Skip to main content

Timeline for Unbounded countable subset

Current License: CC BY-SA 2.5

20 events
when toggle format what by license comment
Jun 15, 2016 at 3:03 review Close votes
Jun 15, 2016 at 6:58
Jun 8, 2016 at 4:14 review Close votes
Jun 8, 2016 at 9:03
Jun 7, 2016 at 20:20 comment added thejoshwolfe Is the OP missing a claim about the size of the set? Isn't the empty set a counterexample for the first sentence?
Dec 7, 2009 at 20:15 vote accept Sune Jakobsen
Nov 27, 2009 at 19:57 comment added Alicia Garcia-Raboso After Sune's explanation I am happy to withdraw my downvote.
Nov 27, 2009 at 16:37 answer added Pete L. Clark timeline score: 4
Nov 27, 2009 at 16:21 comment added Sam Nead I think that your generalization still holds. Perhaps transfinite induction will be useful?
Nov 27, 2009 at 7:04 comment added Sune Jakobsen I haven't solved it. But if the answer to my question was yes, you could find a countable subset $J\subset I$, such that for every $i\in I$ there is a $j\in J: K_j\subset K_i$. If there is a j that works for every i, the intersection would be $K_j$ and thus non-empty. Otherwise you could find a decresing sequence $K_1\supset K_2\supset \dots$ and reduce the problem to the homework problem.
Nov 27, 2009 at 0:40 comment added Sam Nead Well, I apologize in all directions, in that case. <p>Sune: would you like to explain the resolution of your refined intersection problem?
Nov 27, 2009 at 0:16 comment added David E Speyer For the record, I remember inventing and thinking about this question when I was first learning set theory. My motivation went as follows: An equivalent formulation of Zorn's lemma is "In a nonempty poset where every totally ordered subset has an upper bound, there is a maximal element." At the time I found it hard to think about arbitrary totally ordered sets, so I wondered if I could replace this by "In a nonempty poset where every ascending sequence $(a_i)_{i \in Z}$ has an upper bound, there is a maximal element."
Nov 26, 2009 at 18:26 comment added Sune Jakobsen It was not a homework problem, but it was inspired by a homework problem. The problem was: Show that if $K_1\supset K_2\supset \dots$ is a decreasing sequence of non-empty compact sets, the intersection $\cap_{i=1}^{\infty} K_i$ is non-empty. I was wondering if this could be generalized: Let $(K_i)_{i\in I}$ be a system of non-empty compact sets, such that for for all $i,j\in I: K_i\subset K_j \vee K_i\supset K_j$. Is the intersection $\cap_{i\in I} K_i$ non-empty? This is why thought of the problem, but I got interested in the problem for its own rights.
Nov 26, 2009 at 17:54 comment added Sam Nead David - it is presented as a straight problem. Nothing about "I need this for..." Or "I was reading X and thought...". No motivation... However, I will be happy to be corrected by M. Jakobsen.
Nov 26, 2009 at 17:47 comment added David E Speyer I am baffled as to why people think this is a homework problem. It could be assigned in a set theory class, but it is a very natural question and the counter-examples are not elementary. I'll bow to peer pressure and not give an explicit construction, but the basic hint here is to read up on ordinal numbers.
Nov 26, 2009 at 17:46 history edited Sune Jakobsen CC BY-SA 2.5
added 83 characters in body; added 1 characters in body
Nov 26, 2009 at 17:24 comment added Jose Brox You should require $a>s$ (strictly greater than) for some $s$, otherwise you could have S bounded and $a$ a maximum.
Nov 26, 2009 at 17:22 vote accept Sune Jakobsen
Dec 7, 2009 at 20:15
Nov 26, 2009 at 17:13 answer added Kristal Cantwell timeline score: 4
Nov 26, 2009 at 17:06 comment added Alicia Garcia-Raboso I agree, this is a homework problem: -1.
Nov 26, 2009 at 16:36 comment added Sam Nead No. This is a homework problem.
Nov 26, 2009 at 16:31 history asked Sune Jakobsen CC BY-SA 2.5