Talk:Compact Subspace of Linearly Ordered Space/Lemma 1

't Appears to me that the Nec and Suf have been mixed up. Suf means: to prove from the two conditions that the thing is compact. --Lord_Farin (talk) 22:31, 7 February 2013 (UTC)


 * This sort of thing is why I don't approve of mixing "iff" with "necessary and sufficient". There is apparently an understanding on PW of which way it goes by PW convention, but that is not memorable. --Dfeuer (talk) 22:38, 7 February 2013 (UTC)


 * First mentioned goes first. So, if $p \iff q$ is to be proved, $p \implies q$ is necessary condition, $q \implies p$ sufficient. I am working on an alternative, hopefully shorter proof of the necessary condition - so as to comfort you I'm not only thrashing against minor presentational things. --Lord_Farin (talk) 22:40, 7 February 2013 (UTC)


 * We read:


 * $p \implies q$ as $p$ is a sufficient condition for $q$.


 * $p \implies q$ as $q$ is a necessary condition for $p$.


 * IMHO, it is more natural to do it the other way round LF based on how we read it (or at the very least how this wiki says you should read it).


 * (I am quite fond of books that write $\implies$ direction and $\Longleftarrow$ direction but perhaps I have no taste :P) --Jshflynn (talk) 22:57, 7 February 2013 (UTC)


 * Convinced now that there may arise ambiguity. I've been prone to this confusion myself. A solution should be crafted. I also think to have an alternative proof. I'll post it to my sandbox. --Lord_Farin (talk) 23:01, 7 February 2013 (UTC)


 * Proof is up. Please read it and check it for flaws. It's likely to be essentially the same argument but I think I've removed a few subproofs by contradiction. --Lord_Farin (talk) 23:19, 7 February 2013 (UTC)


 * I'm not a fan of $\implies$ direction and $\Longleftarrow$ direction because in heading titles they are unwieldy and ugly. We might want to use "forward direction" and "reverse direction", but I would caution against making too much fuss over this because (a) if we were to impose a change of standard now, there are lots of pages to be amended, and I'll be the one who ends up doing it and I don't want to, (b) someone somewhere will throw his teddy out of the lecture theatre because it's not how he likes it, and (c) I like "necessary condition" and "sufficient condition". --prime mover (talk) 23:25, 7 February 2013 (UTC)


 * Something could be put as an acceptable new alternative so as to not make the "problem" not worse (at least not by the ones considering it a problem). Forward and reverse direction seems a fine solution. We won't be crucified for doing that (at least, no more than we are already anyway). --Lord_Farin (talk) 23:29, 7 February 2013 (UTC)

Converse
I've been banging my head against the converse for a bit, but I haven't managed to get very far. I'm starting to think it's probably necessary to use Zorn's lemma to magick up a minimal subcover and then prove that under these conditions a minimal subcover must be finite. --Dfeuer (talk) 02:57, 8 February 2013 (UTC)


 * I think I have a proof, but as of now it seems to require $\text{AC}_{|Y|}$ (which is an improvement over Zorn nonetheless). --Lord_Farin (talk) 09:04, 8 February 2013 (UTC)


 * The problem lies in that e.g. we can't avoid choice in picking a cover element for $\inf Y$, because there may not exist a greatest element. --Lord_Farin (talk) 09:13, 8 February 2013 (UTC)

Crazy idea that may well not work, but strikes me as reasonably likely to give a clean proof (depending on BPI/UL via Equivalent Definitions of Compactness) is to use the definition of compactness based on ultrafilter convergence. Slightly more specific sketch of the beginning:

Let $\mathcal F$ be an ultrafilter on $Y$.

For $S \in \mathcal F$, let $f(S) = \inf_Y(S)$.

Let $p = \sup_Y f(\mathcal F)$.

Then I hope that $\mathcal F$ converges to $p$, and that this won't be too terribly hard to prove using the usual subbase for the order topology. --Dfeuer (talk) 18:37, 8 February 2013 (UTC)


 * I don't have enough experience with ultrafilters to assess the viability of that strategy. --Lord_Farin (talk) 19:48, 8 February 2013 (UTC)

Looking at this again, I don't think the converse is even correct, unless $X$ is required to be complete. Can you check the following?:

Let $X = [0,1) \cup (2,3) \cup \{4\}$.

Let $Y = [0,1) \cup \{4\}$.

$Y$ is closed in $X$: if $x \in (2,3)$, then $x \in \left(\frac {x+2} 2, \frac{x+3} 2 \right)$.

$Y$ is complete because it is order-isomorphic to $[0,1]$.

$Y$ is not compact:

$\{{\downarrow}x:x<1\} \cup \{{\uparrow 2.5}\}$ is an open cover for $Y$ with no finite subcover. --Dfeuer (talk) 21:14, 8 February 2013 (UTC)


 * 't Appears to be correct. It connects with an issue I glossed over in my purported proof. Yes, I'm quite confident your argument is correct. It is however not necessary to ask that $X$ be complete. I think it suffices to demand $\sup_Y S = \sup_X S$ for all $S \subseteq Y$. --Lord_Farin (talk) 21:30, 8 February 2013 (UTC)


 * You're probably right, but you probably have to require equality of the infima as well if you limit it so. --Dfeuer (talk) 21:38, 8 February 2013 (UTC)


 * Which is to say that by then the theorem statement gets so unwieldy that you're probably better off just requiring $X$ to be Dedekind complete.--Dfeuer (talk) 21:43, 8 February 2013 (UTC)


 * Hm, second thought seems to indicate that modifying the example above by putting $X = Y$ disproves both of our suggested adaptations. What we probably need is a condition on $X$ like $\inf {\uparrow} u = u$ and $\sup {\downarrow} v = v$.


 * Actually, thinking aloud, I propose the conditions that for open $S \subseteq Y$:


 * $\inf_Y S \in S \implies \sup_Y \varnothing \in S$
 * $\sup_Y S \in S \implies \inf_Y \varnothing \in S$


 * which ensure that, except at the ends of $Y$, we won't "mix up open and closed intervals". --Lord_Farin (talk) 21:43, 8 February 2013 (UTC)


 * In my adaptation, $Y$ is still required to be complete. You can't put $X=Y$ in the above example because the topology on $Y$ is not its induced order topology ($\{4\}$ is open in the subspace topology but not the induced order topology). --Dfeuer (talk)


 * Yes it was tacit that $Y$ is complete, having already established the other implication of the theorem. As for the example: you're right. --Lord_Farin (talk) 21:55, 8 February 2013 (UTC)


 * Still, taking $[2..3]$ in place of $(2..3)$ in the middle will obviously render $X$ complete without changing the problem. --Lord_Farin (talk) 22:01, 8 February 2013 (UTC)

No, because then $Y$ is not closed anymore—$2$ becomes a cluster point of $Y$. --Dfeuer (talk) 22:11, 8 February 2013 (UTC)

However, you're probably still right that my condition doesn't work. Let's get a proof together and see what premises it really needs. Dfeuer (talk) 22:16, 8 February 2013 (UTC)


 * I think my proof works, given my premises (which I have tried to construct so that this statement is true). Basically, it goes as follows:


 * Given a cover. Define an ordinal sequence of covering opens $S_\alpha$, each containing $s_\alpha = \inf \complement \left({\bigcup_{\beta < \alpha} S_\beta}\right)$, of length $|Y|$. At limit stage $\lambda$, find a smaller element $y \prec s_\lambda$ in the chosen covering open around $s_\lambda$ (here my premise is used). Easily seen that only finitely many $S_\alpha$ are needed to cover $\bigcup_{\alpha \le \lambda} S_\alpha$. Thus inductively find a finite subcover. Sequence must terminate because each $S_\alpha$ covers a new element by construction. Done, I think.


 * I hope it's clear and correct. --Lord_Farin (talk) 22:27, 8 February 2013 (UTC)


 * What is a covering open? My knowledge of ordinals is primitive at best, so it's not clear to me, but it might be to others. --Dfeuer (talk) 22:36, 8 February 2013 (UTC)


 * An element of the cover. I've deduced however that the argument is correct only in case all elements of the cover are intervals. This is in general not true, so it's back to the drawing board :S. --Lord_Farin (talk) 22:52, 8 February 2013 (UTC)


 * Argument recovered by virtue of Compactness from Basis. --Lord_Farin (talk) 23:03, 8 February 2013 (UTC)


 * I created that page as a lemma for this theorem, so I'm not shocked you found it useful. As soon as I get home I'll try finishing up my proof based on ultrafilters. My primitive notion of what you're writing suggests you're using a (slightly) limited version of the well-ordering principle. If this is so, then that is not strictly comparable to UL, so our proofs should both be offered as alternatives. --Dfeuer (talk) 23:14, 8 February 2013 (UTC)


 * Just when I shut down my computer last night, I conceived of a Zorn-like argument. The interesting part is that I can prove without using Zorn that there exists a maximal element. This means choice is avoided entirely (since it is reduced to an existence claim). Do you want me to write it up? It builds on my previous proof sketch. --Lord_Farin (talk) 09:16, 9 February 2013 (UTC)


 * My vote would be "yes". --prime mover (talk) 10:35, 9 February 2013 (UTC)

Proof is at User:Lord_Farin/Sandbox. Zorn-avoiding construction encountered some flaws, so I went for the easy way. --Lord_Farin (talk) 12:52, 9 February 2013 (UTC)


 * I sandboxed my attempt. I haven't been able to figure out how to complete it, so it might not actually work. Can you glance at it and see if I missed anything stupidly obvious? --Dfeuer (talk) 17:14, 9 February 2013 (UTC)


 * I'm wondering if a useful condition here is that the set of closed rays and intervals form a neighborhood basis (alternative definition) for the topology. --Dfeuer (talk) 19:47, 9 February 2013 (UTC)