To be coherent, an output has to satisfy some inequalities (like if A subset B, P(A) <= P(B), but they can get more complicated). The set of solutions to a system of inequalities is convex (the inequalities correspond to the facets of the polytopes of coherent outputs)

We are not actually proving Sperner's lemma, we are proving a slightly different result. We omitted some details, but because every line can have at most two colors on it, the overall statement still holds (I think, but it's been a while).

And actually, if you look at a continuous function like x*sin(1/x), there are infinitely many points where it is 0 in any interval that contains 0

The interval is closed. If there are infinitely many such arguments, then the sequence of points that all evaluate to the same value must have a limit point (and the problem you discuss is only an issue when ) but by continuity, the function evaluated at the limit point must also evaluate to that same value. xsin(1/x) is only continuous on [0, 1] if you define f(0)=0.

lol glad you liked it

f(I)=I because f(I_k)=I_{k-1}. f(I) is the intersection of all f(I_k) which is the intersection of all l_{k-1} which is precisely I.

@@concavecuboid3253 Why is f(I) the intersection of all the f(I_k)?

Suppose x in I. Then, for all k, x in I_k, so f(x) is in I_k-1. Therefore, f(x) is in I. Suppose x not in I. Then, there is a k so that x is not in I_k. So f^-1(x) is disjoint from I_k+1. Therefore, f^{-1}(x) is disjoint from I and x is not in f(I). I think that I just consists of a single point with period 1.

@@concavecuboid3253 I don't see what you're proving here. In the first part, you've proven the following claim: "Given x in I, f(x) is in I". This means that f(I) is a subset of I. In the second part, you've proven the following claim: "Given x not in I, x is not in f(I)". By contrapositive, this is equivalent to the claim "Given x in f(I), x is in I". This means that f(I) is a subset of I. So you've just proven the same claim two times. I'm not sure myself of how to prove that applying f to I will not make it smaller. Also, I don't think that I generally consists of a single point only. For example, if f is the identity function on some subinterval of [b, c], then our construction of the I_k's will stabilize, so we will keep getting the same subinterval, meaning that I will also be that subinterval instead of just a single point.

@@concavecuboid3253 I think I figured out how to prove that I is a subset of f(I). Take x in I. Then, for all k, x is in I_k, meaning that there is y_k in I_(k+1) with f(y_k) = x. Now, (y_k)_k is a bounded sequence which contains a convergent subsequence by the Bolzano-Weierstrass theorem. Let y be the limit of that convergent subsequence. Then we have by continuity of f that f(y) = x. Additionally, since for any k, the terms of the sequence are contained in I_k from some point on, and I_k is a closed set, y is also contained in I_k. Therefore, y is contained in I. Therefore, for an arbitrary x in I, we have found y in I with f(y) = x. Therefore, I is a subset of f(I). Since we have already proven that f(I) is a subset of I, we can conclude that f(I) = I.

binary shift chaot

2:17 insane functioj

5:23 intermediate val

5:34 baking

Ostrowski

@@alexf2008 I was thinking of Hadwiger. Is Ostrowski a generalization?

increase your framerate to infinite ;)

Im replying for the algorithm. Thx for your repliable comment.

top lad!

Engagement for The Engagement God.

From the Wikipedia article on Sharkovskii's theorem: "Sharkovskii's theorem does not state that there are stable cycles of those periods, just that there are cycles of those periods. For systems such as the logistic map, the bifurcation diagram shows a range of parameter values for which apparently the only cycle has period 3. In fact, there must be cycles of all periods there, but they are not stable and therefore not visible on the computer-generated picture."

If you set f(x)=rx(1-x) and graph y=f^5(x) and y=x, you will see that these lines intersect (apart from where f(x)=x). This means that there is a point with period 5. But the period 5 is unstable, so does not appear in the bifurcation diagram for the logistic map. The same thing is true for all other periods, which is why the bifurcation diagram only has 3 points on it.

Replay

6:09 assuming that lines cannot contain more than two colors Took me a while to catch this myself as he glossed over it bit this crucial remark makes the theorem correct

I am not sure what you mean. Could you give me the time stamp where I say that "v(p)=0 for all primes" 2,3,5,... are not zero."? If p and q are different primes, than v_p(q)=1 because q is divisible by no powers of p, and p^(-0)=1.

I should have been clearer about that. What's being counted is not actually the number of turquoise line segments, (which is 3 in your example) but the sum over all triangles of the number of turquoise segments on the outside of that triangle (which is 4 because the diagonal is on two triangles, so is counted twice).

Sorry about that and thank you for catching it. I should definitely have referred to them p-adic norms or p-adic absolute values instead.

@@concavecuboid3253 No problem, there’s actually a correspondence between valuations and non-Archimedean absolute values (up to equivalence).

Thanks!

This is a fair point. I think the ambiguity lies in calling Sperner's Lemma a result from graph theory, which was incorrect since the exact location of vertices does matter. I am instead using triangulation to mean a partition of a polygon into triangles in a geometric sense which means the result should still hold.

@@concavecuboid3253 what does "three different colored vertices" mean when you have a triangle with four vertices on it, as you do in the Sperner example?

We only care about the 3 vertices on the triangle itself. The colors of the vertices on the edges don't matter.

@@concavecuboid3253 divide each side of a triangle in 3 with 2 false vertices, colour all actual vertices red and, in each side, one false vertex green and one blue. The triangle is not colorful, yet it has an odd number of cyan segments. I think this lemma should respect graph theoretical triangles.