Musings on general topology and orderings

Coloring connected Hausdorff spaces

Posted on 2017/07/17 by dominiczypen

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”.

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

Posted on 2017/07/13 by dominiczypen

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.

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Homogeneous spaces generalized – and an open problem

Posted on 2016/08/13 by dominiczypen

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 $\kappa\geq\aleph_0$ .)

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Two notions of continuity in the space of natural sequences

Posted on 2015/11/19 by dominiczypen

In this presentation by Martín Escardó I came across a nice concept of continuity in the set of all functions $(\mathbb{N} \to \mathbb{N}) := \mathbb{N}^\mathbb{N}$ .

For $n\in \mathbb{N}$ and $\alpha, \beta \in (\mathbb{N}\to \mathbb{N})$ we say that

$\alpha =_{(n)} \beta$ if and only if for all $i\leq n$ we have $\alpha(i) = \beta(i)$ .

Then we define $f: (\mathbb{N}\to \mathbb{N}) \to (\mathbb{N}\to \mathbb{N})$ to be finitely continuous if

$\forall \alpha \in (\mathbb{N}\to \mathbb{N}).\forall n\in \mathbb{N}. \exists m\in \mathbb{N}. \forall \beta \in (\mathbb{N}\to \mathbb{N}).$

$(\alpha =_{(m)}\beta\implies f(\alpha) =_{(n)} f(\beta)).$

So if you want to know the first $n$ positions of the output sequence $f(\alpha)$ , it suffices to know the first $m$ positions of the input sequence $\alpha$ .

It turns out that this definition is equivalent to topological equivalence of functions $f: (\mathbb{N}\to \mathbb{N}) \to (\mathbb{N}\to \mathbb{N})$ , where $\mathbb{N}$ is endowed with the discrete topology and $(\mathbb{N}\to \mathbb{N})$ carries the interval topology.

Nice, isn’t it?

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A reformulation of Hadwiger’s conjecture

Posted on 2015/10/23 by dominiczypen

Let $G = (V,E)$ be a finite, simple, undirected graph. Hadwiger’s conjecture states that

if $n = \chi(G)$ then $K_n$ is a minor of $G$ .

Let’s call that statement (Hadw). We show that it is equivalent to the following statement:

(Hadw’): If $G$ is not a complete graph, then there is a minor $M$ of $G$ such that

$M \not \cong G$ , and

$\chi(M) = \chi(G)$ .

It is clear that (Hadw) implies (Hadw’). For the other direction, take any finite graph $G$ and let $n = \chi(G)$ . If $G$ is complete, we are done. If not, use (Hadw’) and let $M_1$ be a proper minor such that $\chi(M_1) = n$ . If $M_1$ is complete, we are done, otherwise use (Hadw’) again to get a proper minor $M_2$ of $M_1$ with $\chi(M_2) =n$ . And so on. Since $G$ is finite, this procedure is bound to end at $M_k$ for some $k\in\mathbb{N}$ . We
have $\chi(M_k) = n$ , and since the procedure ended at $k$ , the graph $M_k$ must be complete. So we get statement (Hadw).

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Accumulation without converging subsequence

Posted on 2014/10/13 by dominiczypen

“Any sequence with an accumulation point also has a subsequence converging to that point” — right? I believed this until yesterday when I stumbled over the following curious, and countable, Hausdorff space.

Endow $\omega\times\omega$ with the following topology:

$(\omega\times\omega)\setminus\{(0,0)\}$ is given the discrete topology;

$U \subseteq \omega\times \omega$ is a neighborhood of $(0,0)$ if $(0,0)\in U$ and for almost every $n\in \omega$ the set $\bar{U}_n := \{k\in\omega: (n,k) \notin U\}$ is finite.

It is routine to verify that this is a topology on $\omega\times\omega$ .

Observation 1: No sequence in $(\omega\times\omega)\setminus\{(0,0)\}$ converges to $(0,0)$ .
Proof: Let $a:\omega \to (\omega\times\omega)\setminus\{(0,0)\}$ be any sequence. We distinguish two cases:
Case 1: For all $n\in \omega$ the set $\textrm{im}(a) \cap (\{n\}\times \omega)$ is finite. Then we set $U = (\omega\times\omega) \setminus \textrm{im}(a)$ . This is easily seen to be an open nbhood of $(0,0)$ and clearly $a$ doesn’t converge to $(0,0)$ .
Case 2: There is $n_0\in \omega$ such that $V:=\textrm{im}(a) \cap (\{n_0\}\times \omega)$ is infinite. Then $U = (\omega\times\omega) \setminus V$ is an open nbhood of $(0,0)$ , because for all $n\in \omega$ with $n\neq n_0$ we have $\bar{U}_n = \emptyset$ . Moreover, by the conditition described in the statement of Case 2, the sequence $a$ keeps “popping out” of $U$ and therefore does not converge to $(0,0)$ .

Observation 2: There is a sequence $b$ in $(\omega\times\omega)\setminus\{(0,0)\}$ such that $(0,0)$ is an accumulation point of $b$ .
Proof: Since $(\omega\times\omega)\setminus\{(0,0)\}$ is countable, there is a bijection from $\omega$ to $(\omega\times\omega)\setminus\{(0,0)\}$ and we let $b$ be that bijection.

So the sequence $b$ we constructed in Observation 2 has $(0,0)$ as an accumulation point, but no subsequence of $b$ converges to $(0,0)$ because of Observation 1!

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Measuring subsets of N

Posted on 2014/06/18 by dominiczypen

Recently I tried to put a measure on the powerset of the natural numbers $\mathcal{P}(\mathbb{N})$ , but it was pointed out in a comment that the map I defined is not even finitely additive.

Question: Is there a map $\mu:\mathcal{P}(\mathbb{N}) \to [0, 1]$ with the following properties:

$\mu(\mathbb{N}) = 1$ ;
if $A, B\subseteq\mathbb{N}$ such that $A\cap B = \emptyset$ then $\mu(A\cup B) = \mu(A)+\mu(B)$ ;
$\mu$ is translation-invariant, that is $\mu(A+n) = \mu(A)$ for $A\subseteq \mathbb{N}$ and $n\in\mathbb{N}$ where $A+n = \{a+n: a \in A\}$ .