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0

You may be interested in the fact that there exists a "Gromov-Hausdorff-Wasserstein" distance and also a "Gromov-Hausdorff-Prokhorov" distance, each between two metric measure spaces. You'll find some insight in Gromov-Wasserstein distances and the metric approach to object matching. F. Mémoli. Foundations of Computational Mathematics. 11(4), August 2011, ...


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Let $(X,d)$ be a finite metric space, pick $x\in X$. We want to show that $\{x\}$ is open which is the same as saying there is a radius $r>0$ such that $\{y\,:\, d(x,y)<r\}=\{x\}$. How do we show such a radius exists? Well for each $y\in X$ such that $y\neq x$ we know that $d(x,y)>0$. As there are only finitely many such $y$s there is a least $r$ ...


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Hint: show that in any finite metric space, all singletons (sets with a single element) are open. From there, it is easy to show that every subset of a finite metric space is open.


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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 $|k-n|$. Without loss of generality assume ...


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You can also prove this is impossible working directly from definitions. Let $s_n$ be the sequence defined by $s_n=0$ if $n$ is even and $s_n=1$ if $n$ is odd. Suppose that this sequence converges to some $y$ with respect to some metric $d$. Then for all $\delta>0$, there exists $n$ so that $m>n$ implies $d(s_n,y)<\delta$. Either $y \neq 0$ or $y ...


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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 $$\epsilon = {d(0,1)\over 2}>0$$.


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Let the $n$th term of our sequence be $a_n$. Suppose there is a metric $d$ such that $\{a_n \}$ converges on $(\mathbb{R}, d)$ to $p \in \mathbb{R}$. Since $0 \neq 1$, it follows from the definition of a metric that $d(1,0) > 0$. Set $\delta < d(1,0)/2$. Then there exists an $N$ such that for all $n \geq N$ we have $d(a_n,p) <\delta$. Assume ...


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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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If the sequence converges, then any subsequence of it converges to the same limit. This can be proven easily for any metric space. Any constant sequence converges to its repeated element. This too holds in any metric space. It now follows that the sequence you suggest can never converge under any metric on $\mathbb R$. More is true. Given any alternating ...


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Assume that for a certain metric $d$ your sequence converges to $g\in\mathbb{R}$. Then its even subsequence also converges to $g$ but it converges to $0$ thus $g=0$ similarly considering odd subsequence one shows that $g=1$ which gives a contradiction.


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Hint: Look at the odd/even subsequences. What do we know about limits of subsequences of a convergent sequence? Alternate hint: Is it Cauchy?


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$\mathbb{Z}/n\mathbb{Z}$ 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 ...


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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 ...


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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 ...


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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 ...


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Let $k:X\rightarrow Y$ be continuous on $X$, then if $p$ is a limit point of $M\subset X$, then $k(p)$ is a limit point of $k(M)$ or $$k(\overline{M})\subset \overline{k(M)}.$$ $(f-g)(x)$ is continuous hence, let $E$ be a dense subset of $X$, then $$(f-g)(X)\subset (f-g)(\overline{E})\subset \overline{(f-g)(E)}=\{0\}$$


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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 $x\in X$, ...


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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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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: ...


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Hint: Given $x_1, \ldots, x_n$, since $K$ is unbounded, $\displaystyle\exists x_{n+1} \in K \setminus \bigcup_{i=1}^nB(x_i, 1)$. Then $d(x_i, x_{n+1}) \ge 1\ \forall i$.


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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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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 ...


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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 ...


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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 ...


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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 ...


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On the positive side, if $X$ is a goedesic metric space with negative curvature in some sense, for instance if $X$ is locally a CAT$(0)$ space, http://en.wikipedia.org/wiki/CAT%28k%29_space then localy, given two points $x,y$ there is a unique geodesic between them.


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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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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 ...


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If the sets are disjoint then clearly this is satisfied with equality. If one set is a subset of the other (proper or otherwise) then clearly we have inequality. Now given any two sets $A,B$, you can split up $B$into the part which is a subset of $A$ and the part which is disjoint of $A$ (either of these parts may be empty in general). Now apply the two ...


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This question was asked almost a year ago; I wanted, however, to propose an alternate solution. Let $(X,d)$ be a metric space and let $S$ be a finite subset of $X$, say $S=\{x_{1}, \ldots, x_{n}\}$. A set is closed if its complement is open. So, we shall show that $X\backslash S$ is open. Pick an arbitrary point $x \in X \backslash S$. Let ...


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No. Let $$U=\bigcup_{n=1}^\infty B\left((2n+2^{-n},0),1\right)$$ Then $U$ is connected, but for $0<\epsilon<1$, we find that for all $n$, the ball $B((2n+2^{-n}),1-\epsilon)$ is contained in $U_\epsilon$, but for sufficiently large $n$ there is a gap between the $n$th and the $(n+1)$st ball. Indeed, by Pythagoras the width "neck" between consecutive ...


2

Let $A \in \mathcal{C}\big( [0,1] \big)$ be the subset defined by : $$ A = \left\{ f \in \mathcal{C}^{2}\big( [0,1] \big), \; f(0)=f'(0)=1 \quad \mathrm{and} \quad \Vert f'' \Vert_{\infty} \leq 2 \right\}$$ where $\Vert \cdot \Vert_{\infty}$ denotes the norm on $\mathcal{C}\big( [0,1] \big)$ defined by : $\displaystyle \forall f \in \mathcal{C}\big( [0,1] ...


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In a metric space, the following two things are true: any compact set is closed, and any finite set is both closed and compact. Your statement that a finite set might not be closed is not true for metric spaces.


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You've said it yourself. If $A = [0, 1)$ and $B = [1, 2]$ then $A \cap B = \emptyset$ and $\bar{A} \cap B = {1}$. These sets needn't be disjoint. We could make $B = \{0\} \cup [1, 2]$. We can have both sets connected in $\mathbb{R}^2$, if you like.


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Since $ h:Y\rightarrow X $ is a homeomorphism where $(X,d)$ is a metric space and $(Y, \tau)$ is a topological space, the fuction $ \rho:Y\times Y\rightarrow [0,\infty) $ given by $ \rho(y_{1},y_{2})=d(h(y_{1}),h(y_{2})) $, for all $ y_{1},y_{2}\in Y $ is a metric on Y. Let $A\in \tau$. Since $ h^{-1} $ is a continuous function $h(A)$ is d-open in X. Let ...


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I came across with the same question and found no answer in the internet or any lectures yet. But here are some considerations I made by myself: Definition. Let be $\otimes$ a binary operation on $\mathbb{R}_+$ (this is the map $f$, you mentioned), then we call it a catheti product if it satisfies the following conditions: $\otimes$ is ...


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This is not the definition the theory began with; the comparison of geodesic triangles came first. The four-point condition you see has been distilled from that, even shedding the geodesicity assumption. Since a complete CAT(0) space is geodesic, I recommend drawing geodesics between the points you have. The picture will look like this: I added an ...


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They aren't equal only equivalent, which metrics are if they induce the same topology. This means that they have the same open sets, and since a field is also a one-dimensional vector space, this just means they need to have the same unit balls. But then $$\lVert \cdot \rVert_1\le 1 \iff \lVert \cdot\rVert_2\le 1$$ so choose any $y$ such that $0\ne \lVert ...


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As it turns out, the claim is indeed true. The $\sigma$-compactness of the space $R$ ensures that $R$ is still $\sigma$-compact (and thus a Borel-set) in the completion of $R$. Also (which turns out to be equivalent (for separable metric spaces)) it ensures that every finite measure on $R$ is tight. These two (for separable metric spaces equivalent) ...


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Assuming that $p\in X$ and $\delta > 0 $ are fixed (as stated in the comments), it seems clear that $A$ could be open (in fact this is the more "natural" conclusion, unless you can provide more details). To see this, let $p=0\in \mathbb{R}^2=X$, and let $d$ be the usual metric on $\mathbb{R}^2$, namely, $d(x,y)=|x-y|=\sqrt{x^2-y^2}$. The set $A^c$ in ...


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Hint: does the complement of $A$ in $X$ have accumulation points not belonging to it? Recall that $A$ is open in $X$ if and only if its complement in $X$ is closed.


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Sure. Fix a point $x_0\in U$. Let $U_n$ be the set of points reachable from $x_0$ by a path lying in $U$ having length $<n$ and keeping a distance $>1/n$ from the complement of $U$.


1

$A^\varepsilon$ need not be closed. If we take $A = B_r(0)$ in a normed vector space (for example $\mathbb{R}^n$), then $A^\varepsilon = B_{r+\varepsilon}(0)$ is open, but not closed. Even if $A$ is closed, $A^\varepsilon$ need not be closed, take $$A = \left\{ \left(\varepsilon + 2^{-n}\right)\cdot e_n : n \in \mathbb{N}\right\} \subset ...


2

The function $h$ is not necessarily even continuous. Let $X=\mathbb{R}, A=\mathbb{Q}, B= \mathbb{R} \setminus \mathbb{Q}$. Define $f(x) = 1$ for $x \in A$, $g(x) = 0$ for $x \in B$. Then both $f,g$ are uniformly continuous on $A,B$ respectively, but the function $h$ is not even continuous. Note: By taking $A=[0,1], B = \mathbb{R} \setminus A$ we obtain ...


1

Counterexample for the problem as written: $A = (0,1)$, $B = (1,2)$. Take $$ f(x) = \begin{cases} 0 & x \in A\\ 1 & x \in B \end{cases} $$ Then $f$ is not uniformly continuous on $A \cup B$. This will work, however, in the case that both $A$ and $B$ are compact. Proof given compactness: We know the following: for every $\epsilon > 0$, there ...


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Without connectedness we have the following counter example. Let $A = (-\infty,0)$ and $B = (0,\infty)$. Let $f : A \to \Bbb{R}$ and $g : B \to \Bbb{R}$ be given by $f \equiv 0$ and $g \equiv 1$. Then $X = A\cup B = \Bbb{R} \backslash \{0\}$ and $$ h(x) = \begin{cases} 0 & x < 0 \\ 1 & x > 0 \end{cases} $$ which is not uniformly continuous.


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Such issues are encountered when one attempts to glue metric spaces together, even when the parts to be glued are disjoint. For example, let $A$ be the interval $[0,1]$ with the standard metric $d_A(x,y)=|x-y|$, and $B=[2,3]$ with the metric $d_B(x,y) = 10|x-y|$. We could try to put these together into one space $X=[0,1]\cup [2,3] $, extending the metric as ...


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Your method are doomed to fail. Here is why: Totally bounded is equivalent to the condition that the space have finite cover each with radius less than $\epsilon$ for any $\epsilon>0$. Metric subspace of a totally bounded metric space is also totally bounded. Trivially prove from above. By Heine-Borel theorem: every closed and bounded subset in ...


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I cannot recommend this book enough: http://www.amazon.com/Introduction-Metric-Topological-Spaces-Mathematics/dp/019956308X Generally one approaches metric spaces before general topology because metric spaces are in general more concrete/intuitive, and topology is somewhat more of an abstraction. Generally in order to understand concepts in topology, one ...



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