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1

Let $X=\mathbb{R}$ with the discrete metric. Then any function is continuous. Let $Y=\mathbb{R}$ with the standard metric. Define $f:X\rightarrow Y$ to be $f(1)=1$ and $f(x)=\sqrt{2}$ for all $x\neq 1$. Finally set $A=\mathbb{Q}$. Then $\overline{\mathbb{Q}}=\mathbb{R}$ and $f^{-1}(\mathbb{R})=\mathbb{R}$. But, $f^{-1}(\mathbb{Q})=\{1\}$.


3

Since $A\subseteq\bar{A}$ you know that $f^{-1}(A)\subseteq f^{-1}(\bar{A})$, so $$ \overline{f^{-1}(A)}\subseteq f^{-1}(\bar{A}) $$ because $f^{-1}(\bar{A})$ is closed by continuity of $f$. Now, if your claim is true, you'd conclude that $\overline{f^{-1}(A)}= f^{-1}(\bar{A})$ for every $A\subseteq Y$. Can you?


4

Let $Y = \mathbb{R}$ with usual metric and $X = (0,1) \cup \{2\}$ with metric inherited from the standard metric on $\mathbb{R}$. let $f(x) = x$ on $(0,1)$ and $f(2) = 1$ and take $A = (0,1)$. it is then easy to see that for this particular case your statement doesn't hold


2

As I mentioned in my comment, the spheres $S^n$ provide a family of examples. A "sphere" in $S^{n}$ is a copy of $S^{n-1}$. If we normalize the "great circle distance" metric on $S^n$ to have $d(x,-x)=1$, then a sphere around $x$ of radius $r$ is also a sphere of radius $1-r$ around the antipodal point $-x$.


2

By the Smirnov metrization theorem, a space that is locally metrizable and paracompact is metrizable. So as each path component of a Riemannian manifold is metrizable as you explain, local metrizability clearly follows. Paracompactness is often required in the definition of a manifold (or that it is second countable, which implies paracompactness), which is ...


3

+1 for using this book. It's one of my favorites. To your question. Think about $S^1$ in $\Bbb R^2$. Then every 'sphere' is simply 2 points picked from $S^1$. This 'sphere' has 2 centers. The point that is circumferentially central to both points and the point diametrically opposed to that first point. Think about given two points on $S^1$ and either of ...


1

Assume the manifold $M$ is connected. If there exists a piecewise-smooth path from $p$ to $q$ in $M$, then there exists such a path between $p$ and $q'$ for any $q'$ in a small closed neighborhood of $q$, since $M$ is locally path-connected. The set of such $q$ is thus open and closed in $M$, and so must be $M$ itself. It follows that $M$ is path-connected, ...


7

I will answer under the assumption that you refer to sets $S_{\varepsilon}(x) = \{y: \, d(x,y) = \varepsilon\}$ as spheres, and $x$ as a center of the sphere. On $\mathbb{Q}$, given any prime $p$, you can define the $p$-adic metric $|x-y|_p$ to be $p^{-n}$, where $n$ is the unique integer such that $x-y = p^n \frac{a}{b}$ with $a,b$ integers that are ...


2

Let $X$ be topological space and Y be a metric space. We show that X\A is open. Let $x \in$ X\A be arbitrary. Then $f(x)\neq g(x)$ (Otherwise x would be in A). We define $\epsilon:=d(f(x),g(x))$, so the open neighborhoods $U_{\frac{\epsilon}{2}}(f(x))$ and $V_{\frac{\epsilon}{2}}(g(x))$ are disjoint. Hence $W:=f^{-1}(U_{\frac{\epsilon}{2}}(f(x)))\cap ...


7

If $Y$ is a metric space, the set $\Delta =\{(y,y):y\in Y\}$ is closed in $Y\times Y$ -- this holds more generally iff $Y$ is Hausdorff (can you prove this?). Observe now that $$A=\{x\in X:f(x)=g(x)\}$$ equals the set $$B=\{x\in X:h(x)\in \Delta\}=h^{-1}(\Delta)$$ where $h:X\to Y\times Y$ is defined by $h(x)=(f(x),g(x))$. But $\Delta$ is closed, and $h$ is ...


4

Consider the map $F:X\rightarrow \Bbb R$ given by $F(x)=d(f(x),g(x))$. $F$ is continuous (you can check this), and $F^{-1}(\{0\})=A$ so $A$ is closed since $\{0\}$ is closed in $\Bbb R$.


3

It's equal to $$\bigcap \{x\in X: d(f(x),g(x))\le \epsilon\}$$ which is the intersection of closed sets since $Y$ is metric (hence Hausdorff) and $f,g$ are continuous. We know it's closed because $Y$ is Hausdorff, so that the diagonal, $\Delta Y\subseteq Y\times Y$ is closed. This is so because the complement is open: Proof: If $(x,y)\ne (y,y)$ then ...


0

It is not enough to show that there are two disjoint open sets $U$ and $V$ which cover the rationals. The definition of connectedness requires that the union of the disjoint open sets $U$ and $V$ actually equals the set of rationals $\mathbb{Q}$. For this we can take $U = (-\infty, \pi) \cap \mathbb{Q}$ and $V = (\pi, \infty) \cap \mathbb{Q}$. Note that $U$ ...


1

The unit ball around a point $(x_1,y_1)$ is the set of all points $(x_2,y_2)$ of distance $<1$ from $(x_1,y_1)$. It is not a set of pairs of points as you stated. Your drawing appears to be of the unit ball around the origin.


5

I agree with J. Loreaux's commrent above (“Your argument doesn't really work”), and I'd go farther: to me, your argument makes no sense whatever, for several reasons: You say “If the every compact set on a metric space is not bounded, then …”. This is at least confusingly stated. You want to prove that every compact set is bounded. You seem to be trying ...


0

Your argument is essentially correct but pay attention to the essentially. I would phrase it in words like so: If not every compact set on a metric space is bounded then there exists at least one compact set on a metric space that is not bounded. Since it's unbounded we can construct an infinite open covering which doesn't have a finite subcover thus ...


2

$\sin(1/x)$ maps the sequence $x_n=\frac1n$ to $\sin(n)$ which is definitely not Cauchy.


3

$f(x)=\sin\frac1x$ on $(0,1]$. For the disk you just need a bounded holomorphic function on the disk that's discontinuous on the boundary. Such functions do exist, as answers to Continuous Extension of a Bounded Holomorphic Function on the unit disk? show, but it's harder to produce an explicit example.


1

If you are studying CAT(0) spaces, or geometry of metric spaces generally, the book A Course in Metric Geometry by Burago, Burago, and Ivanov should (ideally) always be in your vicinity. The statement is Theorem 2.4.16 there. Here is the proof. Given $a,b$ at distance $L=d(a,b)$, we want an isometry $\gamma:[0,L]\to X$ such that $\gamma(0)=a$ and ...


4

$$ |x_i - z_i|\leq |x_i - y_i| + |y_i - z_i| \quad\forall i \qquad\implies $$ $$ \max_i|x_i - z_i|\leq \max_i\left(|x_i - y_i| + |y_i - z_i|\right) \leq \max_i|x_i - y_i| + \max_i |y_i - z_i| $$


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, ...


2

This metric is called the $p$-adic metric, and has many applications in number theory and other areas of mathematics. For some intuition: given any number $n \in \mathbb N$, we can write $n$ in base $p$ - i.e. $$n = \sum_{i=0}^Na_ip^i \ \ \ \ 0 \le a_i<p$$If $\displaystyle m = \sum_{n=0}^Nb_ip^i$, then $d(m,n)=p^{-j}$ where $j = \min\{i:a_i - b_i \ne ...


6

(i) Consider the sequence $$1, 1+p, 1+p+p^2,\ldots, \sum_{i=0}^np^i,\ldots$$ which is clearly Cauchy, but does not converge to any natural number, $n\in\Bbb N$, so $\Bbb N$ is not complete, hence not compact. Proof of non-convergence: If $n\in\Bbb N$ represent $n$ as a finite sum of powers of $p$, which is possible to do constructively and is called a ...


0

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 $r$ ...


1

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.


2

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 ...


0

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 ...


1

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


0

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 ...


1

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.


0

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 ...


1

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.


0

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


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 ...


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 ...


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

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 ...


0

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 $$(f-g)(X)\subset (f-g)(\overline{E})\subset \overline{(f-g)(E)}=\{0\}$$


1

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 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.


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: ...


0

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 $d(x_i, x_{n+1}) \ge 1\ \forall i$.


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 ...


1

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 ...


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 ...


3

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

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 ...


0

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.


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 ...


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 ...



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