User:Lord Farin/Sandbox

This page exists for me to be able to test out 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.

Improvement of Sequence of Implications of Connectedness Properties
For brevity, let us introduce the following acronyms:


 * $\mathrm{AC}$: Arc-Connected
 * $\mathrm{UC}$: Ultraconnected
 * $\mathrm{PC}$: Path-Connected
 * $\mathrm{HC}$: Hyperconnected
 * $\mathrm C$:  Connected

Then the following sequence of implications holds:


 * $\begin{xy}

<-3em,3em>*+{\mathrm{AC}} = "AC", <3em,3em>*+{\mathrm{UC}} = "UC", <0em,0em>*+{\mathrm{PC}} = "PC", <6em,0em>*+{\mathrm{HC}} = "HC", <3em,-3em>*+{\mathrm C}  = "C",

"AC";"PC" **@2{-} ?><>(1.2)*@2{>}, "UC";"PC" **@2{-} ?><>(1.2)*@2{>}, "PC";"C" **@2{-} ?><>(1.2)*@2{>}, "HC";"C" **@2{-} ?><>(1.2)*@2{>}, \end{xy}$

What do you think? Now that I have created the moulds, it will be easy to adapt to the other 'Sequences of Implication'. --Lord_Farin (talk) 10:21, 31 August 2012 (UTC) It be noted that it will forever be impossible to endow diagrams (and indeed, any TeX rendered with MathJax) with internal links; sorry. --Lord_Farin (talk) 10:22, 31 August 2012 (UTC)


 * No response? :( --Lord_Farin (talk) 21:31, 7 September 2012 (UTC)


 * It's very nice! I don't know the first thing about what it meant, though :) --GFauxPas (talk) 21:53, 7 September 2012 (UTC)


 * Sorry, only just noticed it. Very nice - one caveat: you need to refer to the key to (a) work out what the codes mean, and (b) to get to the link explaining them. The somewhat clumsier page from which the original of this came does have the full map as one self-contained unit. Might be interesting to put the two presentations up on the same page as alternative renditions. --prime mover (talk) 22:21, 7 September 2012 (UTC)

Reading through the MathJax development thread revealed a way to implement internal links, contrary to my rather definitive statement above. Thus, another attempt:


 * $\begin{xy}\xymatrix@C-1em@R-0.5em{

\href{../wiki/Definition:Arc-Connected}{\mathrm{AC}} \ar@2{->}[rd] & & \href{../wiki/Definition:Ultraconnected}{\mathrm{UC}} \ar@2{->}[ld]

\\ & \href{../wiki/Definition:Path-Connected}{\mathrm{PC}} \ar@2{->}[rd] & & \href{../wiki/Definition:Hyperconnected}{\mathrm{HC}} \ar@2{->}[ld]

\\ & & \href{../wiki/Definition:Connected (Topology)}{\mathrm{C}} }\end{xy}$

I like it. Maybe the only downside is that I discovered XyJax to only work with HTML/CSS rendering (but as this is default, I don't think it'll pose a lot of problems). --Lord_Farin (talk) 11:29, 5 November 2012 (UTC)

Also, the links won't turn red when you mistype them (due to their hardcoded, non-wiki nature); it's good to be aware of that quirk. --Lord_Farin (talk) 11:30, 5 November 2012 (UTC)


 * Thoughts? If desired the acronyms can be expanded. --Lord_Farin (talk) 20:51, 9 November 2012 (UTC)

Principle of Induction on WFFs
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.)

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.


 * 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)
 * --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)