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 .)

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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?

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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).

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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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**Question**: Is there a map with the following properties:

- ;
- if such that then ;
- is translation-invariant, that is for and where .

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However, the limit does not always exist (examples?). So here is a modification and we define a map by

for all .

Question: is additive, that is for with do we have ? There is a surprising answer that I’ll post in a few days.

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Friendships, social networks, and the like are often modelled using simple graphs. People are represented by vertices, and each edge denotes a pair of friends. The set of friends, or neighbors, of a vertex is defined to be . The number of friends, or the degree of is set to be and the average number of the friends of a person’s friends is defined by .

We say that a person (or vertex) is *proud* if and . One interesting version of the question above is: Do there exist graphs such that more than half of the people are proud?

The following insight is elementary, but it was still a surprise for me: It turns out that the share of proud people can be arbitrarily close to .. In order to prove this, take any integer and consider the complete graph on points with *one edge removed*. It is easy to see that the 2 people adjacent to the sole edge that was removed are the only ones that are not proud. So the share of proud people is which converges to as grows large.

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Let be the first infinite ordinal; its successor is .We consider a map , or put differently, a -Matrix of real numbers.

We assume that has the following properties:

(1) For all we have , or more informally, we have “convergence to the right”.

(2) For all we have , or more informally, we have “convergence to the bottom”.

(3) , or more informally, the right-hand entries of converge to the bottom right element .

**Question:** Does this imply that , that is, do the bottom entries of also converge to ?

**Answer:** There is, surprisingly (to me, at least) an easy example showing that the answer is **No**. Let be defined by if and otherwise. It is easy to verify that (1), (2), (3) above are satisfied. Note that the right-hand entries are all 0, and they trivially converge to ; but we have for all therefore .

For the kind of two-dimensional convergence we are looking for we need some form of “simultaneous” convergence (conceptually related to uniform convergence), which I might address in a later post.

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**Definition. **Suppose is a set, a totally ordered set, and a function. Then is said to be *a median *of if the sets and have equal size.

It is a reflex of mathematicians to ask about **existence** and **uniqueness** of any concept they stumble upon. (Note that I wrote “a median” and not “the median” above.) Indeed, as much as the definition above seems to make sense: Even for simple example, the median needs not exist. Let and let be defined by and . Then has no median according to the definition above. On the other hand, if and is the inclusion map, then every element of the open interval is a median!

There are many common fixes to the problems of existence and uniqueness, but no definition is really elegant. (Most resort to listing the elements in ascending order and to pick the arithmetical middle of the “middle elements” in the list or something similar.)

Other difficulties arise when we want to pick medians of infinite sample sets. Let be a totally ordered set. We say that is a median *of* if the sets and have equal cardinality. Note that in every element is a median, but has no median at all!

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In Adventswoche (für ) werden Kerzen gewählt, angezündet, und so lange brennen gelassen, bis sie 1cm an ihrer Höhe eingebüsst haben. (Die Kerzen sind initial cm hoch.)

Woche 4 ist ein No-Brainer – da werden immer alle Kerzen angezündet. Ich muss also auf Anfang der Woche 4 “Gleichstand” erreicht haben.

Für sei die Anzahl Male, die ich die Kerzen in Woche anzünde und genau 1cm abbrennen lasse. Für welche Tupel kann ich für den Anfang der Woche 4 Gleichstand erreichen, sodass die Kerzen am Ende der Adventszeit schön gleichzeitig abbrennen?

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