{ERRONEOUS] A Counterexample Against Kakeya's Conjecture

Abstract

I give a counterexample against Kakeya's conjecture for n = 2, signifying that the classic proof was in error.

Full text

[ERRONEOUS] A COUNTEREXAMPLE AGAINST KAKEYA’S
CONJECTURE
VICTOR PORTON, ORCID 0000-0001-7064-7975
Abstract. [I found an error in my proof, see “FIXME” below.] I give a
counterexample against Kakeya’s conjecture for n= 2, signifying that the
classic proof was in error.
1. Intro
It can be seen from the drawing on figure 1, that Kakeya conjecture is false
for n= 2. Below I will present a rigorous proof of this.
Figure 1. Counterexample to Kakeya for n= 2.
The blue is the Cantor set, the red is a Besicovitch set
Therefore the classic proof [1] is wrong. Many articles need to be retracted.
2. The problem
I will point a counterexample against Kakeya’s conjecture for n= 2.
Kakeya’s conjecture for n= 2.A union of unit-length intervals on a plane
has Hausdorff dimension 2, if there are intervals of all possible directions among
them.
I will denote Hausdorff dimension by dim.
2.1. The counterexample. Let x7→ u(x) = (x1, x2,...,) be any of (many) map-
pings from x∈[0; 1] to its binary representation, i.e. xi∈ {0,1},u(x) = P∞
i=1
xi
2i.
Then define c(x) = P∞
i=1
2xi
3i. It is well known that c(x)∈Cwhere Cis the
Cantor set.
The counterexample is the union
(1) U=[
x∈[0;1[
Sx
of intervals
Sx={(c(x)−tcos(πx), t sin(πx)) |t∈]0; 1[}.
2020 Mathematics Subject Classification. 28A80, 51N20, 54H99, 28A78, 54G20, 51M05.
Key words and phrases. Kakeya conjecture, dimensionality, geometry, general topology, long-
standing open problem.
1

[ERRONEOUS] A COUNTEREXAMPLE AGAINST KAKEYA’S CONJECTURE 2
3. The proof
Lemma 1. Each interval Sxis of length 1.
Proof. Because sin2(πx) + cos2(πx) = 1. □
Lemma 2. Ucontains intervals of all directions.
Proof. Because πx has all values in [0;π[. □
Lemma 3. dim U < 2.
Proof. Partition [0; 1[ into 2mintervals Is=k
2m;k+1
2m,k= 0,...,2m−1, each of
length 1/2m. Consider binary digits s= (s1, . . . , sm) (the first mbinary digits of a
binary number 0.s1. . . sm). Define Us=Sx∈IsSx, so by (1) we have U=SsUs.
For x∈Ix, express c(x) = as+r, where
as=
m
X
i=1
2si
3iand r=
∞
X
i=m+1
2xi
3i.
We have r≤1/3mby properties of ternary numbers. So c(x)∈[as;as+ 1/3m].
Approximate Uswith a parallelogram Ts:
Ts=[
c∈[as;as+1/3m]
{(c−tcos(πxs), c +tsin(πxs)|t∈[0; 1]},
where xs=k/2m. This parallelogram has base 1/3mand height |sin(πxs)|with
area O(1/3m).
|cos(πx)−cos(πxs)|and |sin(πx)−sin(πxs)|are (mean value theorem) O(1/3m)
[FIXME: should be O(1/2m)], so Uslies within a δ-neighborhood of Ts, with δ∈
O(1/3m), because:
c(xs)−tcos(πxs), c +tsin(πxs)−c(x)−tcos(πx), t sin(πx)=
c(xs)−c(x)−tcos(πx)−cos(πxs), tsin(πxs)−sin(πx)∈O(1/3m),
because c(xs)−c(x)∈O(1/3m) and |cos(πx)−cos(πxs)|and |sin(πx)−sin(πxs)|
are O(1/3m), |t| ≤ 1.
Further, split Tsinto smaller parallelograms:
Tsl =[
c∈[as;as+1/3m]c(xs)−tcos(πxs), c(xs) + tsin(πxs)|t∈l
2m;l+ 1
2m,
where l= 0,...,2m−1.
Let fix balls of the radius ϵ= 1/3m.
Cover Tsl with balls. Since Tsl lies inside a rectangle of dimensions 1/3m×1/2m,
we can cover it with
O(1) ·O(3m/2m) = O(3m/2m)
balls.
Therefore Tscan be covered by O(2m)·O(3m/2m) = O(3m) and thus Us(as
O(1/3m)-neighborhood of Ts) can also be covered by balls of radius 2ϵwhere ϵ∈
O(3m), that is by O(3m) balls of radius ϵ.
For U, because it consists of 2msets Us, use:
2m·O(3m) = O(2m·3m)
balls.

[ERRONEOUS] A COUNTEREXAMPLE AGAINST KAKEYA’S CONJECTURE 3
I remind, that d-dimensional Hausdorff measure is defined by the formula:
Hd
ϵ(U) = ((diam Ui)d|
∞
X
i=1
Ui⊇U, diam Ui≤ϵ).
The d-dimensional Hausdorff measure sum (the sum in the above formula) can
be as small as:
O(2m·3m)·O(1/3m)d=O(2m·3m·3−md) = O(2m·3m(1−d)).
Obviously, 2m·3m(1−d)= 2m·3m(−1−ϵ)→0 for d= 2 −ϵfor a small ϵ > 0.
So the Hausdorff measure Hd(U) = 0. Therefore
dim U= inf d≥0|Hd(U)=0≤2−ϵ.
□
Theorem 1. For Besicovitch set Uits Hausdorff dimensionality is less than 2.
Proof. Combining the lemmas. □
4. “Political” context
I discovered ordered semigroup/semicategory actions in 2019. The world didn’t
react obeying like a sheep to decision of Russian Orthodoxes to kick people like
me from normal life such as academic career. That’s scary: the most valuable
component is a missing component. Missing ordered semigroup actions amount to
like half of world economy missing [2].
I don’t know how Kakeya conjecture is used in the rest of math, but I use it to
build a bridge between two about halves: academic math and the second lost half,
through my glorification.
The advice: “Solve some famous open problem to prove that you are a good
mathematician.” sounded for me like a mocking. But in the new reality, I indeed
did it.
References
1. Roy O. Davies, Some remarks on the kakeya problem, Proc. Cambridge Philos. Soc. 69 (1971),
no. 3, 417–421, URL:https://doi.org/10.1017/S0305004100046867.
2. Victor Porton, Why algebraic theory of general topology is
super-important, URL:https://science.vporton.name/2022/09/09/
why-algebraic- theory-of- general-topology- is-super- important/.
Email address:porton.victor@gmail.com