User linda brown westrick - MathOverflow

Mutually tangent ellipsoids in 3 space

2012-11-30T00:04:07Z

I recently heard a claim that for any n, it is possible to arrange n ellipsoids in 3 space such that each pair of ellipsoids is kissing. Is this true, and if so, how?

Edit: By kissing, I mean that I would like the interiors of the ellipsoids to be disjoint, but each pair of ellipsoids should intersect at a point.

Answer by Linda Brown Westrick for A question about recursively enumerable sets of rational numbers

2012-10-13T19:23:05Z

The ordering (f(N*),<) that you describe is a recursive ordering, so if it is well-founded, it has the order type of a recursive ordinal. It is known (see e.g. Higher Recursion Theory by Sacks) that the recursive ordinals coincide with the constructive ordinals. Therefore, the order type is constructive.

Answer by Linda Brown Westrick for Musings in set theory: Reverse sets?

2011-05-28T02:09:50Z

The assumption is incorrect because the reverse graph H(X) might not be extensional (i.e. it might not satisfy $(\forall z (z,x) \leftrightarrow (z,y)) \rightarrow x=y$). An example is $X=\{\{1\},\{2\}\}$.

When checking if a harmonic function is continuous on its boundary, is a dense subset enough?

2011-05-24T23:03:12Z

Let $U$ be an open connected subset of $\mathbb{C}$ and let $u:U\rightarrow \mathbb{R}$ be harmonic and bounded on $U$.

Let $f:\partial_\infty U \rightarrow \mathbb{R}$ be a continuous function, and suppose $u$ extends continuously to $f$ on a dense subset of $\partial_\infty U$.

By $\partial_\infty U$ I mean the boundary of $U$ in $\mathbb{C}\cup\{\infty\}$. By "$u$ extends continuously to $f$ on $D$" I mean that for each $a \in D$, $\lim_{z\rightarrow a} u(z)$ exists and is equal to $f(a)$.

Does it follow that $u$ extends continuously to $f$ on all of $\partial_\infty U$?

Characterize where the Dirichlet Problem for the Laplacian is always solvable

2011-05-24T22:28:16Z

Conway's 1978 textbook Functions of One Complex Variable I gives an unsatisfying characterization of the regions for which the Dirichlet Problem can always be solved, and then comments no cleaner characterization is known. Has any progress been made since then? And what simpler characterization is known today, if one is known?

Here is the problem definition:

An open connected set $G\subseteq \mathbb{C}$ is called a Dirichlet Region if for each continuous function $f:\partial_\infty G\rightarrow \mathbb{R}$ there is a continuous function $u:G^- \rightarrow \mathbb{R}$ such that $u$ is harmonic in $G$ and $u(z)=f(z)$ for all $z$ in $\partial_\infty G$.

(The notation $\partial_\infty G$ refers to the boundary of $G$ in $\mathbb{C}\cup\{\infty\}$, and $G^-$ denotes the closure of $G$ in $\mathbb{C}\cup\{\infty\}$.)

The characterization given in the book is:

Given $a \in \partial_\infty G$, a barrier for $G$ at $a$ is a family $\{\psi_r: r>0\}$ of functions such that:
1. $\psi_r$ is well-defined and superharmonic on $B(a;r) \cap G$ with $0\leq \psi_r(z) \leq 1$
2. $\lim_{z\rightarrow a}\psi_r(z) = 0$, and
3. $\lim_{z\rightarrow w} \psi_r(z) = 1$ for $w$ in $G \cap \{w:|w-a|=r\}$.

An open connected set $G$ is a Dirichlet Region iff there is a barrier for $G$ at each point of $\partial_\infty G$.

Meaning of \Subset notation

2010-10-28T07:16:49Z

The symbol $\Subset$ (occurring in places where $\subseteq$ could occur syntactically) comes up frequently in a paper I'm reading. The paper lives at the intersection of a few areas of math, and I don't even know where to begin looking for the meaning of a symbol whose latex code is "\Subset". Do you know what this usually denotes?

Edit: some context follows.

All the sets in question are subsets of $\hat{\mathbb{C}} = \mathbb{C}\cup\{\infty\}$.

Example 1. In a situation where $J$ is closed with empty interior, $U$ and $V$ are closed with $U\subsetneq V$, it is written "Note that $J \Subset U$ and, selecting a neighborhood $W \subset U$ of $J$ which is compactly contained in $V$, ..."

Example 2. In a situation where $R$ is a rational mapping, and where it is assumed that $B\subset \hat{\mathbb{C}}$ is such that $R(B)\Subset B$, it is written "Let $\Omega_0 = \hat{\mathbb{C}}\setminus B$. Define $\Omega_1 = R^{-1}(\Omega_0)$. By the properties of $B$, we have $\Omega_1\Subset\Omega_0$. If we let $U_0$ be any finite union of closed balls such that $\Omega_1 \subset U_0 \subset \Omega_0$, ..."

In both cases I have paraphrased to simplify the notation, so I hope I have not introduced errors into it.

Answer by Linda Brown Westrick for Arithmetic fixed point theorem

2010-07-11T08:17:54Z

Let's start out with the observation that there can be no formula $D$ with the property that for all $\varphi$, $$D([\varphi]) \iff \varphi([\varphi]).$$ For if such a $D$ existed, then defining the formula $E$ by $E(\underline{n}) = \neg D(\underline{n})$, we would have $$D([E]) \iff E([E]) \iff \neg D([E]),$$ a contradiction.

Now, the task is to show that given a formula $F$ of one variable, there is another formula $A$ such that $A\iff F([A])$. Well, if that's not true, then an improbable-looking thing would happen: for every sentence $A$, we would have $$\neg F([A])\iff A.$$

The reason this looks improbable is that the formula $\neg F$ looks fairly similar to the forbidden formula $D$ above. In fact, if I want to juice the similarity for all it's worth, I would explore what happens when $A$ is of the form $A = \varphi([\varphi])$ for some $\varphi$; then we would have $$\neg F([\varphi([\varphi])]) \iff \varphi([\varphi]).$$

But if this holds for all $\varphi$, we can define the forbidden $D$ by $D([\varphi]) = \neg F([\varphi([\varphi])])$. Contradiction.

Now out of this argument, let's extract the specific formula $A$ that we originally wanted. We are looking for an $A$ of the form $\varphi([\varphi])$. Our hint is to use the same formula $E([E])$ which destroyed our hopes about $D$ above. The context is new, but we define $E$ in the same way: $E([\varphi]) = \neg D([\varphi])$. That is $$E([\varphi]) = F([\varphi([\varphi])).$$

Now the only step that remains is the one you've already seen: checking that $A=E([E])$ in fact works.

Sneaky Recursive Non-Well-Orders

2010-07-05T15:45:29Z

Background

An ordinal $\alpha$ is called a recursive ordinal if there is a recursive well-order $R$ on $\mathbb{N}$ such that ordertype($\mathbb{N},R) = \alpha$. For example, $2\omega$ is a recursive ordinal because the ordering of $\mathbb{N}$ as 0, 2, 4, 6, 8, ... 1, 3, 5, 7, ... is computable and has order type $2\omega$.

Kleene encoded the recursive ordinals in the natural numbers in a nifty way which is described at the Wikipedia page on Kleene's O. Now Kleene's $\mathcal{O}$ is a fairly powerful set -- given a Turing machine index for a linear order, $\mathcal{O}$ can decide whether that ordering is a well-ordering or not.

Using Kleene's $\mathcal{O}$, it is possible to describe how to iterate the Turing jump through the recursive ordinals. For each natural number $a\in\mathcal{O}$, we can define a set $H_a$ recursively as follows:

$H_a = \emptyset$ if $a=0$
$H_a = {H_b}'$ if $a=2^b$
$H_a = \{\langle n, x \rangle | x \in H_{\phi_e(n)} \}$ if $a = {3\cdot 5^e}$

For each $a\in \mathcal{O}$, we have $H_a$ <$_T \ \mathcal{O}$ (strict inequality), and no $H_a$ is powerful enough to decide which recursive orders are well-orders.

Question

Among recursive non-well-orders, some hide their descending chains better than others do.

For example, if we only wanted to flag the non-well-orders sporting a recursive descending chain, the full power of $\mathcal{O}$ would not be necessary -- $\emptyset'''$ would do $(\exists e [ \phi_e$ is total and $\forall n [\ \phi_e(n+1)$ <$_R\ \phi_e(n)\ ]]$?). Thus there is a recursive linear non-well-order with no recursive descending chain.

In fact (by similar reasoning), for each $a \in \mathcal{O}$ there must be a recursive linear non-well-order with no recursive-in-$H_a$ descending chain.

I wonder whether we could effectively construct these sneaky recursive non-well-orders.

Is there a recursive function $f$ such that whenever $a\in\mathcal{O}$, $f(a)$ is a Turing index for a linear non-well-order with no $H_a$ -computable descending chain?

When does replacement (accidentally) hold in amenable sets?

2010-06-12T18:49:18Z

A set $M$ is called amenable if it is transitive and satisfies the following conditions:

For all $x,y\in M$, $\{x,y\}\in M$
For all $x\in M$, $\bigcup x \in M$
$\omega \in M$
For all $x,y \in M$, $x\times y \in M$
($\Sigma_0$ comprehension) Whenever $\Phi$ is a $\Sigma_0$ formula of one free variable with parameters from $M$, then for all $x\in M$, $\{z\in x | \Phi(z)\}\in M$

Although the definition of an amenable set does not include replacement, some very limited amount of replacement follows from the axioms given. For example, for all $x,y\in M$, it must be that $\{\{z,w\}|z\in x, w\in y\}\in M$ and $\{\{z\}|z\in x\}\in M$. So just how limited is the replacement in amenable sets? In particular,

If $M$ is an amenable set and $x\in M$, does it follow that $\{\bigcup z | z \in x\} \in M$?

Comment by Linda Brown Westrick

2012-12-01T01:33:13Z

Thanks for the comments and ideas. I added in what I mean by kissing, which, as many guessed, includes disjoint interiors. If I should have used another word, please let me know!

Comment by Linda Brown Westrick

2011-05-27T02:07:24Z

Thank you. I only knew of the Poisson extension for continuous functions, but it seems that it works for $L^p$ more generally. Is there a standard reference for the Poisson extension for $L^p$ functions?

Comment by Linda Brown Westrick

2011-05-25T22:43:07Z

@Yakov Ah, thank you.

Comment by Linda Brown Westrick

2011-05-25T22:36:17Z

@Nilima Yes, I would be very interested in a simple characterization that works only for the plane. Geometric conditions are often the right kind of simple. Now to be more specific about "simple": most of math can be formalized in Second Order Arithmetic; see en.wikipedia.org/wiki/Second-order_arithmetic for its syntax. The formula which says "G is a Dirichlet Region" has two set quantifiers (for all continuous $f$, there exists a harmonic $u$ such that...). Reduction to a single set quantifier would be useful, and an arithmetic characterization would be especially simple.

Comment by Linda Brown Westrick

2011-05-25T21:58:32Z

@Will I think I found the theorem from McKernan you mentioned at math.ucsb.edu/~mckernan/Teaching/06-07/Spring/… and it is a nice sufficient condition, but there is no claim there that it is necessary. (note: Theorem 5.5 is found in a set of McKernan's lecture notes, and no attribution is given there.) It would be interesting to see an example of a Dirichlet Region which did not meet this condition.

Comment by Linda Brown Westrick

2011-05-25T21:39:07Z

@Yakov There is a discrepancy between the cusp failure-example you are offering and Corollary X.4.18 of Conway's Functions of One Complex Variable I, which asserts that "a simply connected region is a Dirichlet Region". I am not familiar enough with potentials to be able to follow the example you referenced from Jost, but since I can see that the region is simply connected, perhaps there is some difference in the context between Jost and Conway.

Comment by Linda Brown Westrick

2011-05-25T20:45:01Z

Thanks to those who have offered elegant sufficient conditions. But as Will suggested, I am indeed interested in exactly classifying the "always" regions. My interest is from mathematical logic. When I come across an unwieldy characterization which has resisted simplification for a long time, I often wonder if that unwieldiness is unavoidable. In logical terms, what I mean by unwieldiness can be made precise: the definition of a Dirichlet Region is syntactically $\Pi^1_2$, and on its face the barrier characterization is also syntactically $\Pi^1_2$; i.e. just as complicated.

Comment by Linda Brown Westrick

2011-05-25T20:20:57Z

@Nilima Thank you for your argument which shows why the boundary regularity condition characterizes the Dirichlet Regions. My question is asking something different, though. The boundary regularity condition, while necessary and sufficient, is a complicated condition. I am wondering if someone has found a simpler condition which also characterizes the Dirichlet Regions.

Comment by Linda Brown Westrick

2010-10-30T19:30:06Z

It sounds like this answer is the consensus. Thanks all!

Comment by Linda Brown Westrick

2010-10-28T18:31:13Z

Taking this answer as a starting point and comparing it with the examples I just added, it seems a related possibility for $U\Subset V$ might be "$U$ has a neighborhood which is compactly contained in $V$". Has anyone seen it used this way?