[Y_Y]

It doesn't, however, allow for non-closed symbols. I can imagine a spiralling brushstroke that gets fainter as it approaches the centre. Maybe the proof can be strengthened, and it certainly won't pass the distinguishability criterion, but we must be rigorous, here if anywhere.

[clintonc]

> here if anywhere

:-D

We can play a lot of games with "what is a symbol", but compactness pervades many of the models that we use to describe reality. The crux of the argument is not necessarily that the symbols themselves are compact as sets, but that the *space of possible descriptions* is compact. In the article, the space of descriptions is (compact) subsets of a (compact) two-dimensional space, which (delightfully) is compact in the appropriate topology.

In your example, the symbols themselves could instead be modeled as a function f:[0,1]^2 -> [0,1] which are "upper semicontinuous", which when appropriately topologized is seen to be compact; in particular, every infinite sequence must have a subsequence that converges to another upper semicontinuous function.

Much of the fun here comes from the Tychonoff theorem, which says that arbitrary products of compact spaces is compact. Since the *measurement space* is compact, the topology of the domain is not as important, as long as the product topology on the function space is the appropriate one. (Mystically, it almost always is.)

[Y_Y]

The proof defines F(X) as the set of compact subsets of X (here the unit square). The compactness of F(X) follows from (necessary?) the compactness of its elements, so we need to find a topology where all our allowable symbols are compact, and this is not the standard topology if you allow the spiral. If you take another topology (equivalently space of symbol descriptions) then you again must show that this compactness still corresponds to the desired notion of "distinguishability".

My topology is rusty and I don't genuinely doubt the validity of the argument, but I'm having fun trying to poke holes.

[clintonc]

Your example centered on a symbol which, if viewed as a subset of the plane, is not compact. I tried to argue that the set of symbols that you describe (ink of varying levels of intensity in the unit square) still is a compact set, even though the symbols themselves are no longer represented by compact sets.

[dinkelberg]

Why should it not allow for your spiralling brushstroke? You can divide the spiral into the strokes corresponding to each turn around the center. We can assume that each of these strokes is measurable (otherwise we would not need the construction of the spiral). Then the entire spiral is just a countable union of those measurable strokes, thus it is measurable itself.

[Y_Y]

For the proof given in TFA they use the assumption that the symbol is compact and hence closed. I argue that the symbol needn't be closed, it may still be measurable.