Two notions of continuity in the space of natural sequences

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

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

  1. M \not \cong G, and
  2. \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

“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

    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\}.
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    Recently I was trying to put a “meaningful” measure on the set \mathbb{N} of natural numbers. The first idea was to measure the density of a subset A \subseteq \mathbb{N} on natural intervals \{1,\ldots,n\} and take the limit of it, that is

    \lim_{n\to\infty}\frac{|\{1,\ldots,n\} \cap A|}{n}.

    However, the limit does not always exist (examples?). So here is a modification and we define a map \mu:\mathcal{P}(\mathbb{N}) \to [0,1] by

    \mu(A) := \lim\inf_{n\to\infty}\frac{|\{1,\ldots,n\} \cap A|}{n} for all A \subseteq \mathbb{N}.

    Question: is \mu additive, that is for A,B\subseteq \mathbb{N} with A\cap B = \emptyset do we have \mu(A\cup B) = \mu(A) + \mu(B)? There is a surprising answer that I’ll post in a few days.

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    Exceptions to the Friendship Paradox

    The Friendship Paradox says that your friends are likely to have more friends on average than you do. I was wondering whether there are settings in which the statement of the Friendship Paradox does not hold. Below we think about this question in precise mathematical terminology.

    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 v is defined to be N(v) = \{w \in V(G) : \{v,w\} \in E(G)\}. The number of friends, or the degree of v\in V(G) is set to be \textrm{deg}(v) = \textrm{card}(N(v) and the average number of the friends of a person’s friends is defined by \textrm{ad}(v) = \textrm{avg}\{\textrm{deg}(w) : w \in N(v)\}.

    We say that a person (or vertex) v is proud if \textrm{deg}(v) > 0 and \textrm{deg}(v) > \textrm{ad}(v). 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 1.. In order to prove this, take any integer m > 4 and consider the complete graph on m 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 \frac{m-2}{m} which converges to 1 as m grows large.

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    Two-dimensional convergence

    Here is something that I was certain would have an easy-to-prove positive answer; but it turns out that there is an exceedingly simple counterexample.

    Let \omega be the first infinite ordinal; its successor is \omega + 1 = \omega \cup \{\omega\}.We consider a map M: (\omega+1) \times (\omega+1) \to \mathbb{R}, or put differently, a (\omega+1) \times (\omega+1)-Matrix M of real numbers.

    We assume that M has the following properties:
    (1) For all k\in \omega we have \lim_{n\to\infty} M(k, n) = M(k,\omega), or more informally, we have “convergence to the right”.
    (2) For all k\in \omega we have \lim_{n\to\infty} M(n, k) = M(\omega, k), or more informally, we have “convergence to the bottom”.
    (3) \lim_{n\to\infty} M(n,\omega) = M(\omega,\omega), or more informally, the right-hand entries of M converge to the bottom right element M(\omega,\omega).

    Question: Does this imply that \lim_{n\to\infty} M(\omega,n) = M(\omega,\omega), that is, do the bottom entries of M also converge to  M(\omega,\omega)?

    Answer: There is, surprisingly (to me, at least) an easy example showing that the answer is No. Let M: (\omega+1)^2 \to \mathbb{R} be defined by (\alpha,\beta)\mapsto 0 if \alpha \leq \beta and (\alpha,\beta)\mapsto 1 otherwise. It is easy to verify that (1), (2), (3) above are satisfied. Note that the right-hand entries M(n, \omega) are all 0, and they trivially converge to M(\omega,\omega) = 0; but we have M(\omega, n) = 1 for all n\in\omega therefore \lim_{n\to\infty} M(\omega,n) = 1\neq 0 =M(\omega,\omega).

    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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