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.

Generated Sigma-Algebras
Let $X$ be a set, and let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a collection of subsets of $X$.

Let $A \subseteq X$ be a subset of $X$.

Then we have the following equality of $\sigma$-algebras on $A$:


 * $\sigma \left({\mathcal G}\right)_A = \sigma \left({A \cap \mathcal G}\right)$

where $\mathcal{A}_A$ denotes the trace $\sigma$-algebra, and $\sigma \left({\mathcal G}\right)$ denotes the $\sigma$-algebra generated by $\mathcal G$.

Comment
I can prove this, but I need a rather technical result (of which I have a reference) that generated sigma-algebras can be obtained by transfinite induction to be able to apply distributivity of intersection. I would rather like to use more elementary means and save the characterisation of generated sigma-algebras for a later moment. Does anyone have an idea (one inclusion is trivial)? --Lord_Farin 06:56, 15 March 2012 (EDT)

Set Intersection Preserves Subsets (General Case)
Let $\left({A_i}\right)_{i \in I}, \left({B_i}\right)_{i \in I}$ be collections of sets.

Suppose that for all $i \in I: A_i \subseteq B_i$.

Then:


 * $\displaystyle \bigcap_{i \in I} A_i \subseteq \bigcap_{i \in I} B_i$

Corollary
Suppose that for all $i \in I: A_i = A$ for some set $A$.

Then:


 * $\displaystyle A \subseteq \bigcap_{i \in I} B_i$

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)

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)