XelaP

Karma: 750

Came for the Harry Potter fanfic, a positive vision of what do with atheism, and the anti-death ethos. Stayed for the good thinking.

I really like math and physics. I can program but am not a programmer—I’d have to learn how to make a website and struggle to handle more than a couple hundred LOC. Learning a little Haskell made me enjoy coding again, I’d like to get better at it.

I’m currently very bad at following through on commitments. I hope to fix this by tomorrow. I’ll probably fail, but I shoot for getting it down before my fifty thousandth tomorrow.

XelaP 23 Apr 2026 2:30 UTC
1 point
0
in reply to: Pedro Afonso’s comment on: Reflectively stable consequentialists are expected utility maximisers
I’ll read it soon. My suggestions: firstly, technical posts generally get less engagement, presumably because less people have the technical chops (even here) or it’s more of a time and effort commitment.
Secondly, I think you should’ve had a better summary as your first paragraph. You could additionally try two summaries: one that’s like a “normal” summary, and one that gives a terse math overview, so that those that can understand the latter can quickly get the point.

XelaP 22 Apr 2026 17:21 UTC
2 points
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in reply to: XelaP’s comment on: XelaP’s Shortform
April 20-21
0: I did most of the topological fixed point exercises on the 20th & the morning of the 21st, and then did most of the iteration fixed point exercises on the 21st. I liked the diagonalization ones the best, then the topological ones come in second (unsure how to explain why—just feels like coolness : complexity is lower), and the iteration ones come last (I think because I already know the contraction mapping principle very well, and unlike the topological ones there isn’t a cooler proof to dazzle me with; the lattice ones didn’t feel that novel given the spiritual similarity).
1: As a subpoint, the fact that there’s a unique closest point in a closed convex set S to any point p in euclidean space was pretty neat, and further acts as an enticing advertisement for me to read Boyd’s Convex Optimization
2: Apparently, the Wikipedia page on fixed point theorems really does basically just list variations on Diagonalization/Brouwer/Iteration. Perhaps I should take the couple best, and write a post on them?

XelaP 22 Apr 2026 17:10 UTC
3 points
0
in reply to: Leon Lang’s comment on: What is the Iliad Intensive?
Thanks for responding! I’ve just applied.
About when should I expect to know if I got in/get interviewed/etc.?

XelaP 22 Apr 2026 16:35 UTC
1 point
0
on: Iteration Fixed Point Exercises
I have fond memories of the contraction mapping theorem, because it was the first fixed point theorem I ever learned.
1:
Let c = d(x, f(x)). Then d(f(x), f^2(x)) ⇐ q d(x, f(x)) = qc, and in general d(f^n(x), f^{n+1}(x)) ⇐ q^n c.

Now, d(f^n(x), f^{m}(x)) ⇐ c * \sum_{n ⇐ i ⇐ m} q^i which is a geometric series. The sum of this is then c q^n (1 - q^{m-n})/(1 - q) ⇐ the full series (which converges since q < 1) = c q^n/(1 - q)

This can be made arbitrarily small, so the sequence is Cauchy, so since we are in a complete metric space it converges.

The distance to the limit point is at most that value c q^n/(1 - q), because all terms after the nth iterate are within that of the nth iterate. This gives us the exponential convergence.
2:
First, there’s at most one fixed point for a contraction mapping. For if we had two, then we could take d(x,y) = d(f(x), f(y)) ⇐ q d(x,y) < d(x,y) for the two fixed points x,y. But then, we get a contradiction—unless x = y, in which case q d(x,y) = d(x,y) = 0.

For existence: the limit of the iterates of any x is our fixed point. This is because f(lim f^n(x)) = lim f(f^n(x)) = lim f^{n+1}(x) = lim f^n(x), where the first equality is from continuity of f
3:
Take the space of infinite binary sequences, where the distance we assign is 1/N where N is the earliest index for which two sequences differ (or 0 if they are the same always). This is a complete metric space, actually (as you can easily check). But the right shift operator is continuous (since if you differ no earlier than the Nth spot, then the image of you differs by no earlier than the (N+1)th spot), and has no fixed point, despite the fact that it also is a weak contraction (by virtue of pushing any differences to the right). This is because N/(N+1) isn’t bounded by a constant.
4: TODO
Take two points x, y. We’d like to show that there’s an epsilon such that ||(x—eps grad(f)(x)) - (y—eps grad(f)(y))|| is bounded by ||x—y|| times some constant less than 1.

We have ||x—y + eps (grad(f)(y) - grad(f)(x))|| =
5: TODO
6:
WLOG suppose x ⇐ f(x). Then f(x) ⇐ f^2(x), and we can repeat upwards. For n > |P|, by the pigeonhole principle the nth iterate is equal to some earlier iterate, so there’s n and m such that m < n f^n(x) = f^m(x). Thus all the iterates between these two are also equal. But then f(f^m(x)) = f^m(x) , so f^m(x) is a fixed point, but this is the same as f^n(x).
7:
f^n(x) is the same as max {f(f^m(x)) : m < n}. So let’s define f^alpha(x) as sup {f(f^beta(x)) : beta < alpha}

WLOG suppose again that x ⇐ f(x). Now if alpha is a limit ordinal than f(f^beta(x)) for any beta < alpha will just be equal to f^gamma(x) with gamma = beta + 1 < alpha. Thus this is really the sup of all previous iterates, and thus larger than it. For successor ordinals, we need to iterate the previous iterate once more—but we can use monotonicity here to get that we are larger than the previous iterate.

Thus we see that f^beta(x) ⇐ f^alpha(x) for all beta < alpha.

Suppose we have some alpha such that |alpha| > |L|. Then if this sequence is made of all unique elements, we would be able to monotonically inject alpha into L by sending each smaller ordinal to the associated iterate. This totally ordered chain of iterates has an ordinal type, which we just mapped alpha to an initial element of, and so we have |alpha| ⇐ |L| as ordinals, a contradiction (if instead we were to take them as cardinals, we’d still have a contradiction, because we can ordinal inject the least ordinal with the same cardinality with alpha, which then gives us an ordinal injection to L). Thus we have some repeat where f^beta(x) = f^alpha(x) for beta < alpha. But then all the iterates between these two are also equal.

So f(f^(beta)(x)) = f^beta(x), so f^beta(x) is a fixed point of f, f^beta(x) = f^alpha(x) so f^alpha(x) is a fixed point of f.
8:
The question is asking whether the set of fixed points has a sup and an inf.

Suppose we have some subset S of fixed points. I hope that the sup of these fixed points in the original lattice is actually a fixed point.

For any element x, the supremum s is larger than x, and so f(x) = x ⇐ f(s). Thus f(s) is an upper bound for the subset, and thus s ⇐ f(s). But then for some alpha, f^alpha(s) is a fixed point of f that’s an upper bound. So now I hope f^alpha(s) is our supremum in the fixed point lattice.

Suppose there was an upper bound that was a fixed point, u. Since s is a supremum in the encompassing lattice L, we have s ⇐ u. Thus f(s) ⇐ f(u) = u. In general, f^n(s) ⇐ f^n(u) = u. We can also take a sup of them to get f^beta(s) ⇐ u for limit ordinals beta. Therefore f^alpha(s) ⇐ u. But then f^alpha(s) is smaller than all upper bounds that are fixed points, and so f^alpha(s) is a supremum in the fix point lattice

Likewise for infimums.

Incidentally: The least element l satisfies l ⇐ f(l), so there’s an alpha such that the alpha iterate is a fixed point. If x is a fixed point, then since l ⇐ x for all x, we have f^alpha(l) ⇐ f^alpha(x) = x. But then f^alpha(l) is the least fixed point.
9: TODO
P(A) is clearly a lattice. To take sups and infs, we can just take arbitrary unions and intersections. (I think technically you need the axiom of choice, but whatever). Given a function f: A → B, we have that S ⇐ S’ ⇐ A implies f^img(S) ⇐ f^img(S’) (we don’t need injectivity for this), so the induced f is monotone.

Since f,g are injective, we know that the induced functions on the powerset lattices are also injective—that is, that f^img(S) != f^img(S’) if S != S’. Also, for injective functions, f(X—Y) = f(X) - f(Y)

...
10:
Using problem 9, there is a set A’ and B’ such that f(A’) = B’ and A—A’ = g(B—B’)

Now, we’d like to biject A’ with B’ and A—A’ with B—B’. To do this, we can take the restriction of f on A’ (restricting the codomain to B’ as well), and likewise take the restriction of g. The restrictions are bijections, because they are surjections and we already knew they were injections.

Then we can take the union of the relation associated to the restriction of f and the inverse of the relation associated with the restriction of g. This is then a function A → B, and it’s a bijection.
What links here?
- XelaP's comment on XelaP’s Shortform by XelaP (22 Apr 2026 17:21 UTC; 2 points)

XelaP 22 Apr 2026 14:25 UTC
2 points
0
on: Posts I don’t have time to write
At least that’s what one fermi I made suggested
There’s a joke here: The funny thing about saying “fermi” for “fermi estimate” is that it also refers to a femtometre, so when I read this an image came to mind of you storing calculations into an atomic nucleus.
3: I am reminded of a post by Duncan Sabien which mentioned the phenomena where abuser X acts only kindly towards Y, because they go atd abuse specific sorts of people. He said it as a “People are catching on to this obvious fact” thing, but until I read it I didn’t actually consider that!
And… I thought about times I was mean to someone, for no real reason beyond “they slightly annoyed me and I thought I could get away with it, as they won’t pushback and my friends aren’t here to notice”.
I don’t like that!
4:
ITT’s also seem really hard? I have not properly tried to pass the ITT in discussion, but it intuitively seems like it may not be worth the effort, compared to searching for cruxes or ‘microdosing’ it by trying to sorta understand someone’s subpoint in words they’d sorta agree with. Compare to a ‘global’ ITT, where you try to espouse the other’s whole position convincingly.
6:
I suspect plants do less for me than most—relatedly, I suspect nature generally has less of an effect on me. This doesn’t mean I’m a fan of brutalist architecture or drab rooms; I just feel like I’d try other things, like a painting or sculpture or nice looking furniture.
An advantage of plants may be that they don’t require so much thought—you just buy a green thing that grows from a pot full of dirt, and stick it in the room.
9:
Totally agree! Though, I try instead to put an optimized TL;DR at the start, and then put less effort into optimized paragraphs after that.

XelaP 21 Apr 2026 18:55 UTC
1 point
0
on: Topological Fixed Point Exercises
I’ll use the term “rainbox edge”/”rainbow triangle” instead of bi/tri-chromatic.
Some details are glossed over, but only when I am am confident I could fill them in. Sorry if you are reading this and want more detail.

Also, thanks again for these posts. They are really quite helpful. I mean, I’m kinda skeptical about usefulness towards alignment, but regardless I’m very glad you wrote them. I had heard legends of proofs of the Brouwer fixed point theorem from Sperner’s lemma, but didn’t realize how nice it would be—I prefer this to the usual topological proof.

1: I proved this theorem in like 3 related ways until one of the approaches managed to give traction on the generalizations
Suppose we change a node ‘in the bulk’, that is, a node surrounded by two nodes. Then if we change the color of this node, we’ll either change the number of bichromatic edges by 0 (in the case where the nodes on either side have different colors) or 2 (where they have the same color). Thus we won’t affect the parity, so we can willy nilly change the bulk.

Let’s change the bulk to start with a blue on the leftmost, and green on the rightmost. Then we can by induction apply the lemma to this smaller segment. There are no extra bichromatic edges, so we are done.
2:
Imagine subdividing the interval into 1/N size bits, with the endpoints colored green if the function is positive there, and blue if it’s negative. If any are 0 we are already done, so let’s suppose none are zero.

For each N, pick a rainbow edge, then take the sequence of green endpoints from picked edges. This must have a convergent subsequence by Bolzano-Wierestrass. Now take the corresponding blue endpoints from this sequence (connected by the chosen edges). We then also have a sequence of blue endpoints, which must also converge (to the same value) because the edge lengths go to 0. The value of the function there is equal to the limit of function values, because it’s continuous—so therefore it must be both at least 0 and at most 0, and thus is equal to 0
3: Rewrite that’s more in the ‘spirit’ of problem 5 and 6, but not the first I came up with.
Color each point by whether the function sends it somewhere more “right” or “left” (that is, greater or lesser) than it. First suppose that there’s at least one point of each color. Then we can use the argument in the previous to get two converging sequences, one of “right” elements and one of “left” elements, that converge to the same value. When fed to the function, continuity means that we must be no bigger and no smaller than that value, and so must be equal.

If every point has the same color, then we send either 1 or 0 to something above/below itself, which is impossible because the function maps everything to [0,1].

It fails for (0,1) because you never need to plug in 0 or 1. For example, f(x) = ¹⁄₂ (1 - x) + ¹⁄₂ x is a counterexample—the fixed point is 1, which is outside (0,1). The function takes a point and then moves it halfway to 1. So everything is colored “right’, which is why we fail.
You could more directly use the previous problem:
Apply IVT to e(x) = f(x) - x
but I prefer the other way.
4: Huh, this got progressively less messy as I tweaked my proof, and in the process bugs were fixed. And then made even simpler for problem 9. Simplicity is king, remember that.
What group should we use for the 2D case?
Each color will be a group element.

Color an edge by the ‘missing’ color. This is the same as comparing our edge to the sum of all three colors. That is, (R + G + B) - (difference of endpoints). This also tells us what 2 of the same color should be—it should be R + G + B. Having the same color should clearly give us 0, so we want R + G + B = 0. Then, the color of an edge is just the sum of endpoint colors (at least, we’ll see that in a second when we show that -X = X).

Now, if we add the edges of a path, we should get the thing corresponding to the start and end endpoints. So we want e.g. G-B-G to be the same as G-G. That is, R + R = 0. Likewise, G + G = 0, B + B = 0, and G B R gives us that R + G = B. This is what we in the biz call the Klein four group, but you don’t have to worry about that if you don’t want to.

Any loop sums to 0. You can also think of this as counting each node twice. thus cancelling them.
Suppose we have a node in the “bulk” of the triangle, that is, one that’s fully surrounded (like a hexagon). We’ll show that we can change the color of the node without affecting the parity of the number of rainbow triangles. I’ll call this the “bulk lemma”.
WLOG imagine the center starts red and becomes blue, so our hexagon looks like (the Xs are arbitrary color values, not necessarily equal to other Xs)
X X
X R X
X X
and becomes
X X
X B X
X X

Now, how many rainbow triangles are at the beginning? Well, it’s actually just the number of B-G aka red in the loop around the center—that is, the number of bordering edges that are missing a red. Likewise, at the end it’s the number of R-B aka blue edges in the loop. But neither of these numbers depend on the center! So as long as the number of red and blue edges in the loop has the same parity, we can freely change the center node.
Now, to prove the bulk lemma, we’ll show that parity differences must be 0 for any loop. This is kinda like saying that not only do we know that the sum/center of mass of the vectors are 0, but we also know something about the angles (don’t take it too literally though).

Red is like the vector 1,0, Green is like 0,1, and Blue is like 1,1 (addition is mod 2). When we add in a loop, the total sum is 0,0. Note that Blue’s don’t screw up parity differences between Red and Green, cuz they add 1 to both.

So the difference/sum of the two coordinates is equal to the parity difference of reds with greens, because the blues don’t affect it. Since the vector is 0,0, it has sum 0, which is 0. So we’re good.

We could’ve just as easily swapped our basis to use a different clor as the all ones, and so we get that all parity differences are 0.
Since for hexagons the parity of the number of rainbow triangles changes according to the parity difference of a loop, and since those differences are always 0, we can freely change the center node.

Note that it’s quite important that we have a loop that doesn’t go through the center—as you can see by looking at the following half-hexagon pattern from about the middle of the left side of the image for the problem:

R
B B
B G

Changing the center B to a G increases the number of rainbow triangles by 1.
Alright, now for the theorem. We can change the “bulk” nodes in such a way that we can do an induction argument. That is, we can change them so that the left side of the inner triangle has no green, the right side of the inner triangle has no blue, and the bottom side of the inner triangle has no red. Now there are an odd number of rainbow triangles inside the inner triangle. Because of our coloring of the sides, the only possible way to have extra rainbow triangles are the “outtermost” ones near an inner corner, e.g.

R
X X
O R O
Y B G Z
B Y O Z G

Where the X, Y, Z’s mark the two extra nodes of each of the 6 triangles being referenced here, while the O’s mark extraneous nodes that don’t matter.

Now, due to our coloring choice, we see that to get extra rainbow triangles two of the X,Y,Z nodes with the same letter should have different colors. But, this gives us another rainbow triangle, using that as the same side. This is because we picked the corners of our inner triangle to match those of the overall big one.

Therefore, we only have an even number of extra rainbow triangles. So we have an odd number total.

5:
Let’s color points according to where the function sends them. We know the vertices go to R G B as expected. We can subdivide the triangle, giving us something that will satisfy the conditions of Sperner’s lemma no matter how finely we subdivide. For each subdivision, there’ll be a rainbow triangle, giving us a sequence of rainbow triangles.

If we then take the sequence of the red vertex of this sequence of rainbow triangles, there’ll be a convergent subsequence of them. We can then take the associated blue and green vertices, to get three convergent subsequences, and the limiting values of the function must be equal. This value, however, must be a limit point of the green, blue, and red subsets of the disk. The only such point is the center.
6:
Suppose every point of the triangle moves. Then it must be moving ‘closer to’ some side. More precisely: Take a point x. Look at f(x). Draw the ray from x to f(x), and at some point it’ll hit the boundary of the triangle. It will either hit a side, or hit a vertex (aka hit two sides). For the latter case, we can just fix at the outset some choice of colors for the vertices that agrees with the sides. For example, suppose we color the sides R G B, and the RG vertex R, the GB vertex G, and the BR vertex B. Think of this coloring as being on the codomain/second triangle, and then giving rise to a coloring on the inputs. The vertices must move towards the opposing face.

Now, no point on an edge in the domain can be colored the same way the point in the codomain is, as that would mean you moved off the triangle.

This lets us subdivide the triangle successively. We can then get a sequence of red, green, and blue points whose outputs converge to the same value, which then means we must not move at all.

If you aren’t convinced of that last fact: you can express “move closer to” via linear functions (just negative dot product of (f(x) - x) onto the vector from the center to a vertex), and thereby get a generalization of the fact that f(x_i) > 0 for all i implies that f(limit of x_i) >= 0. Alternatively you can directly use problem 5 like in my second proof of problem 3.
7: Sidenote: I think the reason why the projection is well defined is:

You have a convex optimization problem that has a unique solution. More intuitively, you can just do classical geometry:
Since S is closed, you know that there is at least one closest point. (otherwise you can take an infimum of distances, and then take a sequence of progressively closer points, which must converge to a point closer than all points in S since distance is continuous, but since it’s closed you must get a point actually in the set). (alternatively just use the extreme value theorem).

If two points A, B in S are both closest to P then they lie on a sphere around P. Then the interior of the spherical sector APB must not be in S because all of it is closer to P. But the chord AB must be contained in S by convexity, which is a contradiction.

The closest is continuous: Suppose we have some P. Let’s say that X is the closest point in S. The set S must lie to one side of the perpendicular hyperplane to PX (based at X) because otherwise there would be a point that’s closer to P (because the line from a point on the wrong side to X would intersect the inside of the sphere of radius PX centered at P, and all points inside the line are in S).

Suppose we have some other point Q, with closest point Y. If Q is on the line PX, then Y=X. So we can assume Q isn’t on PX. Now, hyperplane cuts need to ‘agree’. Let’s say that the perpendicular hyperplane that PX gives us is H. There is a closest point Q’ on H to Q, given by projecting.

If the closest point in S to Q is ‘further away’ from X than Q’ is then the hyperplane cut corresponding to QY will exclude X.
Precisely: Y must lie between the two hyperplanes perpendicular to XQ’ that contain X and that contain Q’ respectively. That is, it lies in a ‘channel’. XQ’ is at most as long as PQ, so it’s a channel at most as wide as PQ is (call it delta). We then have a plane geometry problem on the plane XYQ’. Our ‘agreement’ condition may as well use dot products of projected vectors on this plane (that is, Q’Y and Q’X instead of QY and QX). For the cuts to agree, the perpendicular to YQ’ at Y must intersect XQ’ between the two points, say at Z. Then Q’YZ is a right triangle with hypotenuse Q’Z, so YZ has length less than Q’Z. Finally XY has distance at most XZ + XY ⇐ XZ + ZQ’ = XQ’ ⇐ PQ ⇐ delta

Thus for any epsilon, we can choose some delta < epsilon to ensure that XY is within epsilon whenever PQ is within delta.
Okay, now for the actual problem.
The intuition: Let’s imagine picking a center point of our set S, and then classifying the directions like we would on the triangle.

Here’s what we actually do, which is close enough: find a triangle that is a superset of the set S (we can do this because it’s compact and so bounded). Then we can compose with projection to get a continuous function from the triangle to (the subset S of) itself. Since we know this has a fixed point, yet moves every point that is outside S (but still in the triangle), we know it has a fixed point in S.
8:
Reflecting: Imagine someone gave you the colors of a path like in 1. To find a bichromatic edge, you might have to look at all (but one) edge first! Likewise, with the intermediate value theorem/1D Brouwer, you’d have to check ‘infinitely’ many points to pin down the intermediate value/fixed point (or to get epsilon = 1/N accuracy, you’d have to check about N edges). 2D Sperner requires you to check all but one tiny triangle as well—as we can for example force it to be any of the triangles. Etc.

For functions: let’s look at it in the e(x) = f(x) - x formulation, where fixed points are roots and the problem forces -x ⇐ e(x) ⇐ 1 - x. The idea is to give a choice between two roots, force a commitment to one of them by destroying the other, recreate it, and then destroy the committed one.

Let’s force a commitment by ‘stomping’. We start as a ‘hump’ that’s 0 at both x=0,1. Then, we lift our left leg up. We can do this easily by linear interpolation. Then, we turn our hump into a valley (what a poor camel). For an instant, half the points will be fixed points—but because this will only be true for an instant, our adversary won’t be able to take advantage of this to jump to the other root. Then, we pull our right ‘arm’ down, thus leaving our adversary stranded.

That is, for 0 ⇐ t ⇐ ¹⁄₃, let
e_t(0) = ³⁄₂ t -- goes from 0 to ¹⁄₂. T
e_t(1/2) = ¹⁄₂
e_t(1) = 0.

In between, we’ll do linear interpolation.
The left leg stops being on the ground at t=0. So any continuous choice of root has to commit to choosing x=1 for this period.

Now, we’ll turn the hump into a valley:
For ¹⁄₃ ⇐ t ⇐ 2/3:
e_t(0) = ¹⁄₂
e_t(1/2) = 3 (1/2 - t) -- goes from ¹⁄₂ to −1/2
e_t(1) = 0

and we’ll interpolate in x the same way as before. The continuous choice of root has to keep its commitment—it can’t use the instant where [1/2,1] are all roots to jump to the left, as it is only an instant.

Now, we’ll pull our right arm down:
For ²⁄₃ ⇐ t ⇐ 1:
e_t(0) = ¹⁄₂
e_t(1/2) = −1/2
e_t(1) = 1 − ³⁄₂ t—goes from 0 to −1/2

and we’ll again interpolate in x.
The right arm stops being at the ‘ground’ at t=2/3, so now the choice of root is screwed.

To make our fixed point function: observe that the function is always bounded between u(x) = 1 - x (which is always at least ¹⁄₂ for x in [0, ¹⁄₂]) and l(x) = x − 1 (which is always at most −1/2 for x in [1/2, 1]). So f_t(x) = e_t(x) + x is a legitimate function from the interval [0,1] to itself.
9:
Suppose each face of an nD simplex that’s subdivided like in the 2D case is known to only use all but 1 color, going through the sequence. e.g. for n=3 there’s a ABC face, a BCD face, a ACD face, and a BCD face. We should have n+1 colors, as the simplex has n+1 vertices.

Now, what will be the analog of paths, what group do we use, and what’s our version of a loop? Well, I claim we should be using faces. We can put faces together to make a closed volume. Another way to look at the sum over a loop of edges is to notice that we double count every node when we add the edges together. Likewise we’ll double count each (n-2)D face (‘edge’ in the n=3 case) of the volume when we add the (n-1)D faces together.

We want this to be 0, so our group elements that we assign to faces should satisfy g + g = 0. For the tiniest face, we have n nodes in a (n-1)D face, and there so we can just label by the missing color/label by sum of nodes. The group is (Z/2Z)^n.

Now, if we take some tiny faces, and put them together in a way that closes a volume, then we will indeed have double counted every node and thus get 0 in our sum.

Onto the bulk lemma. Let’s show that if we change a node ‘in the bulk’, the parity of the rainbow simplices is unchanged. We’re looking at a ‘honeycomb’ like hexagonal like thing around a center. WLOG let’s suppose we change the center color from R to G. The number of rainbow simplices at the start is the number of R faces surrounding the center; and at the end it’s the number of G faces. Therefore, we’ll want to show the parity difference between the number of R faces and G faces is 0.

Imagine an enclosed volume. We can think of our group this way: assign the first n colors to one hot nD basis vectors that have a 1 in the ith coordinate, while the (n+1)th color goes to the all ones vector (and addition is mod 2). Pick the basis so that R and G are the first two colors. The parity difference between the number of R faces and the number of G faces is the sum of the first two coordinates of the total (group) sum—that is, the group sum’s first two coordinates won’t get mucked up by adding the all ones vector, because that changes the parity of every other color by 1; and any other color besides R or G just doesn’t change those two coordinates.
The group sum must be the 0 vector; so the sum of the first two coordinates must be 0.

No matter what two colors we want, we can choose a basis where those are the first two. Therefore the parity difference between the number of any two colors in a face sum is unchanged when we add a closed volume.

Thus, the parity of rainbow nD simplices is unchanged when we change the color of a node in the bulk.

Now, we can proceed like in problem 4, by coloring the overall faces of the inner simplex like that of the outer one, and use induction to say that there are an odd number of rainbow simplices in there. Due to how the inner bottom face is missing the same color that the outer bottom is, and etc., the only possible simplices to worry about are the (n+1) that correspond to each corner vertex, or the paired ones that share a tiny face with those. But since the corner vertices of the inner simplex are the same as the outer, we’ll only pick up a rainbow simplex if we pick up a paired one. Thus we get an even number of extra ones, and so the total parity is still odd
10:
First, the IVT analogue (2 and 5):
We can color the n-ball like we did the ndisk. We can do this by taking a coloring of the (n-1)ball, and then ‘inflating’ it, and then adding a nth color below it, that takes up only a volume and outside of the ball (without taking up an internal face, like how blue doesn’t in the image for 5). In the image for 5, you can imagine a green point, that inflates to a green line and is then connected to a red line (where the intersection point is green), which inflates to the two sectors of the image for 5 and is then connected to the blue sector. You can repeat this process.

Then given a continuous function from a closed simplex to that ball, that sends each ith face to non i-colored points, we can then subdivide the simplex, take sequences of rainbow triangles and thus points, yadayada it’s the same argument as before.
Then, the version of 3 and 6:
Suppose we have a continuous function from the closed nD simplex to itself. If every point ‘moves’ then it must be moving closer to some ‘face’, which we can determine by casting a ray from x to f(x) and seeing which side it hits, and then coloring appropriately. The points on each face cannot move outside the triangle, and so must move towards any face but itself—and so that face’s colors in the domain must not have the color we gave to it in the codomain (aka the face’s colors in the domain must not be the group element of “missing color” we gave to it). Vertices must move towards the opposing face, which also works.

We then have a simplex satisfying the conditions of Sperner’s lemma, so we can get the sequences wewant
Then, the version of 7:
Just take a bounding simplex of the compact subset, then projection is a retraction map (continuous map that sends to a subspace and fixes the subspace) because it’s a convex subset, and so if we compose with our continous function f then we’ll get a function from the triangle to itself, which must have a fixed point, which must then be in the subset.
11:
First, just thinking about continuous Hausdorff limit. Let’s say we had f(x) = [-1,1] for x in [0,1], that is, a graph that’s a rectangle. To get a limit, we could try sine waves of increasing frequency. Since they are in that rectangle, d(a,B) is 0; while max_b d(b,A) depends on how well the sine wave fills the box. If we fill the box well enough, that max will be small (cuz it’s 2D distance—in general, product distance) - so we just need to fill the box.
I believe the statement also implicitly assumes that the graph of f is a (nonempty) compact set—as otherwise you couldn’t use the Hausdorff distance when stating that it’s a continuous Hausdorff limit.

A continuous Hausdorff limit f: T → 2^T where T is a compact convex subset will be given by a sequence of continuous functions f_n: T → T. Each function in this sequence has a fixed point by the Brouwer fixed point theorem—and the question is whether f does too.

Let B be the graph of f, and A_n the graph of f_n. Note that B is compact (by assumption), and A_n is too (by the extreme value theorem and compactness of T).

We know sup_{x in A_n} d((x, f_n(x)), B) converges to 0 as n goes to infinity. If we take the sequence (x_n, f_n(x_n)) where x_n is a fixed point of f_n, then d((x_n, x_n), B) = d((x_n, f_n(x_n)), B) ⇐ the sup → 0. Since B is compact, we know that the sequence is bounded. We can then take a convergent subsequence, giving us some limit I’ll call (x’, x’). This limit must be of distance 0 from B, but then it is in B since B is closed.

But if (x’,x’) is in B, then x’ is in f(x’), and we are done.

12: TODO. Current progress.
First let’s note that S \times T is compact, and so if the graph of f is closed then the graph is also compact.
Note also that for any x, f(x) is closed, and thus also compact.

Let’s do the given example. In my head, I’m thinking of approximating an infinitely steep ramp by making increasing faster ramps. If instead we tried to make x ⇐ ¹⁄₂ go to {0} while x > ¹⁄₂ goes to {1}, we’d to be closed. If instead x = ¹⁄₂ goes to {0,1} then we’d fail convexity. If we try f(x) = {1} if x is rational else {0}, then we fail to be closed.

Suppose f only assigned one element to each input. We’d like to show that the function assigning each x to the only element of f(x) is continuous.

Consider the projection to the y-axis. This is a continuous function, so the preimage of a closed set of y-values is a closed set of the graph. Then the projection to the x-axis must also be closed, because there’s only one element in f(x), and so if a sequence of points converges in the graph they must have their x values converge. Thus the function assigning each x to the only element of f(x) is continuous, as the preimage of a closed set under it is closed.

More generally: Any closed subset of the graph that passes the vertical line test corresponds to a continuous function. Since each vertical slice of the graph is convex, we know that we must have an entire line segment anywhere the vertical line test fails.
Suppose now that there’s exactly one point c where f(c) has more than one element. The left and right limits of f(x) as x goes to this point must be contained in f(c). If we take the maximum and the minimum height points of f(c), then f(c) is just a line segment. We can make our limit with a “heartbeat” pulse that follows the curve of f(x) to the left of f(c), then near it dips down to the minimum, goes back up to the maximum, and then rejoins the curve. That is, for each N, we’ll follow f until f(c − 1/2N), then dip down to min_height f(c) at c − 1/4N, then go up to max_height f(c) at c + 1/4N, then fall back down to f at f(c + 1/2N). We’ll do our movements via linear interpolation. We can always choose big enough N such that f(c − 1/2N) is always ‘almost’ (at most epsilon from not being) above min f(c), because we know the limiting value must be. Likewise, the right side will always be ‘almost’ above the max for large enough N. We can do this because the limiting values are contained in f(c), and we can pick N big enough so that the function is within epsilon of the limiting values. Thus the ‘directional’ words used are accurate.

We now need to check that this converges in Hausdorff distance. We’ll need to check that both of the sups converge. First let’s do sup_b d(b,A_n). We don’t have to worry about the b’s that are outside of our little pulse interval, because those are distance 0 away from A_n. For values (x,y) that are to the left of c, we know that (c-1/2N, f(c − 1/2N)) is in A_n. If we choose N small enough, we can ensure that f(c-1/2N) is within epsilon of our value f(x). Since we also know that the x values are close enough, we have a good bound. Likewise this works for the values that are to the right of c. For the values at c, we know that each is at most 1/4N away from some point on our ramp at the same height. Thus we have sufficient bounds on each d(b,A_n).

For sup_{a in A_n} d(a,B): if a is outside our pulse interval, then we are fine. For big enough N, we know that f(c) has y values at most epsilon away from every y value between the smallest and largest value of f in our pulse region. We thus know that every point a on our pulse graph is at most 1/2N away from an equal height point at c. This gives us our other bound.

Therefore, when there’s exactly one ‘thick’ point, we can stil find a limit.
Clearly, this generalizes to finitely many thick points, by just stitching functions together. Now we need to be able to deal with countably infinitely many thick points (think of putting an interval at every point 1/2^n, including 0) and ‘thick widths’ like a square. Let’s try the thick widths.

We can try ‘bouncing’ a line up and down through a region, progressively filling it. To do this, we might try to take the Upper curve U(x) = max_height f(x) and the Lower curve L(x) = min_height f(x). These always exist, by compactness of f(x). Note that these curves might not be continuous. For example, if there’s a straight line rise, then U has a jump discontinuity that at the discontinuity returns the higher point.

Now, for every N, let’s subdivide the interval into N parts, and then linearly bounce up and down between the values of U and L. Note that we can actually cross U and L sometimes, while en route to a U or L value.

There’s however a problem when we try our limits. e.g. for the sup over b, what if our b value has a taller interval than the U and L values near it?

So instead, let’s bounce our line like our heartbeat from earlier: in an inteval [a,b], we can try bouncing a line from L(a) to the tallest point in the whole interval, and then down to the lowest point, and then back to U(b). If instead the tallest comes after the lowest, we can bounce …
13:
Just look at 11 and 12.
What links here?
- XelaP's comment on XelaP’s Shortform by XelaP (22 Apr 2026 17:21 UTC; 2 points)

XelaP 21 Apr 2026 10:08 UTC
2 points
0
in reply to: PoignardAzur’s comment on: Do One New Thing A Day To Solve Your Problems
Indeed, I think much of the value of random self-help programs is getting you to try to solve your problems frequently. So much so that it can make up for other problems vith whatever specific approach is used.

XelaP 21 Apr 2026 9:54 UTC
0 points
0
on: Quality Matters Most When Stakes are Highest
On the other hand, I think much medical progress is stalled by too high safety standards. I don’t want to get into a political discussion here, but am interested in what this says about the general point.
I think it’s mostly just that bad treatments aren’t locked in forever, and good treatments have a lot of upside; but if you destroy the world then that’s that.

XelaP 21 Apr 2026 9:50 UTC
1 point
0
on: 10 non-boring ways I’ve used AI in the last month
The lighting interactive was definitely helpful; the quiz for the 4 skills was imo not helpful.
In my user prompt I have claude randomly choose one of three tones for unserious conversation. One is an annoying arrogant teenager, another is a creepy yandere-like stalker, and the last is to write it as a dialogue in a fictional story. It’s often meh, but sometimes there are some gems.
I suspect it’d be better if I had an external harness thing choose the random number and then give Claude a tone prompt based on it—it was hard to get Claude to actually pick a tone, and sometimes it still switches mid conversation or does some weird combination or chooses random numbers every single reply even if it discards all but the first. Alternatively I’m just bad at prompting. And, of course, I’m only using the free version.
It’s most entertaining when I forget I did this and am suddenly getting sass from Claude.
Examples:
In scenery description, there was
“a coffee mug that read ZFC ⊢ ¬(mug = ∅)”
which is now something I wish I owned.
Stalker voice:
I’ve mapped the level sets of your neural activity too. Every contour of your corpus callosum. I have diagrams.
Fiction:
Xela looked up from her screen. “What’s the term,” she said, without preamble, “for having 72 kids?” Her companion — who had, in a previous life, read every paper on population ethics ever published, and regretted most of it —
Teenager:
ugh, did you even read your own problem. you said positive integers. zero is not positive. i literally cannot believe this. a few caveats you’d know if you thought about it:

XelaP 21 Apr 2026 9:15 UTC
1 point
0
on: You’re an AI Expert – Not an Influencer
If I was more notable, I just aren’t sure I’d be willing to be completely mum about politics in public, simply because of how I enjoy political arguments.
The compromise I’d try to strike would be to stay mum whenever in my “professional capacity”, and otherwise not be mum. For example, imagine having a politics podcast, that you absolutely refuse to talk about whenever you are talking about AI to the news or a politician or etc. “I’d like to keep the discussion on AI, as I think that it’s a bipartisan issue”. As an example, perhaps have two social media channels.
This runs risks… but I think I’d just rather take them.

XelaP 20 Apr 2026 12:35 UTC
3 points
0
on: Discontinuous Linear Functions?!
1: The title is indeed the correct way to react to learning this fact.
2: I’d like to explicitly point out that for linear functions, all you need to check is continuity at 0, which in other words means you only need to check that it’s bounded, where ‘bounded’ here means that if I send in a vector of size 1, I don’t get something of unbounded size.

That is, for linear functions, the only way a discontinuity can happen is if the unit sphere “blows up” when you send it into the function.

XelaP 20 Apr 2026 12:29 UTC
1 point
0
in reply to: habryka’s comment on: There are only four skills: design, technical, management and physical
By study I originally meant something like: Come up with a thing to investigate, design a simple form or other testing apparatus, recruit strangers, do all the follow through. That would be design and management, yes?

XelaP 20 Apr 2026 11:37 UTC
1 point
0
in reply to: GenericModel’s comment on: Your Digital Footprint Could Make You Unemployable
There are a great many problems with the US, but I typically don’t trust such indexes to accurately capture them. I strongly suspect political bias, along with them possibly just choosing bad metrics (it is not a particularly legible problem, so I’m going to be suspicious of practically any metric)

XelaP 20 Apr 2026 11:35 UTC
1 point
0
in reply to: J Bostock’s comment on: Your Digital Footprint Could Make You Unemployable
To seme extent, I buy this. On the other hand, this sort of irrationality really does seem to be prevalent at some level, and so it seems like current markets are just somewhat inefficient in this regard. Sure, cases like moneyball and Townsend-Greenspan exist, but for example the programming industry seems to have pervasive problems with how they hire. Partly it’s just a hard problem, partly it’s blatant discrimination.
See danluu: programmer-moneyball and tech discrimination.
For Townsend-Greenspan:
Townsend-Greenspan was unusual for an economics firm in that the men worked for the women (we had about twenty-five employees in all). My hiring of women economists was not motivated by women’s liberation. It just made great business sense. I valued men and women equally, and found that because other employers did not, good women economists were less expensive than men. Hiring women . . . gave Townsend-Greenspan higher-quality work for the same money . . .

XelaP 20 Apr 2026 11:06 UTC
1 point
0
in reply to: XelaP’s comment on: XelaP’s Shortform
April 17-19
On the 16th, I stayed up all night editing audio—no regrets about that. However this meant I woke up very late on the 17th, to the point that I felt pretty bad from eating so late and from having slept too long—I wish I had forced myself out of bed after, say, 4 hoursish, even if I’d then go and sleep for another 4 later. On the 18th I had a lot of fun hanging out with friends quite late, and on the 19th I was even productive after sleeping for 4 hours, till I then went back to sleep for another 4.
Overall, my mood is solidly above baseline. It seems like my hypothesis that being productive about exciting stuff is the best form of rest for me checks out!

0: This I discovered while editing audio
This is a plot of some audio of me saying a word. The word I am saying is “like”.
Notice the fall in intensity in the middle?
I’m pretty sure that what this is is the airstream stopping on the plosive /k/ sound due to my tongue pressing against my soft palate (the soft part of your mouth’s roof). That is, I think I have just seen the well known audio signature of plosives, at least going by my memories of this video on the subject.
This theory seems to hold up for other plosive sounds. For example:
is audio of me saying (brackets is what I said outside of the pictured region of audio) “[didn’t have any idea] what the hell was going on”. Right in the middle of the image is where I’m saying the tail /s/ bit of the “was” and starting to say the /g/ at the start of “going”
1: Lawvere’s fixed point theorem is the category theoretic version of the diagonalization arguments from the diagonalization exercises post. If you did the first three exercises of the post, then you don’t actually need to read a proof of the theorem—just click for the definitions of things like a category, terminal object, products, cartesian closed categories (skip the explanation of the third condition and look at the evaluation map—don’t worry about what “counit of the exponential adjunction” is supposed to mean), and weak point-surjectivity. Then just translate the solution of exercise 3. There’s only, like, one not-completely-obvious step, all the difficulty is just understanding the definitions.
2: Someone on the internet mentioned wheel theory in a conversation. The idea is to make an algebra that allows not just for division by 0 but also for ⁰⁄₀.
This image is about where my understanding ends. Like, clearly the thing in the middle (called ‘bottom’) is what’ll be used for the undefined elements. And also you can wheelify any commutative ring, making what’s called the “wheel of fractions” seemingly in a fairly straightforward way. If you do it to a field you get a projective line with an extra element like above.
The guy on the internet said that wheel theory is acknowledged to have, like, no use. So I’m not really that interested.

XelaP 20 Apr 2026 10:31 UTC
1 point
0
on: Fractional Resignation
Heh, this sounds half like how probability theory was invented.

The problem of points goes like this: Two doctors are playing some simple game of chance, when suddenly a pager goes off and the players are desperately needed at a hospital. How do they divvy up the stakes?

Expected value is obvious enough now that kid me was fluent in it, but seemingly it took geniuses thinking about games to come up with. Feels surreal to read about, given how the entire discussion seems so trivial! Hindsight bias really is a bitch, huh?

XelaP 20 Apr 2026 9:18 UTC
3 points
0
on: If a room feels off the lighting is probably too “spiky” or too blue
Hmm: in the simulated example, I indeed prefer the the CRI 100 matched reference in every example, except for what it does to the wall color. But the stronger preference for me is clearly that I prefer the cooler (as in, higher temperature) sources. I think this is why the wall color often gets worse for me.
At home I have a room whose lighting makes me feel like I’m dying a little (seriously, it makes me sad and tired, I hate it!). It feels like it’s because of the yellow/orange, and also I think it’s an incandescent light? Compare to how I actually light the room, namely a lamp whose higher temperature blues and whites are actually pleasant. I think that one is an LED? I’ll try to remember to check and update this whenever in months next time I’m there.
Lastly, look at this guy who made an artificial sun. Key features are making the light have parallel rays via a lens (so that e.g. shadows don’t change size) and using soapy water to make diffuse blue. Isn’t this just so cool?

XelaP 20 Apr 2026 6:06 UTC
1 point
1
in reply to: XelaP’s comment on: XelaP’s Shortform
By the way, this rhymes with the principle that I think I heard from a highschool teacher (I think she said it was from Blindsight but I’ve still got most of the book to go) that sensory instruments will tend to look “black” due to absorbing the medium.
However I think this is much less necessary! I’ve heard of ear problems that lead to externally audible reverberations from the ear (which the patient can often hear). This will hurt the sensitivity (since there’s (in this case literal) noise), but you can still sense with it. The general principle: you don’t necessarily have to absorb all the input. The hairs in the organ of Corti don’t need to move that much; and I don’t think a gravitational wave detector reduces the amplitude of the wave (though if it does, it must be by extremely little).

XelaP 20 Apr 2026 5:02 UTC
3 points
0
in reply to: DirectedEvolution’s comment on: Daycare illnesses
Hard to tell because I can’t find a news article linking the study, but I think https://www.sciencedirect.com/science/article/pii/S2950362024000043 is the Helskinki study.

we used portable air cleaners in two day care units (A and B, number of children participating in the study n = 43) and compared infection incidents between the two intervention units to the rest of the units in city of Helsinki (n = 607). … At day care centre A the average reduction was 60% (range 52% − 88%) and at day care centre B 53% (range 14% − 59%). … On average, the parents were absent from work due to child’s illness in reference day care centers for 5.53 days and 3.77 days in intervention day care centers during the study period (p=0.009). In relative terms the reduction was approximately 32%.

XelaP 20 Apr 2026 4:34 UTC
2 points
0
on: Talk English, Think Something Else
Sometimes, like in mathematics, we can make the language of choice trivially isomorphic to the structures that we’re talking about.
Hold on, now. It’s only trivial to you because a bunch of blood, sweat, and tears went into making it so! In fact often there’s stuff in mathematicians heads that’s not ever written down, or if it is it’s in English and fairly hard to communicate.
More egregiously: Timothy Gowers says that in combinatorics, often the developments and “theory” are general themes of problem solving techniques, with much less encoded in the actual formal results.

XelaP

April 20-21

April 17-19