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4

In $\mathbb R^4$ pick regular $n$-gons around $0$ with one in the $(x,y,0,0)$ plane, and the other in the $(0,0,z,w)$ plane. So essentially the isometries of one do not affect the others. A "natural" space with $\mathbb Z/n\mathbb Z$ symmetry is a pyramid with base the regular $n$-gon[*]. In general, if there are spaces $U\subseteq \mathbb R^n$ and ...


3

Alternatively you can use the equivalently definition that : $K$ is compact $\iff $ Every open cover of K has a finite subcover Let $U_1(x)$ be a ball with radius $1$ around x. Then cover your set with those balls and use the compactness to get a finite cover of those balls with radius 1. Its easy to conclude now that your set is bounded above. Edit: ...


2

Since the Hausdorff metric does not distinguish between a set and its closure, we can take the quotient of $2^Y$ by the equivalence relation "$A\sim B$ if $\overline{A}=\overline{B}$", reducing the problem to the study of the space of closed subsets. This is usually called a hyperspace, though hyperspaces come in different flavors: e.g., in normed spaces one ...


2

Problem i) $\forall x,x'\in A$ and $\forall y,y'\in B$, by the triangle inequality we have: $$d(x,y)\leq d(x,x')+d(x',y')+d(y',y)$$ Since $d(A)\geq d(x,x')$ and $d(B)\geq d(y,y')$, we can say: $$d(x,y)\leq d(A)+d(x',y')+d(B)\qquad(*)$$ from which we deduce that $$\sup_{x\in A\; y\,\in B}d(x,y)\leq d(A)+d(x',y')+d(B)\qquad(**)$$ This follows from the fact ...


2

A simple example of why this is headed in a wrong direction: If $X=\mathbb [0,1]$ and $Y_1=[0,1]$, then the integral might be defined as normal. If $Y_2=[-1/2,1/2]$, then the integral defined as normal also exists. But, as metric spaces without knowing their "real number" structure, $Y_1$ and $Y_2$ are essentially the same metric space, but the integral ...


2

I like to think of that $n$ as the dimension of the underlying manifold (I guess this is also the reason behind this). So you can remember, that $S^n$ is defined as the unit vectors in $\mathbb{R}^{n+1}$, since it is an $n$-dimensional "object". This is also, since it is the boundary of $D^{n+1}$ which is $(n+1)$-dimensional. Therefore I would definitely ...


1

I don't think that you need to treat odd and even cases separately, but the parity observations are good ones. Assume that $m, k$ and $n$ are distinct integers, since the other cases are easily dealt with. Suppose that $l_1!$ is the maximal factorial dividing $|m-k|$ and $l_2!$ is the maximal factorial dividing $|k-n|$. Then $\min\{l_1!,l_2!\}$ divides ...


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I think no. Say distance is d, then forall $N$ you have d distance between members, so no N for $n>N$ shorter than thje distance $\epsilon$. So it not convergent.


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In greater detail than my comment. A metric space is Hausdorff, so given $x_n\to 1$ you know for $n>N(\epsilon)$ that $d(1,x_n)<\epsilon$, so choose $$\epsilon = {d(0,1)\over 2}>0$$.


1

To complete the proof indicated (although I Balla's approach is generally better) show that $x_k$ is not cauchy, given $n_0$ for any $m>n_0$ we have $$d(x, x_{m}) \leq d(x, x_{n_0}) +d(x_m, x_{n_0})$$ and so $$d(x, x_{m}) -d(x, x_{n_0}) \leq d(x_m, x_{n_0})$$ and by chosing $m$ large enough we have that $d(x_m, x_{n_0})$ is large.


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A sufficient condition is that any two points of $X$ can be joined by a local geodesic (it does not have to be globally minimizing), and geodesics do not split: if two geodesics share an arc, they coincide. In the literature, the latter condition is usually stated simply as "geodesics do not split". Suppose the above hold. Given any point $x\in X$, ...


1

To keep in touch with your previous work, if you suppose that $x_{n_k}\to y$, note that the function $h\to d(x,h)$ is continuous. Therefore, the sequence $d(x,x_{n_k})_k$ converges to $d(x,y)$. But since $d(x,x_{n_k}) > n_k$ and the sequence $(n_k)_k$ is increasing, the sequence $d(x,x_{n_k})$ is unbounded and convergent. Contradiction. A ...


1

$\mathbb{Z}/n\mathbb{Z}$ is the fundamental group of the 3-dimensional Lens space $L(n,1)$. There is a universal covering map $S^3 \mapsto L(n,1)$ and a deck transformation action of the group $\mathbb{Z} / n \mathbb{Z}$ on the space $S^3$ which acts by isometries of the standard metric on $S^3$. This descends to a metric on $L(n,1)$ which is locally ...


1

The Ascoli Arzela theorem tells you exactly what you need: from a sequence of bilipschitz homeomorphisms $f_i$ for which $dil(f)$ approaches 1, you extract a subsequence converging to one with dilation equal to 1. It's not true that the set of homeomorphisms $X\to Y$ is compact, but the magic of Ascoli Arzela is a very useful criterion on sequences of ...


1

The short answer is no. In practice, for a generic metric space one must make specific requiriments (like: goedesics exist, geodesics are unique and so on...) You can build many counterexamples to the local uniqueness of geodesic by simple means. Example 1) the $L^1$ metric. No local uniqueness of geodesics between two points. Take $\mathbb R^2$ with the ...


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Suppose that $(X,d')$ is compact (so $(X,d)$ is compact as well). Given $\epsilon>0$, we have the cover $\left\{B_d(x,\frac{\varepsilon}{2}):x\in X\right\}$ of $X$ by $d'$-open sets. If $\epsilon'$ is a Lebesgue number for that cover, then $\epsilon'$ satisfies the condition you want. Notice that the conjecture means that the identity $(X,d')\to ...


1

First I will summarize the question using more definitions, to ease the discussion: Curves and Lengths (1) Curves are continuous functions from an interval of reals into the metric space. (2) A polygonal path is a sequence of points (vertices) in the metric space. (3) The length of a polygonal path is the sum of the point-to-point distances going from ...



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