User:Lord Farin/Sandbox/Archive

This page contains stuff that was in my sandbox, sitting around doing nothing, but I didn't want to delete it.

= Weak/Strict Upper/Lower Closure =

These four concepts (cf. Weak Upper Closure) give rise to very similar results.

I suspect it would be most consistent to put them up separately for all four; what do you, reader, think?

Is there a page where it can be expressed that posets admit a duality principle (by passing to the opposite poset, reversing the associated relation)? --Lord_Farin 04:28, 6 April 2012 (EDT)


 * Same thing as we did for upper bound, strict upper bound, lower bound, strict lower bound, etc. Messy and tedious to do, but it looks quite good and works well.


 * There does exist a page somewhere about reversing the ordering, but only in the fairly general terms of inverse relation, and that an inverse ordering is also an ordering. If there's a need for more precision for specific purposes, then feel free to expand it. --prime mover 05:38, 6 April 2012 (EDT)


 * That is Inverse of Ordering is Ordering. --Lord_Farin 10:38, 6 April 2012 (EDT)

Ordering Duality
Let $\Phi$ be a theorem in the language of order theory.

Let $\Phi^{\text{op}}$ be the formula resulting from $\Phi$ by reversing all $\preceq$ signs into $\succeq$ signs.

Then $\Phi^{\text{op}}$ is also a theorem in the language of order theory.

Caution
When higher symbols like $\max$ are used, the $\preceq$ in their definitions also have to be reversed.

Fortunately, this may be accomplished by processing the following changes:


 * $\max \leftrightarrow \min$
 * $\sup \leftrightarrow \inf$
 * Greatest Element $\leftrightarrow$ Smallest Element
 * Maximal Element $\leftrightarrow$ Minimal Element

= Carathéodory's Theorem =

For every $A \subseteq X$, denote with $\mathcal C \left({A}\right)$ the collection of countable $\mathcal S$-covers $\left({S_n}\right)_{n \in \N}$.

Next, define the mapping $\mu^*: \mathcal P \left({X}\right) \to \overline{\R}$ by:


 * $\forall A \subseteq X: \mu^* \left({A}\right) = \inf \ \left\{{\displaystyle \sum_{n \in \N} \mu \left({S_n}\right): \left({S_n}\right)_{n \in \N} \in \mathcal C \left({A}\right) }\right\}$

Here, it is understood that the infimum is taken in the extended real numbers.

Hence, by Infimum of Empty Set, $\inf \varnothing = +\infty$.

Lemma 1
$\mu^*: \mathcal P \left({X}\right) \to \overline{\R}$ is an outer measure.

Lemma 2
For all $S \in \mathcal S$, have $\mu^* \left({S}\right) = \mu \left({S}\right)$

Proof
Next, define a collection $\mathcal A^*$ of subsets of $X$ by:


 * $\mathcal A^* := \left\{{A \subseteq X: \forall B \subseteq X: \mu^* \left({B}\right) = \mu^* \left({B \cap A}\right) + \mu^* \left({B \setminus A}\right)}\right\}$

Let $S,T \in \mathcal S$. Then reason as follows:

Would the result follow from Induced Outer Measure is Outer Measure, Induced Outer Measure Restricted to Semiring is Pre-Measure, Elements of Semiring are Measurable with Respect to Induced Outer Measure, and Outer Measure Restricted to Measurable Sets is Measure (once the pages are finished)? Are those the lemmas needed? –Abcxyz (talk | contribs) 14:47, 23 March 2012 (EDT)
 * Yes, and that Measurable Sets of Outer Measure form Sigma-Algebra or whatever it is called, which is already up. --Lord_Farin 18:43, 23 March 2012 (EDT)
 * It is Measurable Sets are a Sigma-Algebra of Sets. Should the page name be changed to what you (Lord Farin) wrote to have the explicit reference to an outer measure? –Abcxyz (talk | contribs) 19:16, 23 March 2012 (EDT)
 * Not at this point. That may be justified once the foundations and refactorisations are in place and we can get to properly naming pages. As of now, it would only at best be replacing the one idiosyncrasy with the other. Good job breaking this proof into multiple stages, each with merit for their own page. Do you mind posting them, too? --Lord_Farin 19:36, 23 March 2012 (EDT)
 * I wouldn't mind posting them. (I believe I have the proofs, unless I messed up somewhere.) Of course, I wouldn't mind anybody else posting them either. –Abcxyz (talk | contribs) 20:18, 23 March 2012 (EDT)
 * We're also going to have to include the part with uniqueness. By the way, I won't edit ProofWiki tomorrow because I'll be out of town. –Abcxyz (talk | contribs) 20:31, 23 March 2012 (EDT)

Uniqueness is just an application of Uniqueness of Measures; no problem there. --Lord_Farin 03:26, 24 March 2012 (EDT)

= Extended Reals as Two-Point Compactification =

There will be text here explaining this idea, which puts a compactifying topology on $\overline{\R}$, making the notions of diverging to $\pm \infty$ precise, and also allows for more rigorous treatment of, for example, convergence issues in measure theory.

Some search suggest that the required topology is the Definition:Order Topology, which apparently doesn't exist; I will try and write something down. Plainly, it's a topology on a toset generated by all the segments. Whoa, ideas start tumbling in, about $\max$ and $\min$, $\sup$, $\inf$ etc. etc continuous, morphisms in associated categories etc etc. I'd better be satisfied with Characterization of Measures as the milestone for today, or I won't sleep at all tonight. --Lord_Farin 18:59, 16 March 2012 (EDT)


 * Research done, searching for time to post the stuff. --Lord_Farin 14:42, 20 March 2012 (EDT)


 * Update: this project is on a stall as I lack a source of any kind; I will move it to the archive unless someone else has a source or is also knowledgeable on this, to avoid idiosyncrasies. --Lord_Farin 17:22, 30 March 2012 (EDT)

To be written:


 * Euclidean Space Subspace of Extended Real Number Space
 * Definition:Extended Real Addition
 * Definition:Extended Real Multiplication
 * Definition:Extended Real Subtraction
 * Definition:Order Topology
 * Definition:Order-Compatible Topology (better name desired)
 * Definition:Order Completion
 * I would like to write Completion Theorem (Posets), but I haven't verified that my construction works yet; it may appear here.


 * Definition:Complete Poset (every ascending/descending chain has upper/lower bound)
 * Definition:Extended Real Number Space
 * Ordering on Extended Real Numbers is Total Ordering
 * Extended Real Numbers form Commutative Monoid under Multiplication
 * Extended Real Numbers form Monoid under Multiplication
 * Extended Real Addition is Commutative, Extended Real Addition is Associative
 * Extended Real Multiplication is Commutative, Extended Real Multiplication is Associative
 * Extended Real Number Space is Compact
 * Infimum of Empty Set is Greatest Element
 * Supremum of Empty Set is Smallest Element
 * Loads of pages concerning divergence to infinity can be made rigorous
 * An infinitude awaits

= On continuous functions =

Let $\left({X, \left\Vert{\cdot}\right\Vert_X}\right)$ be a Banach space, and let $\left({Y, \left\Vert{\cdot}\right\Vert_Y}\right)$ be a normed vector space.

Let $f: X \to Y$ be a continuous function.

Let $\left({x_n}\right)_{n\in\N}$ be a bounded sequence in $X$.

Suppose that $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = y$, with $y \in Y$.

Let $f^{-1} \left({y}\right) := \left\{{x \in X: f \left({x}\right) = y}\right\}$ be the preimage of $y$ under $f$.

Assume that it is nonempty.

Then:

$\forall \epsilon > 0: \exists N \in \N: \forall n \in \N: n \ge N \implies \displaystyle \inf_{x \in f^{-1} \left({y}\right)} \left\Vert{x_n, x}\right\Vert_X < \epsilon$

For an example where the statement does not hold, consider the function $f : \Q \to \Q$ defined by $f\left({x}\right) = x^2$ if $x \le 0$ and $f\left({x}\right) = 2 x^2$ if $x \ge 0$. Then for any Cauchy sequence $\langle {a_n} \rangle$ of rational numbers that converges to $-\sqrt 2$, we have $\displaystyle \lim_{n\to\infty} f\left({a_n}\right) = 2$, but $f^{-1}\left({2}\right) = \left\{ {1} \right\}$.

I have a feeling that the statement is still false even if $X$ is a Banach space. As of now, I can’t prove or disprove it yet. Abcxyz 11:14, 10 March 2012 (EST)


 * Thanks for the comment. I will now try to write a proof. --Lord_Farin 11:23, 10 March 2012 (EST)

Disproof
Indeed, the statement is false even if $X$ is a Banach space. Here's the (dis)proof:

Consider the normed vector space $X$ over $\R$ given by the set of all continuous functions $\alpha : [0, 1] \to [0, 1]$, equipped with the supremum norm $\displaystyle \left\Vert {\alpha} \right\Vert_{\infty} = \sup_{x\in [0, 1]} \alpha\left({x}\right)$.

We now show that $X$ is a Banach space over $\R$. It remains to show that $X$ is a complete metric space.

Let $\alpha_1, \alpha_2, \alpha_3, \ldots: [0, 1] \to [0, 1]$ be a Cauchy sequence of continuous functions. (Here, the metric used is the metric induced by the supremum norm.)

Let $\displaystyle \alpha = \lim_{n\to\infty} \alpha_n$. It remains to show that $\alpha$ is continuous.

Let $\epsilon > 0$.

Then there exists an $N$ such that for all $n > N$, $\left\Vert \alpha_n - \alpha \right\Vert_{\infty} < \epsilon$.

By the definition of supremum norm, for all $x \in [0, 1]$, $\left\vert \alpha_n\left({x}\right) - \alpha\left({x}\right) \right\vert < \epsilon$.

Hence $\alpha$ is continuous by the uniform limit theorem.

Now, consider the function $f : X \to [0, 1]$ defined by $\displaystyle f\left({\alpha}\right) = \int_0^1 \alpha\left({x}\right) \,\mathrm{d}x$. We now show that $f$ is continuous.

Let $\alpha_0 \in X$, and let $\alpha \in X$ be such that $\left\Vert \alpha - \alpha_0 \right\Vert_{\infty} < \epsilon$.

Then:

Hence $f$ is continuous.

(Now for the counter-example to the statement.)

The pre-image of $\left\{ {0} \right\}$ under $f$ is the zero function.

Consider the sequence of continuous functions $\alpha_1, \alpha_2, \alpha_3, \ldots : [0, 1] \to [0, 1]$ defined by $\alpha_n\left({x}\right) = \max \left\{ {0, 1 - nx} \right\}$.

A straightforward calculation yields $\displaystyle \lim_{n\to\infty} f\left({\alpha_n}\right) = 0$.

However, $\left\Vert \alpha_n \right\Vert_{\infty} = 1$ for all $n \in \N$.

Abcxyz 12:21, 10 March 2012 (EST)

Remark
The problem (or whatever you want to call it) with the sequence $\alpha_1, \alpha_2, \alpha_3, \ldots$ used in the above (dis)proof was that it does not have a convergent subsequence. Maybe instead of assuming that the sequence $\langle {x_n} \rangle$ is bounded, we can assume that it has a convergent subsequence? Perhaps this could be the next guess? Abcxyz 12:53, 10 March 2012 (EST)


 * Thanks for this very detailed disproof of my assertion. Assuming a convergent subsequence won't work because we can riffle in the zero function on the even places in $x_n$ and repeat the argument above. Assuming a convergent subsequence does in fact allow to prove that $\displaystyle\liminf_{n\to\infty}\left\Vert{x_n-f^{-1} \left({y}\right)}\right\Vert = 0$; but that's a very weak statement. More thought is needed. --Lord_Farin 14:20, 10 March 2012 (EST)


 * I think I have the best statement possible: if the subsequence $x_{n_r}$ converges, then $\displaystyle\lim_{r\to\infty} x_{n_r} \in f^{-1}\left({y}\right)$. Verification is straightforward; now to think of uses. --Lord_Farin 14:23, 10 March 2012 (EST)

Specific cases
How about the case where $X$ is a Euclidean space? Does the statement hold in that case? Any thoughts? (I have a feeling that the statement does actually hold in that case, but of course I might be dead wrong.) Abcxyz 15:39, 10 March 2012 (EST)


 * Interesting thought. I thought of an example that shows it is at least required, even in this case, to assume that the sequence is bounded. Namely, define:


 * $f \left({x}\right) = \begin{cases}0 & \text{if } x \in \N \text{ or } x\le 0\\

\frac1{n+1} & \text{if }x = n + \dfrac12 \text{ with } n \in \N\end{cases}$


 * and subsequently by straight line segments between these points. The sequence forming a counterexample is $x_n = n+\dfrac12$.


 * Returning to the case where the sequence is bounded, I think it may be possible to create a 'partition of subsequences' which covers the whole sequence and whose limits are of course the preimages of $y$. To avoid the axiom of choice and cases where the preimage set isn't discrete, probably the sequences (even their indices) shouldn't be assumed disjoint. I will need to think about this some more, but it appears to work (assuming the function to be defined on some closed subset of $\R^n$ to ensure no problems occur at the boundaries with nasty asymptotic behaviour). --Lord_Farin 18:11, 10 March 2012 (EST)

Actually, I think the statement holds in the case that every closed ball in $X$ is sequentially compact.

Let $\left(X, \left\Vert \cdot \right\Vert \right)$ be a Banach space satisfying the above property, and let $\left(Y, d\right)$ be a metric space.

Since we only concern ourselves with bounded sequences $\langle {x_n} \rangle$ in $X$, we may assume the function $f$ is defined on some closed ball in $X$ which the sequence $\langle {x_n} \rangle$ is constrained to be in.

Suppose that the statement is false. Then there exists an $\epsilon > 0$ such that there exist arbitrarily large $n$ with $\displaystyle \inf_{x \in f^{-1}\left({y}\right)}\left\Vert {x_n - x}\right\Vert \ge \epsilon$.

Therefore, there exists a subsequence $\langle {x_p} \rangle$ of $\langle {x_n} \rangle$ such that for all $p \in \N$, $\displaystyle \inf_{x \in f^{-1}\left({y}\right)} \left\Vert {x_p - x} \right\Vert \ge \epsilon$.

By the definition of sequential compactness, there exists a convergent subsequence $\langle {x_q} \rangle$ of $\langle {x_p} \rangle$.

For all $q \in \N$, $\displaystyle \inf_{x \in f^{-1}\left({y}\right)} \left\Vert {x_q - x} \right\Vert \ge \epsilon$.

Let $\displaystyle x^\star = \lim_{q \to \infty} x_q$.

By the continuity of $f$, we have $y = f\left({x^\star}\right)$.

By the definition of pre-image, $x^\star \in f^{-1}\left({y}\right)$.

But then there exists a $q \in \N$ such that $\left\Vert {x_q – x^\star} \right\Vert < \epsilon$, contradicting the assumption that $\displaystyle \inf_{x \in f^{-1}\left({y}\right)} \left\Vert {x_q - x} \right\Vert \ge \epsilon$ for all $q \in \N$.

Hence the statement holds.

Abcxyz 21:41, 10 March 2012 (EST)

So, if $\left( {X, \left\Vert \cdot \right\Vert} \right)$ is a Banach space, $\left( {Y, d} \right)$ is a metric space, and the terms of the sequence $\langle {x_n} \rangle$ are members of a sequentially compact subset of $X$, then the statement holds. Abcxyz 22:59, 10 March 2012 (EST)


 * I came to the rather similar conclusion that the property holds for any sequence in a complete, sequentially compact metric space. My proof is quite analogous and also is a proof by contradiction. I haven't ever seen any similar result even stated; however, I can imagine that it's sometimes useful when the preimage is a single point. Anyone else ever saw a result similar to this? --Lord_Farin 10:56, 11 March 2012 (EDT)


 * Note that any sequentially compact metric space is automatically complete. --Lord_Farin 18:14, 13 March 2012 (EDT)


 * Hope you don't mind - I updated the links to Preimage so as to be consistent with the refactoring. --prime mover 05:11, 18 March 2012 (EDT)

= Graph Theory axiomatisation =

I thought of a sort of viable system; it might seem cumbersome, but I think it provides first-order rigidity (up to a point, of course).

Let $\mathsf{Graph}$ be the following language:


 * It has three relations:
 * The unary relation $\mathsf V$, intended to mean $\mathsf V x$ iff $x$ is a vertex
 * The unary relation $\mathsf E$, intended to mean $\mathsf E x$ iff $x$ is an edge
 * The ternary relation $\mathsf{Edge}$, intended to mean $\mathsf{Edge}(x,y,z)$ iff $x$ is an edge with end points $y, z$ (be it in specified order or not, that can be chosen by adding axioms)


 * It has no function symbols


 * It is subject to the following axioms:
 * $\forall x: \mathsf V x \lor \mathsf E x$ (everything is a vertex or an edge)
 * $\forall x: \neg \left({\mathsf V x \land \mathsf E x}\right)$ (nothing is both a vertex and an edge)
 * $\forall x,y,z: \mathsf{Edge} \left({x, y, z}\right) \implies \left({\mathsf E x \land \mathsf V y \land \mathsf V z}\right)$ (expressing what each component of $\mathsf{Edge}$ is)
 * $\forall x,y,z,\tilde y, \tilde z: \mathsf{Edge} \left({x, y, z}\right) \land \mathsf{Edge} \left({x, \tilde y, \tilde z}\right) \implies \left({ \left({y = \tilde y \land z = \tilde z}\right) \lor \left({y = \tilde z \land z = \tilde y}\right) }\right)$ (an edge concerns itself with at most two vertices)
 * $\forall x: \mathsf E x \implies \exists y,z: \mathsf{Edge} \left({x, y, z}\right)$ (there are no void edges)

In principle, this will be enough (I think). However, one can add:


 * $\forall x,y,z: \mathsf{Edge} \left({x, y, z}\right) \implies \mathsf{Edge} \left({x, z, y}\right)$ (signifying an undirected graph)
 * $\forall x,y,z: \mathsf{Edge} \left({x, y, z}\right) \implies \neg \mathsf{Edge} \left({x, z, y}\right)$ (a directed graph)
 * $\forall x,y,z: \mathsf{Edge} \left({x, y, z}\right) \implies \neg \left({y = z}\right)$ (a loop-free graph)
 * $\forall x,y,z,\tilde x: \mathsf{Edge} \left({x, y, z}\right) \land \mathsf{Edge} \left({\tilde x, y, z}\right) \implies x = \tilde x$ (a non-multigraph)

I have discovered that it is hard to formulate the notion of a path in first-order logic. This isn't strange, it's like saying that a set has cardinality $n$. An $n$-path can be achieved though.

I have put a moment's thought into dropping axiom 4. It may be interesting for investigation of, for example, non-injective maps (described by an edge with multiple starts... well, you get it).

The unary symbols $\mathsf E, \mathsf V$ combined with axiom 1,2 are effectively splitting the set that is a model into two parts, similar to the ordered pair used in the definition at the moment. What do you think? --Lord_Farin 17:59, 20 February 2012 (EST)
 * That seems like a reasonable axiomatization, though I'm not sure if it's really necessary. It seems at first blush like it would make a number of proofs much more annoying. I'll try to find a good book on digraphs and see how they manage it, and perhaps talk to someone. Scshunt 23:21, 20 February 2012 (EST)
 * What proofs exactly? One only needs to take care about what $\mathsf{Edge}$ does differently from normal, if it does any such thing. I'm just curious; you have of course every right to criticism. --Lord_Farin 02:56, 21 February 2012 (EST)