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Yes, it is possible, using a modified version of the grid construction from the question (thanks to jackdean for the more elegant version of my argument). Firstly, let $N=5^{24}$, so that $x^2+y^2=N^2 …

Here are triangulations of a side $1$ square with vertices at a arbitrarily high distance of all the four vertices of the square. The sides of the side $1$ square are not edges but it is easy to see t …

$K_t$ need not be a translate of $K$. Let $A=[-4,4]\times[-1,1]\subseteq\mathbb{R}^2$ and consider the convex cone $K=\{t\cdot v;t\in[0,\infty),v\in A\times\{1\}\}\subseteq\mathbb{R}^3$. Note that $\p …

Yes, the almost partition exists. Instead of letting $\mu(E)\geq\frac{\mu(\Omega)}{2}$, I let $\mu(E)\in(0,\mu(\Omega))$ be arbitrary and proved that you can divide $\Omega$ into $N$ open simply conne …

Yes, $\mathbb{R}^n\setminus E$ has to be path-connected. Let $x,y\in\mathbb{R}^n\setminus E$, we will prove that there are many paths going from $x$ to $y$ inside $\mathbb{R}^n\setminus E$. We can sup …

I think (see (??) below) there is no connected complete Riemannian manifold $(M,g)$ which is not flat and such that there is an isometry $f:(M,g)\to(M,\lambda^2g)$, where $\lambda\neq0,1$. I will assu …

It seems $\gamma:\mathbb{R}\to\mathbb{R}^2$ (I assume $\gamma$ is injective and continuous) is indeed a line. My argument is very similar to the one by Ilkka Törmä (I thought I could write a shorter o …

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 …

Not necessarily. Consider the sphere $(\mathbb{S}^2,g)$ with its usual metric and give it a new metric $hg$, where $h\leq1$, $h$ has three local minima $h(p_1)=0.7,h(p_2)=0.8,h(p_3)=0.9$ (where $p,q,r …

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$. Below I show it for $p>2$, but this is enough because if we have convex sets $A\subseteq B$, …

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, …

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 …

This is true if $\Omega\neq\mathbb{R}^2$ (so that every path component of $\Omega$ has nonempty boundary). Firstly, $F^*e=e$ means that $F$ is a local isometry. We know that two local isometries $f,g: …

If I have understood the definitions correctly, $f(K)$ need not be doubling. For example consider a map $f$ from $[0,1]$ to the Hilbert space $\mathbb{R}\times l^2$ defined in the following way. Let $ …

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 …