Talk:Compactness Theorem

Wikipedia claims this theorem is actually equivalent to BPIT and Gödel's Completeness Theorem. --Dfeuer (talk) 07:58, 13 December 2012 (UTC)


 * Before posting up any more proofs using this claim, it would be good to post both BPIT and GCT up on this site. As it stands, with the crucial information missing, it makes us look embarrassingly lame. It makes us look like a bunch of amateurs pretending to be clever, which is one of the main criticisms which is brought against this site. --prime mover (talk) 22:55, 13 December 2012 (UTC)


 * Sure. I can try to do some, but a lot of the set theory is over my head. It disturbs me, however, to see "this theorem depends on AoC" when in fact it only depends on BPIT, AoCC, AoDC, etc., even if I don't personally know how to prove that. --Dfeuer (talk) 22:59, 13 December 2012 (UTC)


 * I hate to say this, but if all this is over your head, should you not leave well alone? Is all you know about this what you've picked up from wikipedia? --prime mover (talk) 07:25, 14 December 2012 (UTC)

Substantial error in proof with ultraproducts
As I noted with a questionable template, $\{\uparrow(E) \mid E \in T \}$ isn't actually a filter on $\Sigma$ as claimed. I just glanced over at Wikipedia and see that rather than looking at finite sets containing individual statements, their proof considers finite sets including finite sets of statements, which makes a lot more sense. Since I don't know anything about ultraproducts yet, I don't feel comfortable trying to fix this proof, but someone really should. --Dfeuer (talk) 21:08, 24 December 2012 (UTC)


 * I went back to look at the original page put up by some dude with the username qedetc who didn't stay around long (put off by my rudeness, I believe). His original posting did not provide any links (and consistently refused to do so, one of the reasons for my rudeness to him). (Or her, coulda been f, nobody knows.) So it's possible that "filter" as originally used here is not the same as the definition we provide on the website.
 * Please feel free to take all pages written by qedetc with suspicion. --prime mover (talk) 22:40, 24 December 2012 (UTC)


 * Rewrote that section. Proof is almost the same; it's just using finite collections of sentences rather than individual sentences. From this you can justify that the set defined is contained in an ultrafilter. (As it stood, the proof was nearly correct - $\{\uparrow(E) \mid E \in T \}$ is contained in an ultrafilter, but it's not clearly exactly why this should be. As I've rewritten it this is justified using the finite intersection property.) Bakkot (talk) 08:41, 26 December 2012 (UTC)


 * Thanks - this page and much of the stuff which was added at the time is up for a reorganisation so as to ensure that it adheres to house style. Your input is welcome. --prime mover (talk) 10:17, 26 December 2012 (UTC)