User:Lord Farin/Sandbox

This page exists for me to be able to test features I am developing. Also, incomplete proofs may appear here.

Feel free to comment.

Over time, stuff may move to User:Lord_Farin/Sandbox/Archive.

Subpages of this one may exist; they are listed at this PW special page.

Equivalence of Boolean Lattice Definitions
Here, a Definition:Boolean Algebra is to be defined as what is currently on Definition:Huntington Algebra (equivalently, what is on Definition:Lattice/Definition 3 without the ordering, but with complement and distributivity). I have plans for this sections ready in my head, just want to verify that the argument works before I post it up.


 * Definition 1 implies Definition 2 is trivial by expanding definitions of Boolean algebra, join and meet.
 * Definition 2 implies Definition 1 is trickier. An attempt:


 * First off, I think one of the first things that would clarify matters is to unify notation between Huntington and Boolean algebras. Personally, I'd say just make everything use $\wedge$, $\vee$, $\neg$, $\top$, and $\bot$, whether it wants to or not. Trying to keep track of $\top$ and $\wedge$ and $e_*$ and $\circ$ and all that is driving me a bit batty.
 * I think among the first things we need is to establish that in a Huntington algebra, $\top$ and $\bot$ are unique, complements are unique,and $\top=\neg \bot$.
 * Huntington absorption laws. For a lattice, we need $a \vee (a \wedge b) = a \wedge (a \vee b) = a$. Can we get those from a Huntington algebra? Well, we have $a \vee (a \wedge b) = (a \wedge \top) \vee (a \wedge b) = a \wedge (\top \vee b)= a \wedge ((b \vee \neg b)\vee b)=a \wedge ((b \vee b)\vee \neg b)=a\wedge (b \vee \neg b)=a\wedge\top=a$.


 * Much of that is already up. Please keep in mind that the area is going to be restructured soon. What is on those pages now is largely irrelevant, though information will of course be incarnated in some form in the new set-up. I deliberately asked you for looking at the math (use Definition:Lattice/Definition 3) but all I got was comments (in themselves valuable, that be noted, but most of these things I've already created a plan for). Thanks for looking, though. --Lord_Farin (talk) 08:30, 18 January 2013 (UTC)


 * Sorry, I wasn't sure just what you wanted. Use definition 3 for what, exactly?--Dfeuer (talk) 14:19, 18 January 2013 (UTC)


 * Definition 3 for the lattice, together with distributivity and complements for meet and join is the (or a) definition for Boolean algebra I intend to use (effectively just dropping the ordering from a complemented distributive lattice; haven't settled on how I will present all the different axiomatisations for Boolean algebras). Effectively the below tries to prove that in Def 3 of a lattice, we may replace the rather annoying to verify "$\preceq$ is as on Semilattice Induces Ordering" by simply "$a \wedge b \preceq a \vee b$" (which I personally find more elegant). --Lord_Farin (talk) 14:35, 18 January 2013 (UTC)


 * We probably want both alternative conditions. I took a look at the current Boolean algebra definition. It looks to me like the current definition has a definition of a bounded lattice embedded in it, so it would probably be best to factor that out. But you probably already are. —Dfeuer (talk) 14:48, 18 January 2013 (UTC)

It is a lattice
Section heading will be adjusted in due time.

Recall Definition 3 (currently) of a Boolean algebra.

This means $\left({S, \vee, \wedge, \preceq}\right)$ is a lattice as soon as we can establish:


 * $a \preceq b$ iff $a \vee b = b$ iff $a \wedge b = a$

(second iff by the argument that 3 implies 2 on Equivalence of Lattice Definitions)

The latter two imply the former, since we know that:


 * $a \wedge b \preceq a \vee b$

Conversely, suppose $a \preceq b$.

By associativity and idempotence of $\vee$:


 * $a \vee b = a \vee (b \vee b) = (a \vee b) \vee b$

whence also $b \preceq a \vee b$.

Conversely, by compatibility of $\preceq$ with $\vee$:


 * $a \vee b \preceq b \vee b = b$

hence $a \vee b = b$ by antisymmetry.

Thus $\left({S, \vee, \wedge, \preceq}\right)$ is a lattice.

It is distributive and complemented
Is demanded (i.e. tedious brainless writing).

Hence Definition 2 implies Definition 1.

Principle of Induction on WFFs/Structural Induction
Let $\mathcal L$ be a formal language in an alphabet $\mathcal A$.

Let $\mathcal F$ be a bottom-up grammar for $\mathcal L$.

Let $\mathcal A^*$ be the set of all words in the alphabet $\mathcal A$.

Let $\phi \left({\mathbf A}\right)$ be a propositional function on $\mathcal A^*$.

Suppose that the rules of formation of $\mathcal F$ preserve truth of $\phi$, i.e., suppose:


 * $(1): \phi \left({p}\right)$ is true for all letters $p$ of $\mathcal L$
 * $(2):$ If a WFW $\mathbf A$ of $\mathcal L$ is obtained from WFWs $\mathbf B_1, \ldots, \mathbf B_n$ by a rule of formation from $\mathcal F$, then:


 * If $\phi \left({\mathbf B_1}\right), \ldots, \phi \left({\mathbf B_n}\right)$ are all true, then so is $\phi \left({\mathbf A}\right)$

Then $\phi \left({\mathbf A}\right)$ is true for all WFWs $\mathbf A$ of $\mathcal L$.

Proof
Postponed until theorem statement is accredited. Note that the statement in current form assumes that all words are finite, as are all rules of formation. This is fine for our current examples, but may need generalisation in the future. Comments/thoughts? --Lord_Farin (talk) 20:51, 9 November 2012 (UTC)


 * You're inching closer to User:Lord_Farin/Sandbox/Archive. This is quite exciting to me.


 * I saw something similar today. Namely, the induction principle for words (in Simovici and Tenni). It's much simpler than this though. I will think about it some more. --Jshflynn (talk) 21:30, 9 November 2012 (UTC)


 * A formal language accepting all words should reduce the theorem statement to that for words only (e.g. take concatenation as the only rule of formation, singletons as letters (that's pretty natural :) )). Considering the reference you put up: Indeed. A proof of that would most likely proceed by induction on WFFs. It could also be used to give a syntactic proof of Duality Principle (Category Theory)/Formal Duality (basically the current proof is a semantic one, referring to model-theoretic constructions: "undefined" is the key word). --Lord_Farin (talk) 21:37, 9 November 2012 (UTC)


 * Out of curiosity, does it matter if this theorem or proof is expressible in $\mathcal L$? --GFauxPas (talk) 15:11, 15 November 2012 (UTC)


 * $\phi$ needn't be expressible in $\mathcal L$; the global theorem statement can't, since quantification over WFWs is not allowed. I'd say the theorem is phrased in some meta-language which we agree to comprehend already (almost always, natural language plays this rôle). --Lord_Farin (talk) 15:32, 15 November 2012 (UTC)

Proof of Joining Arcs makes Another Arc?
I have the following idea for a proof. I'm not sure about what it needs to actually work just yet, but I feel that it's in the right direction (the correct ansatz, Germans would say).

Define $\mathcal R = \left\{{\left({t, u}\right) \in I \times I: f \left({t}\right) = g \left({u}\right)}\right\}$.

Then $\mathcal R$ is non-empty as $\left({1, 0}\right) \in \mathcal R$.

Next, we define the relation $\preceq$ on $\mathcal R$ by:


 * $\left({t, u}\right) \preceq \left({t', u'}\right)$ iff $t \le t'$ and $u' \le u$

(Include lemma proving $\preceq$ is an ordering; this is almost immediate.)
 * This is what I would like to be called the product ordering, as Kelley and at least a couple others do. --Dfeuer (talk) 19:29, 20 December 2012 (UTC)

Suppose now $\preceq$ has a minimal element $\left({t, u}\right)$.

Then $\left({t, u}\right) \ne \left({0, 1}\right)$, for $a \ne c$.

(Elaborate on why patching $f$ up to $t$ and $g$ from $u$ on yields arc.)

By continuity (I think I use Hausdorffness or something similar here), one can prove that every downward chain in $\mathcal R$ has a lower bound, so an application of Zorn's Lemma shows the minimal element exists.

When eventually putting up the argument, I'll inverse the order sign to make Zorn's lemma more intuitively applicable.

I'd be delighted if someone could avoid AC-equivalents, or point to a resource mentioning its necessity. --Lord_Farin (talk) 14:36, 15 November 2012 (UTC)


 * The words and symbols are blurring before my eyes. Headspace minimal at the moment. Can do little but routine mechanical maintenance at the moment, till this project I'm on is truly finished and (most importantly) the client has paid. --prime mover (talk) 22:05, 15 November 2012 (UTC)


 * How about this (without AoC):


 * Since $T$ is Hausdorff, [...] the diagonal set $\Delta_T$ is closed in $T \times T$.


 * Let $F: I \times I \to T \times T$ be the mapping defined by $F \left({x, y}\right) = \left({f \left({x}\right), g \left({y}\right)}\right)$.


 * By [some theorem], $F$ is continuous.


 * Not sure what that theorem might be called, but it's trivial: $f^{-1}(U) \times g^{-1}(V) = F^{-1}(U \times V)$.


 * By Continuity Defined from Closed Sets, the preimage $\mathcal R = F^{-1} \left({\Delta_T}\right)$ is closed in $I \times I$.


 * From the Heine–Borel Theorem, $\mathcal R$ is compact in $I \times I$.


 * From Continuous Image of Compact Space is Compact, $\left\{{u - v: \left({u, v}\right) \in \mathcal R}\right\}$ is compact in $\R$.


 * Note that $\mathcal R$ is non-empty, as $\left({1, 0}\right) \in \mathcal R$.


 * [...] So $\exists \left({u_0, v_0}\right) \in \mathcal R: \forall \left({u, v}\right) \in \mathcal R: u_0 - v_0 \le u - v$. (By the way, $\left({u_0, v_0}\right)$ is a $\preceq$-minimal element of $\mathcal R$.)


 * Let $\lambda = 1 + u_0 - v_0$.


 * Since $a \ne c$, [...] we have $\lambda > 0$.


 * Let $h: I \to T$ be the mapping defined by:
 * $h \left({x}\right) = \begin{cases}

f \left({\lambda x}\right) &: \lambda x \le u_0 \\ g \left({\lambda x - u_0 + v_0}\right) &: \lambda x > u_0 \end{cases}$


 * [...] Then $h$ is an arc in $T$.


 * --abcxyz (talk) 01:01, 16 November 2012 (UTC)


 * By the way, does anyone know why "Category:Proofread" shows up so big on this (specific) page? --abcxyz (talk) 01:03, 16 November 2012 (UTC)


 * Nice proof (sketch)! I really like the topological approach. It appears to be correct - it's reassuring that this argument again uses Hausdorffness, strengthening the intuition that this is the right condition on $T$. As is so often the case, when specific knowledge of the ordering is at hand, one can avoid Zorn. Here it is no different. --Lord_Farin (talk) 12:32, 16 November 2012 (UTC)


 * Thanks; as you can see, a couple of results need to be put up first.
 * It seems like the double is causing the unusually big size. I've removed one of them. --abcxyz (talk) 17:50, 17 November 2012 (UTC)

Although Steen and Seebach define an arc as ProofWiki does, some others use a narrower definition, requiring an arc to be homeomorphic to $[0,1]$. See, e.g.,. Note that attempts to point out a serious error in S&amp;S with regard to arc connectedness, but I think may misinterpret one aspect of S&amp;S's definition. A response to that post,, states that
 * every Hausdorff path-connected space is arc-connected
 * The path is compact and metrizable (Hausdorffness helps); it is also locally connected, being the quotient of [0,1], now follow the proof of the Hahn-Mazurkiewicz theorem to construct an arc in the path that connects the begin and end points.

Assuming this is right, assuming Hausdorff in this theorem renders it silly. --Dfeuer (talk) 02:50, 21 December 2012 (UTC)


 * As follows from the assumption that ref. [2] is correct, the theorem is plainly false if we drop Hausdorffness (as the teleophase topology points out (which I also covered on the talk of that page)). The whole section of the site appears to be up for review. But not by me, I don't have the sources or the patience, nor do I enjoy it. I do however feel that any reworking should take place from a thorough investigation of the literature, which is to be referenced. --Lord_Farin (talk) 15:18, 21 December 2012 (UTC)

Remaking of Meet/Join theory
Currently the following subpages of this page are concerned with this (the italicized sentences these contain are instructions for what to do upon moving to main wiki):


 * User:Lord_Farin/Sandbox/Definition:Boolean Algebra

Further things to be done:
 * Appropriate categories holding results about these concepts need to be named and created.

List will be amended as work progresses. Pages listed here are subject to continuous and significant changes. --Lord_Farin (talk) 19:43, 19 December 2012 (UTC)


 * Don't forget about full lattices! They can also be seen as algebraic structures with meet and join, where they're join semilattices under join, meet semilattices under meet, and meet and join obey absorption laws: $a \vee (a \wedge b) = a \wedge (a \vee b) = a$. --Dfeuer (talk) 04:08, 20 December 2012 (UTC)


 * Let me put that differently: $(L, \preceq, \wedge, \vee)$ can be declared a lattice in one of two ways:
 * $(L, \preceq, \wedge)$ is a meet semilattice and $(L, \preceq, \vee)$ is a join semilattice, or
 * $(L, \wedge)$ and $(L, \vee)$ are semilattices, $\wedge$ and $\vee$ obey the absorption laws $a \vee (a \wedge b) = a \wedge (a \vee b) = a$, and $\preceq$ is defined in terms of join or meet.
 * What's annoying is that it seems like most folks either include join/meet or $\preceq$ as part of the structure, not both, and accommodating both options gets a tad awkward. --Dfeuer (talk) 04:24, 20 December 2012 (UTC)


 * I wasn't done yet, but thanks. I'll try to conceive of a structure that handles all these matters to a satisfactory extent. --Lord_Farin (talk) 08:02, 20 December 2012 (UTC)


 * This is my best take at it. What do you think? --Lord_Farin (talk) 18:36, 21 December 2012 (UTC)

Definitions pushed to main. Will spend time on streamlining this with existing material. --Lord_Farin (talk) 15:13, 31 December 2012 (UTC)


 * Things blend in quite naturally. Haven't gotten round to Boolean/Huntington Algebras; this will pose a considerable challenge. For now, I'll stick to the equivalence of the definitions I posted. Auxiliary results towards this goal are:
 * Ordering in terms of Join
 * Ordering in terms of Meet
 * probably more
 * It is very tempting to step down further and develop rigidly the duality principle for orderings, but I haven't made up my mind about how to structure this, and also don't want to be bogged down even more in the order theory department. In due time, it'll come - probably (hopefully?) accompanied by a proper approach to duality and dual theorems. The call for ideas in this regard remains open. --Lord_Farin (talk) 13:08, 4 January 2013 (UTC)


 * Oh screw. I'm going in and cover the duality stuff. --Lord_Farin (talk) 18:56, 4 January 2013 (UTC)


 * Hm. Only four days have passed and I've covered the essential parts. Many proofs can now be cut to two simple statements: Remarking a statement is dual to another; subsequently invoking the Duality Principle (Order Theory). Enjoy! --Lord_Farin (talk) 00:02, 8 January 2013 (UTC)