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Here is an example of a metric $d$ on $\mathbf R$ such that $(\mathbf R,d)$ is not complete. Set $d(x,y)=\bigl\lvert\mathrm e^{-x}-\mathrm e^{-y}\bigr\rvert$. It trivially satisfies the axioms of a metric. Now the sequence of natural numbers iss a Cauchy sequence. Indeed choose $\varepsilon> 0$ and let $N$ be an integer such that $\;\mathrm ... 1 Let $$m_0 := \{t \in \mathbb{R}^\mathbb{N} : \{t_1, t_2, \ldots \} \text{ is finite} \}$$ be the sequences of real numbers that only take on finitely many values. Equip it with the sup-norm, i.e. $$\| t \|_\infty := \sup_{i \in \mathbb{N}} |t_i|.$$ This is an incomplete normed vector space (so it's also a metric space). To see this, consider the sequences ... 1 For a non-metrizable but locally compact Hausdorff counterexample, consider$\omega_1$with its order topology. I will show that$\mathcal{S}$does not even contain all the open sets. Let$\mathcal{B}$be the collection of all sets$B \subset \omega_1$such that either$B$is countable or$B$contains a club set. (We consider "countable" to include ... 3 For the answer to the first question: let$\mathcal S' = \{A \in \mathcal S : A^C \in \mathcal S\}$. See that$\mathcal S'$contains all open sets, that$\mathcal S'$is closed under taking complements, and that$\mathcal S'$is closed under countable unions. 6 The reason is that the intersection of infinitely many open sets need not be open: you need the set of$V_{q_k}$to be finite in order to ensure that$V$is actually a neighborhood of$p$, rather than merely some set containing$p$. 3 Hint: For any$m \in \mathbb N,\Gamma$is contained in $$\bigcup_{k=1}^m \ B(f(k/m), K/m),$$ where$B(a,r)$is the closed ball centered at$a$of radius$r.$0 I will try my best to motivate that definition. Let's say$y$is "close" to$x$if the distance between$x$and$y$is less than$1$. Now, we know that — with this definition of "close" — everything in$\Bbb R$is "close" to something in$\Bbb Z=\{\dots,-2,-1,0,1,2,\dots\}$. Also, it's quite easy to see that we can't replace$\Bbb Z$with a finite set. With ... 0 From what I remember taking Analysis, compactness is a generalization of a closed interval$[a, b] \subset \mathbb{R}$under the standard Euclidean metric to general metric spaces with general metrics. A closed interval on$\mathbb{R}$is a closed set that's "small" in some sense. Think about comparing$[a,b]$to$(a,b)$under the Euclidean metric;$(a,b)$... 0 Theorem. (Heine-Borel.) A metric space is compact iff it is complete and totally bounded. So you can think of compactness as a strengthening of completeness in which a certain smallness condition (namely, total boundedness) is added to the mix. Okay, so why bother with this weird mix of completeness and smallness? The nice thing about compactness is ... 0 A compact set is "the next best thing" to a finite set. Almost all statements about finite sets are true of compact set- they are closed, bounded, etc. 2 Here is one way I like to think of it: Suppose you're trying to cover an infinite compact set, and you really want to give it an infinite cover that doesn't have a finite subcover. So you get a collection of infinite sets that looks like it nearly covers everything - maybe you've left behind a countable subset of an uncountable set or something. You must ... 2 Your examples: The discrete metric space on a finite set is compact. Closed bounded sets in$\mathbb{R}^n$are compact. The discrete metric space on an infinite set is not compact. Many examples in$\mathbb{R}^n$are available here, but open balls are probably the most easily visualized. If your goal is to study metric spaces rather than topological ... 5 Consider the following as subspaces of$\mathbb R\{0,1\}$or in fact any finite set is compact and discrete$[0,1]$is compact but not discrete.$\mathbb Z$and$\left\{\frac1n:n\in\mathbb N\,\right\}$are discrete but not compact. (But$\left\{\frac1n:n\in\mathbb N\,\right\}\cup\{0\}$is compact and not discrete)$(0,1)$and$\mathbb Q$are neither ... 2 No,$\overline{\mathbb{R}}$with the usual distance is not a metric space. It is better understood as a topological space (with the order topology). Although it is not a metric space with the usual distance, it is metrizable. Think about a way to put a metric on it. But I will disagree with one of the comments (the one that says that ... 0 Of course$H(S))$had better be the class of nonempty compact subsets of$S$... Say$S$is compact. Two proofs: First proof: Since$S$is compact,$S$is totally bounded: Let$\epsilon>0$. There exist$x_1,\dots,x_N\in S$with $$S=\bigcup_{n=1}^NB(x_n,\epsilon).$$ Now for every$F\subset\{1,\dots\,N\}$let$H_F$be the set of all$K\in H(S)$such that ... 1 There are different ways to do that. You could use euclidean distance (or any other metric applicable to vector spaces) if the alphabet is in a metric space. As a special case if you use$L^1$distance and your alphabet is${0, 1}$with usual metric you will get the hamming distance. Since you're talking permutation distance, then yes it's a metric too ... 0 This is just an application of the law of sines: $$\frac{d(u,v)}{(\rho - \delta)} \geq \frac{d(u,v)}{d(p,v)} \geq \frac{d(u,v)}{2 d(p,v)} \geq \sin(\frac{\angle_p(x,y)}{2})$$ 0 Think about it this way: If you want$d(x,a) \leq 1$, i.e.,$|x_{1} - a_{1}| + |x_{2} - a_{2}| \leq 1$, then draw out your$XY$plane, and label some point$a$. Now put another point somewhere on the$XY$plane, and connect the two points by a straight line. Now, put your pencil on$a$, and get to$x$by first going only horizontally (parallel to$x$-axis ... 0 Yes, the proof is correct. To make the style consistent across the proof, you could have written "Let$x,y\in\mathbb R$" in$2$., like you did in$1$. and$3$. 2 In general in analysis it is desirable to have objects of interest such as functions classified in spaces with "good properties". This allows one to use some standardised proof techniques for some problems and is very convenient. Completeness turns out to be one of the most important of such properties. Also, integration is one of the fundamental operations ... 3 The map$f \mapsto \int_a^b f(x)\,dx$is continuous with respect to the metric$d$. So if you have a sequence$f_n \to f$converging in the metric$d$, you know that$\int_a^b f_n(x)\,dx \to \int_a^b f(x)\,dx$. In other words, convergence in the metric$d$is exactly the right thing to guarantee that you can pass the limit under the integral sign. 2 You were doing fine until you got to the point of showing that$\sup E\in E$. Picking the sequence$\langle r_n:n\in\Bbb Z^+\rangle$in$E$converging to$\sup E$is find but then you write this: For each$r_n$there is some$U_\alpha$such that$B(x, r_n) \subseteq U_\alpha$The set$U_\alpha$must surely depend on$r_n$and on$x$, but these ... 3 The group of real numbers$(\Bbb R,+,0,-)$with its usual topology is a topological group: The addition$+:\Bbb R\times\Bbb R \to \Bbb R$(where$\Bbb R\times \Bbb R$has the product topology) and the negation$-:\Bbb R \to \Bbb R$are continuous maps. These properties of$\Bbb R$are exactly what leads to formulas like $$\lim(x_n+y_n) = x+y\quad \text{ and ... 1 Indeed the algebra of limits holds (for sums and scalar multiples) in any normed space. The proof is identical. The remaining algebra of limits hold in any normed algebra. Note here that the norm on an algebra imust satisfy \lVert xy\rVert\leq\lVert x\rVert\lVert y\rVert for all x,y in the algebra. Also note that non-zero must be replaced by invertible ... 1 The first statement is true both ways. Specifically, suppose (X, ||\cdot||) is a normed linear space. Then the norm ||\cdot || is induced by an inner product iff the parallelogram law holds in (X,||\cdot||). For the second statement, this is not true. Call the condition d(x,y)=d(x+a,y+a) translation invariance, and the condition d(x,y)=d(ax,ay) ... 3 You might consider the following image: Each of the black circles represents the minimum distance points must be away from each other. The blue circle is the unit circle. Notice that moving any single point out of this configuration necessarily either places a point into the unit circle or increases the sum of distances to the origin. Edit: There is ... 1 The target function is continuous and we can restrict the domain to points in the compact annulus 1\le r\le 3. Hence the minimum of the function is attained. Therefore it suffices to show for any configuration that is not a regular decagon centered at O and with side length 1/\sqrt 2, there exists a better configuration. This way you can readily show ... 5 If the sequence of z differs from the sequence of x before it it differs from the sequence of y (i.e N(x,z) \leq N (x,y)) one has d(x,y) \leq d(x,z) so the triangle inequality will be fullfiled trivially. Therefore it remains to look at the case N(x,y) < N(x,z) this means, that the digits x_n,y_n,z_n are the same for n < N(x,y). But by ... 1 You can proof that they are equivalent definitions. 2 Both definitions are equivalent. Note$$B_r(a) \subset \overline{B_r(x)} \subset B_{r+1}(x)$$3 See the first paragraph after the Remark at the top of page 9: Since X is a separable metric space, by Tychonoff’s embedding theorem, X is homeomorphic to a subset of a compact metric space. Thus X admits an equivalent metric \rho with respect to which it is totally bounded. I would appeal instead to the Urysohn metrization theorem, but ... 0 Yet another answer: You can show that x \mapsto x + a is continuous. Now O is an open set, and O = (O+a) - a is open. Next: what is the definition of continuity in general? 0 The Heine-Borel theorem plays a crucial role in the development of both Riemann and Lebesgue integration on \mathbb{R}^d. Perhaps the most important result in Riemann integration is that continuous functions on closed intervals [a,b] are integrable. The reason is that [a,b] is compact by Heine-Borel, so continuous functions on it are uniformly ... 1 Assume f_n is a Cauchy sequence in C([0,1]). You said you showed that f_n(x) is a Cauchy sequence of real numbers. The standard norm is complete on \mathbb R, so for any x \in [0,1] we know f_n(x) converges to some real number \alpha_x. Now, define a function f on [0,1] by f(x)=\alpha_x. We need to show that f \in C([0,1]) and that ... 0 Your guess at using \epsilon = x_1^2 + x_2^2 - 1 is not correct, but perhaps it is going in the correct definition. Instead, you should use \epsilon = \sqrt{x_1^2 + x_2^2} - 1. The geometric/intuitive reason for this choice of \epsilon is that it is equal to the shortest distance from x to the set A^c=\overline B(O,1): draw the segment ... 2 First, \mathbb{H}^n is a complete Riemannian manifold of constant sectional curvature -1. Second, for each \kappa<1 there exists \delta>0 such that every complete Riemannian manifold M whose sectional curvatures are all \le \kappa is \delta-hyperbolic; by definition means that every geodesic triangle T \subset M is \delta thin, ... 1 Let d' be the metric restriced to A, B_d(x,\epsilon)=\{y\in S:d(x,y)<\epsilon\} and B_{d'}(x,\epsilon)=\{y\in A:d'(x,y)<\epsilon\} \underline{A\cap(B\cup A^c)^\circ\subseteq B} Suppose x\in A\cap(B\cup A^c)^\circ. So x\in A and x\in(B\cup A^c)^\circ. Hence there is an \epsilon>0 such that B_d(x,\epsilon)\subseteq B\cup A^c. ... 0 I may be interpreting your question wrong but it seems to me you are asking if X is closed or open in the metric space M=(X, d). And the answer to that is: yes. The entire metric space is always closed AND open within itself. This can be confusing. We know (0, 1) is open in \mathbb{R} but it is true that (0, 1) is also closed when the metric space ... 1 Closed, not open. Define$$f(0)= 0,\ f(1/4) = 1/2, \ f(1/2) = 0, \ f(1) = 1.$$(Good to draw a picture.) So we have four points on the graph of f. Connect these points in order with line segments. We then have a piecewise-linear, continuous and surjective f: [0,1]\to [0,1]. If E\subset [0,1] is closed, then E is compact, hence f(E) is compact ... 1 Draw a picture of the circle that has centre the origin O, and radius 1. Let P=(a,b) be such that (OP)^2=a^2+b^2=r^2\gt 1. It is obvious that there is an open disk D with centre P such that D lies entirely within A. Now it's over. But if we want to be explicit, the disk D with radius \epsilon=\frac{r-1}{2} will do the job. For suppose ... 0 Let \epsilon = d(x, 0) - 1. Thus d(x, 0) - \epsilon = 1 where d is the standard metric. Then if y \in B_\epsilon(x) (open ball of radius \epsilon centered at x), we want to show d(y, 0) > 1 (since d(y, 0) > 1 if and only if y_1^2 + y_2^2 > 1). By the triangle inequality$$d(y, 0) + \epsilon > d(y, 0) + d(x, y) \ge d(x, 0) ... 1 Consider the monomials $$f_n(x) = \frac{x^{n^2}}{n}.$$ 1 One classic example that can help you here is the set of functions of the form $$f_n(x) = \sqrt{x^2 + 1/n}$$ for$n \in \Bbb N$. Note that the functions$f_n(x)$form a Cauchy sequence, but their derivatives do not. 2 Since it seems that you want to solve it yourself, I'll just give you a hint. If you want a more accurate answer, just leave a comment. HINT: In a normed space the unit open balls is the set of all elements which have a norm strictly less than 1. Now, what does it mean that$\left\Vert f \right\Vert_1 < 1$? What does it mean that$\left\Vert f ...

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Yes there are a lot of related papers that I came across couple of weeks ago, which talk about converting Pearson Correlation to Euclidean distance, when data is z-normalized. Your question was back in 2013, I hope you are still interested: StatStream Statistical Monitoring of Thousands of Data streams in real time (First paper to proof the relation and ...

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Let me start with the following definition: In a metric space $(M,d)$, we can say that $S$ is an open set (with respect to the topology induced by $d$) if for every element $s\in S$, there exists $\epsilon >0$ such that the ball $$B(s,\epsilon)=\{ x\in M\mid d(x,s)<\epsilon\}\qquad \text{ satisfies }\qquad B(s,\epsilon)\subset S.$$ This means that if ...

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A subset $A \subset M$ is open in the metric space sense, if for every $x \in A$ there is a $\epsilon > 0$ so that the open Ball $B_\epsilon(x)$ is a subset of $A$. This basically means that a set is open iff every point admits a small neighborhood that lies completely in $A$. Now $\mathcal{T}$ is meant to be the collection of all subsets of $M$ that are ...

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Let $X_i = (X,d_i)$. Assume without loss that $X_1$ is complete and let $\{x_n\}$ be a Cauchy sequence in $X_2$, i.e. for every $\epsilon >0$ there is $N$ such that for $n,m \ge N$ we have $d_2(x_n,x_m) \le \epsilon$. If we now apply the inequality that we assumed on the distance functions we get d_1(x_n,x_m) \le \frac{1}{\alpha}d_2(x_n,x_m) \le ...

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The answer is yes. It's fairly easy to see that equivalent metrics define the same topology on the space (i.e. the open sets are the same); also the Cauchy sequences in $(X,d_1)$ and $(X,d_2)$ are also the same. Suppose $(X,d_1)$ is complete, then any Cauchy sequence $\{x_n\}$ has a limit $x$ in $X$. This means that for an arbitrarily small $\epsilon >0$, ...

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Such metrics are also called Lipschitz equivalent, the inequality states that the identity map from the space endowed with one metric to the space endowed with the other is a Lipschitz map (in both directions). That ensures that the space is complete in one metric if and only if it is complete in the other metric. In fact, that holds for a weaker concept of ...

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