## A Funny Field

Let $\omega$ denote the first infinite cardinal – that is, the set of non-negative integers. Let $p_0 = 2$ be the smallest prime number, and let $(p_n)_{n\in\omega}$ enumerate all prime numbers in ascending order.

Let $\mathcal{U}$ be a free ultrafilter on $\omega$. We consider the field

$F = \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{\mathcal U}.$

This uses the standard notation for considering the equivalence relation $\simeq_{\mathcal U}$ on $\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}$ where we have $x\simeq_{\mathcal U}y$ for $x,y \in \prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}$ if and only if $\{i\in\omega: x(i) =y(i)\} \in \mathcal U$. (It is easy to verify that this is an equivalence relation.) On $\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}$ we use component-wise addition and multiplication. It is a standard exercise to show that whenever $x\simeq_{\mathcal U} x'$ and $y\simeq_{\mathcal U} y'$ then $(x+y)\simeq_{\mathcal U} (x'+y')$, and the same holds for multiplication. So the operations are well-defined on the quotient $F = \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\mathcal U} =\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{\mathcal U}$, and it is another standard exercise to show that $F$ is indeed a field (as opposed to $\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}$, which has lots of zero divisors). Moreover, $F$ is uncountable and has characteristic 0.

I don’t know if and where this field has been studied, or if there is a well-known field that is isomorphic to $F$.

There are several questions I cannot answer, and I would be grateful for any hints on them.

1. If we take different free ultrafilters ${\mathcal U}_1, {\mathcal U}_2$, can it happen that $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{{\mathcal U}_1}$ is not isomorphic to $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{{\mathcal U}_2}$?
2. If yes: Suppose ${\mathcal U}_1, {\mathcal U}_2$ are free ultrafilters such that $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{{\mathcal U}_1}$ and $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{{\mathcal U}_2}$ are isomorphic fields. What can be said about ${\mathcal U}_1, {\mathcal U}_2$? For instance, do they have to be in relation with respect to the Rudin-Keisler ordering?
3. Can $F$ be made into an ordered field?
4. Are there surjective group homomorphisms from the additive group of $F$ to the additive group of $\mathbb{R}$, or vice versa? What about the multiplicative group of $F\setminus \{0\}$?

## Making critical graphs as regular as possible

Suppose you want to have a graph $G = (V,E)$ with chromatic number $\chi(G)$ equaling some value $k$, such that $G$ is minimal with this property. So you end up with a $k$-(vertex-)critical graph.

It is easy to construct critical graphs by starting with some easy-to-verify example like $C_5$ and then adding points and connecting them to all the vertices already present. But the graphs you get are highly non-regular: some vertices have quite low degree, and the one vertex you added has maximum degree.

I was wondering where the limits for regularity for $k$-critical graphs are. Here’s an attempt to make this a bit more formal:

For a finite, simple, undirected graph $G=(V,E)$ let $\delta(G)$ and $\Delta(G)$ denote the minimum and maximum degree of $G$, respectively.

Is there a global constant $K\in\mathbb{N}$ such that whenever $n,k$ are integers with $n\geq 4, k\geq 1$ and $n>k$, there is a $k$-vertex critical graph $G=(V,E)$ with $|V|=n$ and $\Delta(G)-\delta(G) \leq K$?

I asked this question on MathOverflow a while ago, but it has not been answered yet.

## Generalizing the T_0 separation axiom

The starting point of this blog post is a slight reformulation of the $T_0$ separation axiom: A topological space $(X,\tau)$ is $T_0$ if for all $x\neq y\in X$ there is a set $U\in \tau$ such that

$\{x,y\}\cap U \neq \emptyset \text{ and } \{x,y\}\not\subseteq U.$

Given a cardinal $\kappa \geq 2$, we say that a space $(X,\tau)$ is $T^{\kappa}_0$ if for all subsets $S\subseteq X$ with $|S|=\kappa$ there is a set $U\in \tau$ such that $U$ “splits” $S$, or more formally

$S\cap U \neq \emptyset \text{ and } S\not\subseteq U.$

Obviously, if $\lambda\geq \kappa\geq 2$ and if $(X,\tau)$ is $T^\kappa_0$, then $X$ is also $T^\lambda_0$. We say, the space $(X,\tau)$ is minimally $T^\kappa_0$ if it is $T^\kappa_0$, but for all cardinals $\alpha<\kappa$ with $\alpha\geq 2$, the space $(X,\tau)$ is not $T^\alpha_0$.

Question. Given cardinals $\lambda\geq\kappa\geq 2$, is there a topological space $(X,\tau)$ such that $|X|=\lambda$ and $(X,\tau)$ is minimally $T^\kappa_0$?

## Basics on towers on the natural numbers

For $A, B \subseteq \omega$ we write $A \subseteq^* B$ if $A\setminus B$ is finite, and we write $A\simeq^* B$ if $A\subseteq^* B$ and $B\subseteq^* A$.

A tower is a collection ${\cal T}$ of co-infinite subsets of $\omega$ such that for all $A\neq B\in {\cal T}$ we have $A\not\simeq^*B$ and either $A\subseteq^* B$ or $B \subseteq^* A$. ($A\subseteq \omega$ is co-infinite if $\omega\setminus A$ is infinite.)

If ${\cal S}, {\cal T}$ are towers, we say that ${\cal S}\leq_t {\cal T}$ if $A\in {\cal S}$ and $B\in{\cal T}$ with $B\subseteq^* A$ jointly imply that $B\in{\cal S}$. (In other words, this means that ${\cal S}$ is a down-set or initial segment of ${\cal T}$ with respect to $\subseteq^*$). It is easy to prove that $\leq_t$ is a partial order on the collection of all towers on $\omega$.

The remainder of this post is about maximal towers with respect to $\leq_t$. The proof of the following lemma is routine.

Lemma 1. If ${\frak T}$ is a collection of towers, such that for all ${\cal S}, {\cal T}\in {\frak T}$ we have either ${\cal S}\leq_t {\cal T}$ or ${\cal T}\leq_t {\cal S}$, then

1. $\bigcup {\frak T}$ is a tower, and
2. ${\cal T}\leq_t \bigcup{\frak T}$ for all ${\cal T}\in{\frak T}$.

Corollary 1. Zorn’s Lemma and Lemma 1 imply that there is a tower in $\omega$ that is maximal with respect to $\leq_t$.

Lemma 2. If ${\cal A}$ is a countable tower then ${\cal A}$ is not maximal.

Proof. Let $(A_n)_{n\in\omega}$ be a sequence of co-infinite subsets of $\omega$ such that for all $n\in\omega$ we have $A_n \subseteq^* A_{n+1}$ . We want to show there is $A\subseteq \omega$ co-infinite with $A_n\subseteq^* A$ for all $n\in\omega$.

Step 1. If $k\in \omega$, then $\bigcup_{i=0}^k A_i$ is co-infinite.

Step 2. There is $f:\omega\to\omega$ strictly increasing such that $f(n) \in \omega\setminus\big(\bigcup_{i=0}^n A_i\big)$ for all $n\in\omega$.

Step 3. Set $A = \bigcup_{k\in\omega}\big(A_k\setminus [0,\ldots,f(k)]\big).$

Then it follows that

1: $A_k\subseteq^* A$ for all $k\in \omega$ since $(A_k\setminus A) \subseteq [0,\ldots,f(k)]$ which is a finite set.

2: $A$ is co-infinite, since for all $k\in \omega$ we have $f(k)\notin A$, so $A\cap \{f(k):k\in\omega\}=\emptyset$, and since $f$ is strictly increasing we have $\{f(k):k\in\omega\}$ is infinite, so $A$ is co-infinite.

Letting ${\cal A} =\{A_n:n\in\omega\}$ we get that ${\cal A}' = {\cal A} \cup\{A\}$ is a tower with ${\cal A} \leq_t {\cal A}'$ but ${\cal A}' \not\leq_t {\cal A}$, so the countable tower ${\cal A}$ is not maximal. $\Box$

## A definition of minor (in graph theory)

Many people I talk to about graph theory feel some uneasiness when it comes to the notion of “minor”. I want to try to alleviate this feeling by providing the definiton of minor that I work with.

First an easy definition. If $G$ is a simple, undirected graph and $S, T\subseteq V(G)$ are non-empty and disjoint, we say that $S, T$ are connected to each other if there are $s\in S, t\in T$ such that $\{s,t\}\in E(G)$.

Let $G, H$ be simple, undirected graphs. We say that $G$ is a minor of $H$ if there is a collection ${\cal S}$ of non-empty, mutually disjoint, and connected subsets of $V(H)$ and a bijection $\varphi:V(G) \to {\cal S}$ such that

whenever $v,w\in V(G)$ and $\{v,w\}\in E(G)$ then the sets $\varphi(v)$ and $\varphi(w)$ are connected to each other in $H$.

## Any graph or its complement is connected

This is a short note of something that I wasn’t sure whether it is true for infinite graphs. Let $G = (V,E)$ be any simple, undirected graph, finite or infinite, such that $V \neq \emptyset$. By $\bar{G}$ we denote the complement of $G$.

Proposition. At least one of $G, \bar{G}$ is connected.

Proof. If $V$ has only 1 element, the statement is trivially true. So suppose that $|V| > 1$. Let $\kappa = |V|$ and let $\varphi:\kappa\to V$ be a bijection. Suppose that neither $G$ nor $\bar{G}$ is connected. It is easy to see that this implies that there is a smallest $\alpha_0\in\kappa$ such that the induced subgraph $G_{\alpha_0}$ on the set $\text{im}(\varphi|_{\{0,\ldots,\alpha_0\}}) = \text{im}(\varphi|_{\alpha_0\cup \{\alpha_0\}}) \subseteq V$ has the property that neither $G_{\alpha_0}$ nor its complement $\bar{G}_{\alpha_0}$ is connected.

Case 1. $\varphi(\alpha_0)$ has no neighbors in the graph $G_{\alpha_0}$. But then $\varphi(\alpha_0)$ is connected to every vertex in $\bar{G}_{\alpha_0}$, so $\bar{G}_{\alpha_0}$ is connected, contradicting our assumption.

Case 2. $\varphi(\alpha_0)$ has no neighbors in the graph $\bar{G}_{\alpha_0}$. Same contradiction as Case 1.

Case 3. $\varphi(\alpha_0)$ has neighbors in the graph $G_{\alpha_0}$ as well as in the graph $\bar{G}_{\alpha_0}$. We know that at least one of $G_{\alpha_0}\setminus\{\varphi(\alpha_0)\}$ and $\bar{G}_{\alpha_0}\setminus\{\varphi(\alpha_0)\}$ is connected. We may assume that $G_{\alpha_0}\setminus\{\varphi(\alpha_0)\}$ is connected. But since there is an edge from some point in $G_{\alpha_0}\setminus\{\varphi(\alpha_0)\}$ to $\varphi(\alpha_0)$ by assumption of Case 3, we know that $G_{\alpha_0}$ is connected, contradicting our choice of $\alpha_0$.

Remark. The above proof used the well-ordering principle, which is equivalent to the Axiom of Choice (AC). It would be interesting to know whether the statement of the proposition above implies (AC).

Update. Will Brian gave a neat proof of the proposition above without using the Axiom of Choice, so the proposition does not imply (AC).

## Ode to ed (Unix standard text editor)

Encouraged by Eric S. Raymond’s book The Art of Unix Programming I have started using ed. My affection for it keeps growing. To begin with, it is the prime example of minimality, a property that is cherished all over the Unix world (Windows people don’t care about minimality). Also, ed is part of the POSIX standard, so you will find it on virtually every Un*x-like system. It is the standard text editor of Un*x.

Its famously terse interface teaches you a mental discipline that I find helpful in creating files, and I find that my .tex files (math texts, letters, or otherwise) do get more concise when I use ed to edit them.

Moreover, ed is (almost) orthogonal, which means that for many tasks there is *one* way (or very few ways) to do the task. In other editors, such as emacs or vi you have dozens of ways to do things, and you have no chance of remembering all commands. In ed, there are just essentially 24 commands (and you can do almost everything with a subset of 13 commands) every one consisting of a single letter, plus some arguments. You can’t get terser than that!

Also, using ed primarily, you are outside the raging editor wars (usually vi vs emacs), and, in some ways, *above* them. Oh – just one more thing: have you ever wondered what editor the source code of Unix itself was written in…?

## Coloring connected Hausdorff spaces

Motivation. I stumbled over the following hypergraph coloring concept when reading about an old (and open) problem by Erdos and Lovasz. Let $H=(V,E)$ be a hypergraph such that for all $e\in E$ we have $|e| > 1$, and let $Z \neq \emptyset$ be a set. Then a map $c:V\to Z$ is said to be a (hypergraph) coloring if for all $e\in E$ the restriction $c|_e$ is not constant (that is, the vertices pertaining to any edge are colored with more than 1 color). Trivially, for any nonempty hypergraph with no singleton edges, the identity map $\text{id}:V\to V$ is a coloring. The chromatic number of a hypergraph with no singleton edges is the least cardinal $\kappa$ such that there is a coloring $c:V\to \kappa$.

Equivalent topological formulation. If $(X,\tau)$ is a connected Hausdorff space such that $|X| > 1$, we can use this to color the associated hypergraph $(X,\tau\setminus\{\emptyset\})$. (Indeed we can apply it to any topological space without isolated points.)

We can reformulate this in topological terms in the following way: Let $(X,\tau)$ be a connected Hausdorff space. We define the nowhere dense covering number $\nu(X)$ to be the minimum cardinality of a set ${\cal N}$ of nowhere dense subsets of $X$ such that $\bigcup {\cal N} = X$. It is not hard to see that $\nu(X)$ equals the chromatic number of the hypergraph $(X,\tau\setminus\{\emptyset\})$.

For many standard connected Hausdorff spaces $X$ we have $\nu(X) = 2$. For instance, if $X = \mathbb{R}$ with the Euclidean topology, let ${\cal N} = \{\mathbb{Q}, \mathbb{R}\setminus\mathbb{Q}\}$. It took some effort to see that there are connected Hausdorff spaces $X$ with $\nu(X) > 2$.

Question. For which cardinals $\kappa > 2$ is there a connected Hausdorff space $(X,\tau)$ such that $\nu(X) = \kappa$?

Many more natural questions arise in this context, such as how does $\nu(\cdot)$ behave with topological products, and so on. So far I haven’t found a reference studying this concept of “Hausdorff space coloring”.

## Ihm ging ein Licht auf: Grieche entdeckte vor 2500 Jahren die Sonne

Ein Ereignis ist in diesen Sommertagen etwas in Vergessenheit geraten: Vor 2500 Jahren hat der Grieche Heliotides die Sonne entdeckt. Aus heutiger Sicht schwer zu glauben, dass diese Entdeckung eines grossen Kopfes bedurfte – aber schliesslich hatte es auch einen Pythagoras gebraucht, damit wir heute Zahlen haben, mit denen wir Rechnungen im Restaurant begleichen koennen.

Heliotides wurde uebrigens fuer seine Entdeckung in den Stand eines Gottes erhoben. Dies hat das Gewicht von etwa 3 heutigen Nobelpreisen. Schon die Griechen konnten bahnbrechende Entwicklungen in der Grundlagenforschung richtig einordnen.

Mit einem “Cheers!” an den Postillon.

## Homogeneous spaces generalized – and an open problem

Homogeneous spaces are topological spaces $(X,\tau)$ that locally look everywhere the same. Put in mathematical terms, $(X,\tau)$ is homogeneous if for any $x,y\in X$ there is an isomorphism
$\varphi:X\to X$ such that $\varphi(x)=y$.

We can generalize this by not restricting ourselves to a pair of points $(x,y)$ in the following way:

Let $\kappa>0$ be a cardinal and let $(X,\tau)$ be a topological space. We say that $X$ is
$\kappa$homogeneous if

1. $|X| \geq \kappa$, and
2. whenever $A,B\subseteq X$ are subsets with $|A|=|B|=\kappa$ and $\psi:A\to B$ is a bijective map, then there is a homeomorphism $\varphi: X\to X$ such that $\varphi|_A = \psi$.

A natural question is now to ask whether for cardinals $\alpha<\beta$ we have
that there is a space that is $\alpha$-homogeneous, but not $\beta$-homgeneous.

Joel David Hamkins gave this wonderful partial answer, pointing out that the disjoint union of 2 circles is 1-homogeneous but not 2-homogeneous; and moreover that $\mathbb{R}^2$ is
2-homogeneous but not 3-homogeneous.

Interestingly, neither Joel nor another mathematician, Andreas Blass, who wrote several comments on Joel’s post, could figure out how to continue from there – which gives rise to the following

Open problem: For integers $n\geq 5$, are there spaces that are $n$-homogeneous, but not $(n+1)$-homogeneous?

(The question has a positive answer given in Joel’s post for cardinals $\kappa\geq\aleph_0$.)