actually it is not very hard to add one more to a countable family, the surprising part is that it carries over to the uncountable case. For example, every countable ordinal can be embedding into the rationals, but $\omega_1$ cannot be embedded into it (even without preserving the order). | |
| Mar 24, 2011 at 15:46 | comment | added | Tony Huynh | @Asaf: Thanks, I see what David is trying to say now. For a proof of his claim, let $(X_i)$ be an enumeration of the family. For each $i$ choose $x_i \in X_i$ such that $x_i$ is not in the union of the previous $i−1$ sets. We can then extend the family by adding $X:=(x_i)$ to it. | |
| Mar 24, 2011 at 8:18 | comment | added | Asaf Karagila♦ | @Tony: I believe this discussion is only about infinite subsets of $\mathbb{N}$. | |
| Mar 24, 2011 at 0:26 | comment | added | Tony Huynh | @David: I think this is false. Let the family consist of all finite subsets of $\mathbb{N}$, together with $\mathbb{N}$ itself. One cannot extend this family while maintaining the finite intersection property. | |
| Mar 23, 2011 at 21:04 | comment | added | David Fernandez-Breton | For me, even more surprising than the existence of such an uncountable family, is the fact that any countable family (of subsets of $\mathbb N$ such that the intersection of any two elements of the family is finite) can still be further extended, i.e. there is another infinite subset of $\mathbb N$ whose intersection with any set from the previous family is still finite. | |
| Mar 23, 2011 at 4:51 | comment | added | Philip Brooker | @Andres: Indeed, as I progressed through my graduate studies I noticed precisely what you claim above. As someone who hopes to make further contributions to Banach space theory, learning more set theory, combinatorics and descriptive set theory is now high on my priority list. | |
| Mar 23, 2011 at 3:33 | comment | added | Andrés E. Caicedo | @Philip: In fact, this is but one of a family of results in infinitary combinatorics that one encounters with some frequency in the theory of Banach spaces. Recent years have seen a steep increment in the sophistication of the set theory being directly applied in this area. | |
| Mar 23, 2011 at 0:08 | comment | added | Tony Huynh | @Tim and Philip: Thanks for the references! | |
| Mar 22, 2011 at 23:27 | comment | added | Philip Brooker | Tony's example here appears, interestingly enough, in a book on Banach space theory; in Albiac and Kalton's Topics in Banach Space Theory, it appears on page 45 as Lemma 2.5.3. There it is used to show (Theorem 2.5.4) that there is no continuous linear projection of $\ell_\infty$ onto its closed subspace $c_0$. | |
| Mar 22, 2011 at 21:25 | comment | added | Asaf Karagila♦ | I believe the answer Tim is looking for is "Dedekind cuts". | |
| Mar 22, 2011 at 21:10 | comment | added | mathahada | Tim: I thought of a solution but perhaps its different from yours because its not a two word phrase. A good way is to enumerate the rationals and for every real number $r$ let $A_r$ be the set of all those $n$ such that $q_n < r$. I thought of two candidate two word phrases: omega-one and "enumerating rationals", but the first doesn't work and the second is very far from a complete solution | |
| Mar 22, 2011 at 18:03 | comment | added | Timothy Chow | @Hans: Tony's problem appears in Donald Newman's book A Problem Seminar and was also problem B-4 on the 1989 Putnam exam. @KConrad: "there is" or "there isn't" is a two-word answer but there is a two-word phrase that gives a complete solution to the problem I posed. | |
| Mar 22, 2011 at 17:53 | comment | added | Andrés E. Caicedo | @Hans: Given an infinite sequence of 1s and 2s, its initial segments are numbers (written in decimal notation, for example), so any such sequence corresponds to an infinite subset of ${\mathbb N}$, and any two of these sets have finite intersection. | |
| Mar 22, 2011 at 17:12 | comment | added | KConrad | Tim, I realize one of "there is" or "there isn't" is a two-word solution. | |
| Mar 22, 2011 at 16:28 | comment | added | Hans-Peter Stricker | @Tony: Please give me a hint where I can learn more about this! | |
| Mar 22, 2011 at 13:03 | comment | added | Timothy Chow | Along similar lines, is there an uncountable family of subsets of $\mathbb N$ that is totally ordered by inclusion? This can cause quite a bit of head-scratching until you realize that there's a two-word solution. | |
| Mar 22, 2011 at 12:55 | history | answered | Tony Huynh | CC BY-SA 2.5 |