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This tag is used if a reference is needed in a paper or textbook on a specific result.

11 votes
1 answer
468 views

Uncountable families of measurable sets with pairwise positive intersections

Let $(X,\mathcal{B},\mu)$ be an arbitrary finitely additive probability measure space, let $a>0$ and let $(A_i)_{i\in I}$ be an uncountable family of subsets with measure $\geq a$. Is there an uncount …
5 votes
0 answers
101 views

Non-monotileable amenable groups

This is crossposted from MSE. We say a subset $A$ of a group $G$ is a monotile for $G$ if $G$ is a disjoint union of right translates of $A$. In his article Monotileable Amenable Groups, B. Weiss give …
1 vote
0 answers
182 views

Does every amenable group $G$ admit a two-sided Folner sequence?

By two-sided Følner sequence I mean a sequence $(F_N)_N$ of subsets of $G$ which is both a left-Følner and a right-Følner sequence. Context: I just came up with this question and surprisingly I haven' …
3 votes
1 answer
210 views

Maximum cardinality of separated sets in the Hamming distance

This question is motivated by section 15.1 (Codes) of Alon and Spencer's The probabilistic method. Fix $\alpha<\frac{1}{2}$ and for each $n\in\mathbb{N}$ let $\{0,1\}^n$ be the length $n$ binary strin …
4 votes
1 answer
281 views

Is every 1-Lipschitz homeomorphism $f:X\to X$ from a compact metric space to itself an isome...

I found a statement involving a homeomorphism $f:X\to X$ of a compact metric space $X$, with Lipshitz coefficient 1, i.e., a non-expansive map, and cannot think of an example where $f$ is not an isome …
1 vote
Accepted

Bisector of two points in a Riemannian manifold has measure $0$

Here is a short proof supposing that $M$ is complete (if not the statement is false, see the last paragraph) using the idea from Leo Moos' answer of using Rademacher's theorem. Suppose $\mathcal{B}(p, …
Saúl RM's user avatar
  • 10.6k
7 votes
2 answers
177 views

Bisector of two points in a Riemannian manifold has measure $0$

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$? I was thinkin …
4 votes
Accepted

What is the minimum-curvature curve interpolating a given set of points in the plane?

As Matt F. says, his answer is not optimal, but the optimal solution, for most polygons (see (!!) below) comes from a similar construction using just arcs of circumference and segments. This answer gi …
Saúl RM's user avatar
  • 10.6k
9 votes

What is the minimum-curvature curve interpolating a given set of points in the plane?

(This answer was posted before the convexity condition on the curve $\gamma$ was added to the question) Suppose you have any finite set of points in $\mathbb{R}^2$, and rotate $\mathbb{R}^2$ so that t …
Saúl RM's user avatar
  • 10.6k
2 votes
Accepted

First visit of intervals for an irrational rotation

No, it is not possible. In the following I will use $I_n=(a_n,b_n)$ instead of $[a_n,b_n)$ (this is not a problem, you can just increase $a_n$ a bit so that the statement with $I_n=(a_n,b_n)$ is stron …
Saúl RM's user avatar
  • 10.6k
22 votes
Accepted

The $9$th tetration of $-\sqrt2$

This is not a huge coincidence: the idea is that the sequence $a_n={}^{n}(-\sqrt{2})$ has small norm until $n=6$, then it gets out of hand for $n=7$ ($a_7\sim-33+29i$), so that $a_8=e^{a_7\ln(-\sqrt{2 …
Iosif Pinelis's user avatar
1 vote
0 answers
74 views

Keller's cubing conjecture but with arbitrary cubes of side $1$

These days I have been reading about Keller's cube tyling conjecture, which asks if in any covering of $\mathbb{R}^n$ by translates of $[0,1]^n$ with disjoint interiors there are two cubes sharing one …
6 votes

What is the smallest size of a shape in which all fixed $n$-polyominos can fit?

It seems $S_n$ is $\geq\displaystyle\Theta\left(\frac{n^2}{\log(n)}\right)$. In the following, I will consider polyominos as subsets of $\mathbb{Z}^2$ (so, a polyomino is represented by the set of cen …
Saúl RM's user avatar
  • 10.6k
40 votes
2 answers
2k views

Can the nth projective space be covered by n charts?

That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$? I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t b …
5 votes
0 answers
112 views

Stronger form of countable dense homogeneity

I am completing my undergrad thesis about topological properties of some subspaces of the real numbers, and CDH spaces are one of the topics I´ve covered (I know almost nothing about it, I only prove …