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Let $p$ be the closest equidistant point to your set of points, $\{p_1,\dots,p_N\}$. Then $p$ is in the affine subspace $X$ generated by the points: if not, the orthogonal projection of $p$ in $X$ is …

Any curve $\gamma:[a,b]\to M$ parametrized by arc length is an $\varepsilon$-geodesic for any $\varepsilon>0$. The inequality $(1-\varepsilon)dist_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y) …

There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R …

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 …

$K_t$ 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 …

The result is not necessarily true. Consider for $n\geq1$ the open ball $R_n$ with diameter from $(\frac{1}{n},0)$ to $(\frac{1}{n+1},0)$. Let $L_n$ be the reflection of $R_n$ with respect to the $y$ …

To give a positive answer to the question it is enough to, for a fixed $\varepsilon$, give a collection of disjoint balls in $C[0,1]$ of radius $\varepsilon$ which is dense in $C[0,1]$. Indeed, then f …

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 …

I am following Clara Löh's Geometric Group Theory. An Introduction, and in remark 5.1.12, she defines the category QMet whose objects are metric spaces and whose morphisms are quasi-isometric imbeddin …

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 …

This construction (the picture has the case $N=6$) seems to work. It is obtained in 2 steps: -Begin with a convex set formed by $N$ equal pieces, each having a boundary formed by two segments and a pi …

As Sam Hopkins commented, 8 vertices are enough. Let $Q$ be the pentagon from the picture and let $\pi$ be the plane containing it. Now we can define the triangle $P$ as a triangle of less diameter th …

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 $ …

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 …

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: …