I think this this points to the strategic supremacy of relevant infrastructure in these scenarios. From what I remember of the battleship era, having an advantage in design didn't seem to be a particularly large advantage - once a new era was entered, everyone with sufficient infrastructure switches to the new technology and an arms race starts from scratch.
This feels similar to the AI scenario, where technology seems likely to spread quickly through a combination of high financial incentive, interconnected social networks, state-sponsored espionage etc. The way in which a serious differential emerges is likely to be more through a gap in the infrastructure to implement the new technology. It seems that the current world is tilted towards infrastructure ability diffusing fast enough to, but it seems possible that if we have a massive increase in economic growth then this balance is altered and infrastructure gaps emerge, creating differentials that can't easily be reversed by a few algorithm leaks.
Apologies if this is not the discussion you wanted, but it's hard to engage with comparability classes without a framework for how their boundaries are even minimally plausible.
Would you say that all types of discomfort are comparable with higher quantities of themselves? Is there always a marginally worse type of discomfort for any given negative experience? So long as both of these are true (and I struggle to deny them) then transitivity seems to connect the entire spectrum of negative experience. Do you think there is a way to remove the transitivity of comparability and still have a coherent system? This, to me, would be the core requirement for making dust specks and torture incomparable.
I've realised that you've gotta be careful with this method because when you find a trichromatic subtriangle of the original, it won't necessarily have the property of only having points of two colours along the edges, and so may not in fact contain a point that maps to the centre.
This isn't a problem if we just increase the number n by which we divide the whole triangle instead of recursively dividing subtriangles. Unfortunately now we're not reducing the range of co-ords where this fixed point must be, only finding a triad of arbitrarily close points that map to a triangle surrounding the centre. You can, for example, take the centre point of the first of these triangles (with some method of numbering to make the function definite) for each value of n=1,2,3.. as a sequence in R2. This must have a convergent sequence which should converge to a point that maps to the centre but I can't prove that last stage.
Cleanest solution I can find for #8:
Also, if we have a proof for #6 there's a pleasant method for #7 that should work in any dimension:
We take our closed convex set S that has the bounded function h:S→S . We take a triangle T that covers S so that any point in S is also in T .
Now we define a new function h′:T→T such that h′(x)=h(cs(x)) where cs(x) is the function that maps x to the nearest point in S.
By #6 we know that h′ has a fixed point, since cs is continuous. We know that the fixed point of h′ cannot lie outside S because the range of h′ is S. This means h′ has a fixed point within S and since for x∈S, h(x)=h′(x), h has a fixed point.
Yeah agreed, in fact I don't think you even need to continually bisect, you can just increase n indefinitely. Iterating becomes more dangerous as you move to higher dimensions because an n dimensional simplex with n+1 colours that has been coloured according to analogous rules doesn't necessarily contain the point that maps to zero.
On the second point, yes I'd been assuming that a bounded function had a bounded gradient, which certainly isn't true for say sin(x^2), the final step needs more work, I like the way you did it in the proof below.
Here's a messy way that at least doesn't need too much exhaustive search:
First let's separate all of the red nodes into groups so that within each group you can get to any other node in that group only passing through red nodes, but not to red nodes in any other group.
Now, we trace out the paths that surround these groups - they immediately look like the paths from Question 1 so this feels like a good start. More precisely, we draw out the paths such that each vertex forms one side of a triangle that has a blue node at its opposite corner. Note that you can have multiple paths stemming from the same group if the group touches the side of the larger triangle, or if it has internal holes.
Now we have this set of paths we can split them into three kinds. The first is loops, which arise when you have a group which never touches the edge of the larger triangle, or inside 'holes' in large groups. These can be seen as a path starting and finishing at the same node. They therefore have an even number of b-g vertices. The second kind is those that begin at the edge of the large triangle and end at the same edge. These paths begin and end on the same colour and therefore also have an even number of b-g vertices. Finally and most importantly there is a kind of path that goes from one edge to the other -in the case of the reds, the left edge to the right edge. This will happen once with the group that includes the top red node, and if any other group spans the larger triangle then it will generate two more of these paths. Sperner's lemma tells us that these will have an odd number of b-g vertices and we know that there will be an odd number of such paths, so this final type generates an odd number of total b-g vertices.
By the way that we have defined these paths, the total number of r-g-b triangles is equal to the number of g-b vertices on the paths in the set generated above. This number is the sum of an odd number from the spanning paths and a series of even numbers from the other paths, giving an odd overall number of r-g-b vertices, proving number 4 (as long as I haven't made an error in categorizing the paths).
I hope this makes sense, let me know if it doesn't or has errors :)
I was able to get at least (I think) close to proving 2 using Sperner's Lemma as follows:
You can map the continuous function f(x) to a path of the kind found in Question 1 of length n+1 by evaluating f(x) at x=0, x=1 and n-1 equally spaced divisions between these two points and setting a node as blue if f(x) < 0 else as green.
By Sperner's Lemma there is an odd, and therefore non-zero number of b-g vertices. You can then take any b-g pair of nodes as the starting points for a new path and repeat the process. After k iterations you have two values of x - only one where f(x) is below zero - that are 1/(n^k) away from each other. We thus can find arbitrarily close points that straddle zero. By taking the sequence f(x) of initial nodes x we get a sequence that, by B-W, has a sub-sequence which converges to zero. By continuity we have proved the existence of an x such that f(x)=0.
We can be sure that the sub-sequence does in fact converge to zero, rather than any other value because if it converges to any number |a|>0, the gradient of f(x) would have to be arbitrarily high to dip back below/above 0 for a value of x arbitrarily close by and therefore would not be a continuous function.
Comments to tighten up/poke holes in the above appreciated :)