# Infinitely Many Blue-Eyed Logicians

This is an interesting generalisation of the Blue-Eyed Logicians puzzle. If you are not familiar with the original problem, you may wish to attempt that first!

The problem. Let $\kappa$ be an infinite cardinal. A population of $\kappa$ logicians, of potentially assorted eye colours, living on an eye-land island. Everyone can see everyone else’s eyes, but not their own.

In this society, it is an awful taboo to have blue eyes. Anyone who can deduce that they have blue eyes leaves the island on a midnight ferry, never to return – but because it’s such a taboo, there can be no discussion of eye colour.

Unbeknownst to them, every single one of the logicians has blue eyes. Oh dear….

But because there can be no discussion, no logician can ever deduce this, and they all live in perfect harmony.

One day, an oracle proclaims that ‘There is someone on the island with blue eyes’. Call this day 0.

Question: What happens next? We assume that the logicians are immortal, and allow times to run over $t\in\text{ON}$ – that is, all ordinal-valued times. So after days $1, 2, ... n, ...$, there’s a day $\omega$. This is followed by days $\omega+1, \omega+2, ...\omega+ n, ...$, and then $2\omega$, etc.

I (believe that I) have a solution, which will be posted here in due course.

DH

# Space-Filling Curves in Infinite Dimensions

“Space filling curves shouldn’t exist in infinite dimensions. That’s just rude” – Ben.

Bizarrely enough, they do! But only in a certain sense, as we will see below. Throughout, we equip $\mathbb{R}$ with the usual (Euclidean) topology.

Problem 1. Let $X$ be an infinite dimensional Banach space. Show that there is no continuous surjective map $f: \mathbb{R} \rightarrow (X, \| \|)$.

Problem 2. Let $1, and consider the sequence space $l^p=L^p(\mathbb{N})$. Show that there exists a continuous surjection $f: \mathbb{R} \rightarrow (l^p, w)$.

Just in case that wasn’t counterintuitive enough for you, we could also show:

Problem 2′. Show that there exists a continuous surjection $f: \mathbb{R} \rightarrow (L^p(\mathbb{R}),w)$.

So there are space-filling curves on function spaces! This is very conter-intuitive, since $L^p(\mathbb{R})$ naïvely seems to be much bigger than $\mathbb{R}$.

Problem 3. What happens when we replace $l^p$ by $(l^\infty, w)$? What about $(l^\infty, w^*)$, or $(l^1, w)$?

Once you’ve done these, it’s instructive to see what general results you can get out of this.

# Redecorating

I have a wallpaper! I wanted something decorative, but also unobtrusive, to put as a wallpaper here – but preferably something that would fit with the mathematical theme of the site.

The wallpaper is bond percolation on a square grid, with $p=0.4$. This is below the critical point of two dimensions, $p_c=0.5$, so we don’t expect to see very big clusters emerging.

I simulated this in mathematica – I’ll put up the code, so that you can experiment with different probabilities, at some point in the near future. I may also write a generally accessible blog post about percolation to explain what’s going on!

Edit: Here’s the code!:

n = 50;
Edges = {};
Vertexes = Flatten[Table[i + n*j, {i, 1, n}, {j, 1, n}], 1];
Do[U = RandomReal[];
If[U > 0.6, AppendTo[Edges, i + (n)*j <-> i + 1 + (n)*j],];
V = RandomReal[];
If[V > 0.6, AppendTo[Edges, i + (n)*j <-> i + (n)*(j + 1)],], {i, 1,
n – 1}, {j, 1, n – 1}];
Result = Graph[Vertexes, Edges,
VertexCoordinates ->
Flatten[Table[{i, j}, {i, 1, n}, {j, 1, n}], 1]]

This should be run in mathematica. Since not everyone has access to this, I will try to build an interactive version that I can put up here.

# Overkill? Me?

It’s an elementary fact of linear analysis that a normed space being finite dimensional is equivalent to its unit ball being compact under the norm topology. The forwards implication here is fairly clean, and boils down to repeated applications of Bolzano-Weierstrass. The converse is also elementary, but a bit ugly.

This is a nice application I found of the theory of weak topologies, which exports all of the work to the general theorem of topology, and makes a ‘cleaner’ proof of the converse implication.

Lemma Suppose $(B_X, \|\hspace{0.1cm} \|)$ is compact. Then $X$ is finite dimensional.

Proof. Equip the unit ball with the weak topology $w$, and consider the identity map  $\iota: (B_X, \|\hspace{0.1cm}\|)\rightarrow (B_X, w)$.

$\iota$ is clearly a bijection, and since the weak topology is coarser than the norm topology, $\iota$ is continuous. Moreover, since the weak topology is always Hausdorff, and we assumed that $(B_X, \|\hspace{0.1cm}\|)$ is compact, $\iota$ is a continuous bijection from a compact space into a Hausdorff space, and so is a homeomorphism. Hence

$(B_X, \|\hspace{0.1cm}\|)=(B_X,w).$

In particular, the half-ball $U= \frac{1}{2}B_X^o$ can be written as $V\cap B_X$, for some weakly open set $V\subset X$.

Suppose, for a contradiction, that $\text{dim } X=\infty.$ Since $V$ is a weakly open neighbourhood of the origin, it contains a finite co-dimensional subspace $Y$. Since $X$ is infinite-dimensional, and $Y$ has finite codimension, $Y$ cannot be the trivial subspace: $Y \neq 0$, and so contains a point $y\in Y$ of norm $\|y\|=\frac{3}{4}$. But this is absurd, since $V\cap B_X = \{x\in X: \|x\|<\frac{1}{2}\}$. This gives us a contradiction, and so we can conclude that $\text{dim } X<\infty$.