A **tower** is a collection of co-infinite subsets of such that for all we have and either or . ( is co-infinite if is infinite.)

If are towers, we say that if and with jointly imply that . (In other words, this means that is a down-set or initial segment of with respect to ). It is easy to prove that is a partial order on the collection of all towers on .

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

**Lemma 1.** If is a collection of towers, such that for all we have either or , then

1. is a tower, and

2. for all .

**Corollary 1.** Zorn’s Lemma and Lemma 1 imply that there is a tower in that is maximal with respect to .

**Lemma 2.** If is a countable tower then is not maximal.

*Proof.* Let be a sequence of co-infinite subsets of such that for all we have . We want to show there is co-infinite with for all .

*Step 1.* If , then is co-infinite.

*Step 2.* There is strictly increasing such that for all .

*Step 3.* Set

Then it follows that

1: for all since which is a finite set.

2: is co-infinite, since for all we have , so , and since is strictly increasing we have is infinite, so is co-infinite.

Letting we get that is a tower with but , so the countable tower is not maximal.

]]>First an easy definition. If is a simple, undirected graph and are non-empty and disjoint, we say that are *connected to each other* if there are such that .

Let be simple, undirected graphs. We say that is a **minor** of if there is a collection of non-empty, mutually disjoint, and connected subsets of and a bijection such that

]]>whenever and then the sets and are connected to each other in .

**Proposition.** At least one of is connected.

*Proof.* If has only 1 element, the statement is trivially true. So suppose that . Let and let be a bijection. Suppose that neither nor is connected. It is easy to see that this implies that there is a smallest such that the induced subgraph on the set has the property that neither nor its complement is connected.

*Case 1.* has no neighbors in the graph . But then is connected to every vertex in , so is connected, contradicting our assumption.

*Case 2.* has no neighbors in the graph . Same contradiction as Case 1.

*Case 3.* has neighbors in the graph as well as in the graph . We know that at least one of and is connected. We may assume that is connected. But since there is an edge from some point in to by assumption of Case 3, we know that is connected, contradicting our choice of .

**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).

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…?

**Equivalent topological formulation**. If is a connected Hausdorff space such that , we can use this to color the associated hypergraph . (Indeed we can apply it to any topological space without isolated points.)

We can reformulate this in topological terms in the following way: Let be a connected Hausdorff space. We define the *nowhere dense covering number* to be the minimum cardinality of a set of nowhere dense subsets of such that . It is not hard to see that equals the chromatic number of the hypergraph .

For many standard connected Hausdorff spaces we have . For instance, if with the Euclidean topology, let . It took some effort to see that there are connected Hausdorff spaces with .

**Question**. For which cardinals is there a connected Hausdorff space such that ?

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

]]>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.

]]>such that .

We can generalize this by not restricting ourselves to a pair of points in the following way:

Let be a cardinal and let be a topological space. We say that is

–*homogeneous* if

1. , and

2. whenever are subsets with and is a bijective map, then there is a homeomorphism such that .

A natural question is now to ask whether for cardinals we have

that there is a space that is -homogeneous, but not -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 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 , are there spaces that are -homogeneous, but not -homogeneous?

(The question has a positive answer given in Joel’s post for cardinals .)

]]>For and we say that

if and only if for all we have .

Then we define to be **finitely continuous** if

So if you want to know the first positions of the output sequence , it suffices to know the first positions of the input sequence .

It turns out that this definition is equivalent to topological equivalence of functions , where is endowed with the discrete topology and carries the interval topology.

Nice, isn’t it?

]]>if then is a minor of .

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

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

- , and
- .

It is clear that (Hadw) implies (Hadw’). For the other direction, take any finite graph and let . If is complete, we are done. If not, use (Hadw’) and let be a proper minor such that . If is complete, we are done, otherwise use (Hadw’) again to get a proper minor of with . And so on. Since is finite, this procedure is bound to end at for some . We

have , and since the procedure ended at , the graph must be complete. So we get statement (Hadw).

Endow with the following topology:

It is routine to verify that this is a topology on .

**Observation 1:** No sequence in converges to .

*Proof:* Let be any sequence. We distinguish two cases:

*Case 1:* For all the set is finite. Then we set . This is easily seen to be an open nbhood of and clearly doesn’t converge to .

*Case 2:* There is such that is infinite. Then is an open nbhood of , because for all with we have . Moreover, by the conditition described in the statement of Case 2, the sequence keeps “popping out” of and therefore does not converge to .

**Observation 2:** There is a sequence in such that is an accumulation point of .

*Proof:* Since is countable, there is a bijection from to and we let be that bijection.

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

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