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$(0,0)$ is a limit point but $(0,0)\notin M$ (Put $x_n=\frac{1}{2n\pi} \Rightarrow x_n\to 0 , \sin x_n=\sin \frac{1}{\frac{1}{2n\pi}}=\sin 2n\pi=0$ Theorem Subspace of Complete Metric Space is Closed if and only if Complete.

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If $X$ is a connected metric space such that every real valued continuous function on $X$ is uniformly continuous, then $X$ is compact. In particular, it is bounded. Proof: If $X$ is not compact, there is a sequence $S = \{x_n\}_{n=1}^\infty$ with no convergent subsequence. Then for each $n$ there is $\epsilon_n > 0$ such that the ball ...

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First of all, note that \begin{align}d(x,y) \\&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\&\le \sqrt{2\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&=\sqrt2 \max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt2 d^*(x,y)\end{align}. On the other hand, \begin{align}d^*(x,y)\\&=\max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt{\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&\le ... 3 If we are allowed to take steps in any of the eight directions you describe, the number of steps it takes to get from (x_1,y_1) to (x_2,y_2) is the maximum of |x_1-x_2| and |y_1-y_2|. This is often called the L^\infty metric. 0 For the reverse implication suppose every positive function has positive infimum. Then the reciprocal of every positive function (being positive) has a positive infimum, say \epsilon. So the original function was bounded by 1/\epsilon. Therefore every positive function is bounded. Now let g \colon M \to \mathbb R be an arbitrary function and consider ... 2 If f has no infimum > 0, then f^{-1}((\epsilon, \infty)) \ne M for any \epsilon > 0. 1 It is proven by Swenson in E. Swenson, A cut point theorem for CAT(0) groups, J. Differential Geom. 53 (1999), no. 2, 327–358. that each infinite CAT(0) group contains an infinite order element. 2 In short, yes if something in math has a property that has a definition, then the definition can be used to prove that the property. It would be a bit odd for a set to be closed, yet you would be unable to prove it is closed using the definition. That being said, there are multiple equivalent definitions of what it means to be closed and you mentioned two of ... 1 It's perfectly fine to show that a finite set is closed because it's equal to its own closure. But to compute the closure, you have to take the set of all boundary points of A union A, not the set of interior points! So you need to show that a finite set has no boundary points, in other words, that no sequence contained in a finite set converges outside ... 0 Hint: All metric spaces are Hausdorff. 0 My inequality chain was: \begin{align*} d(x,z)+d(z,y)={}&\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\sum_{n=1}^\infty2^{-n}\frac{\rho_n(z-y)}{1+\rho_n(z-y)}={} \\ {}={}&\sum_{n=1}^\infty2^{-n}\left(\frac{\rho_n(x-z)}{1+\rho_n(x-z)}+\frac{\rho_n(z-y)}{\rho_n(z-y)+1}\right)={} \\ ... 2 Hint: Fixing notation, I assume you meanm((p,p'), (q,q')) = \sqrt{ d(p,q)^2 + d'(p',q')^2}$$This is a generalization of a Euclidean norm on \mathbb R^2. You can prove the triangle inequality the same way as we do there. Namely, write$$m((p,p'), (q,q'))^2 = \langle (d(p,q),d'(p',q')), (d(p,q),d'(p',q')) \rangle_{\mathbb R^2}^2$$Your proof will ... 3 it is just$$ \left| \frac {1}{x}-\frac{1}{z} \right|=\left|\frac {1}{x}-\frac {1}{y}+\frac{1}{y}-\frac{1}{z}\right| \leq \left|\frac {1}{x}-\frac{1}{y}\right|+\left|\frac {1}{y}-\frac{1}{z}\right|$$using the Module properties 2 take the sequence x_n=(1,1/2,....,1/n,0,0.....) then it doesn't converge in l_1 as \sum (1/n) is not finite but it converges in l_\infty to (1/n)_{n\geqslant1}^{\infty} 2 Firstly we notice C_0 is a complete metric space hence it is a closed subset of L^{\infty} Now observe that L^{\infty} is connected therefore the only closed and open sets are the empty set and the whole space. Hence C_0 cannot possibly be open. 0 The answer is negative. The always vanishing sequence \mathcal O belongs to C_0 and for all \epsilon >0 the sequence whose terms are all equal to \epsilon doesn't belong to C_0 but it's distance to \mathcal O is equal to \epsilon. 2 Hint: The constant \varepsilon sequence is in the 2\varepsilon-ball around 0. 0 The metric on L^\infty is usually taken to be$$d(x,y) = \sup_k |x_k - y_k|.$$Note that$$\vec 0 = (0,0,0,\ldots) \in C_0,$$but for any \epsilon > 0,$$\vec \epsilon = (\epsilon,\epsilon,\epsilon,\ldots) \notin C_0,$$even though d(\vec 0,\vec \epsilon) = \epsilon. Thus \vec 0 is not an interior point of C_0. 1 You are right. The result holds in every locally convex space (i.e., a vector space X with a family \mathcal P of semi-norms: (X,\mathcal P) and (X,\sigma(X,X')) have the same bounded sets. 0 As Ian said, for small distances, the two metrics are essentially the same. They only differ for large distances, which are not relevant to Cauchiness, convergence, separability, etc. It may be instructive to compare \rho to the metric \tilde \rho(x_1,x_2)=\min(1,d(x_1,x_2)). This definition makes it clear that \tilde \rho is identical to d at ... 13 With two different metrics? Yes, obviously. (But with two different metrics it is not the same metric space, by definition -- the concept of a metric space includes which metric we're using). For example take \mathbb R with respectively the standard metric, and the metric$$ d_2(a,b)=|f(a)-f(b)| \quad\text{where }f(x)=\begin{cases} \pi & \text{if }x=0 ...

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I guess you want something more rigorous to justify the intuitively obvious choice of radius. This can be done: Let $d$ be the usual metric; let $x \in \mathbb{R}^{n} - \{ 0 \}$; then $$d(y,0) \geq d(x,0) - d(x,y) > 0$$ if $d(x,y) < \frac{d(x,0)}{2}$, by the triangle inequality and the symmetry of metrics; therefore, the open ball ...

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What part are you not sure about? You should show it's impossible for 0,0... to be in that ball, if you choose epsilon small enough. It looks like you're on the right track so far.

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Your proof is okay. More generally, we can define any topological space to be $T_1$ if its singleton sets are closed (or equivalently, for any points $x \neq y$, there is an open set containing $x$ and not containing $y$). $T_1$ is an example of a 3[separation axiom]. You've shown that metric spaces are $T_1$, but much stronger separation axioms hold for ...

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If you consider $\ell^2$ with the $c_0$ metric, then $\ell^2$ is not closed in $c_0$. In fact, it is dense in $c_0$, because the set $$c_{00} = \{(x_n) : \exists N \in \mathbb{N} \text{ such that } x_n = 0 \quad\forall n\geq N\}$$ is a subset of $\ell^2$, which is dense in $c_0$. However, note that $\ell^2$ with the euclidean distance is complete. ...

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At the end of your proof, you set $\alpha = \phi(\bar{z}) - \epsilon$, $x^*=\bar{z}$, but $x^* \not\in G : = \{ \phi(x) < \phi(\bar{z}) - \epsilon\}$. Edit: I believe you want to set $\alpha = \phi(z) + \epsilon$, that way at least $z=x\in G$ and for each $B_\delta(x)$, there exists $t\in B_\delta(x)$ such that $$\phi(t) \not\in (\phi(z) - \epsilon, ... 1 Let A be any set that isn't closed. Define \phi(x) = \alpha if x is in A; = \alpha - 1 is x is not in A. Then \{ x \ | \ \phi(x) \geq \alpha \} = A which is not closed. 0 Suppose f is not continuous, then there is x_0\in X and a sequence x_n\in X such that x_n\to x_0, but |f(x_n)-f(x_0)|>\varepsilon>0,\forall n\in\mathbb{N}. Without loss of generality, let f(x_n)\geq f(x_0),for each n\in \mathbb{N}(else we could work with a subsequence or work with \leq \alpha). let \alpha=f(x_0)+\varepsilon/2, the set ... 0 In \mathbb{R}^5, (2,0,0,0,0)=(1,1,1,1,1)+(1,-1,-1,-1,-1), but d(2,0,0,0,0)=1 and d(1,1,1,1,1)=d(1,-1,-1,-1,-1)=\frac{1}{\sqrt 5}. So triangle inequality doesn't hold. 0 The real map f: t \to \frac{t}{1+t} is continuous, strictly increasing on [0,+\infty), f([0,+\infty))=[0,1), f(1)=\frac{1}{2} and$$f^\prime(t)=\frac{1}{(1+t)^2}.$$Hence for u \in [0,\frac{1}{2}], you have using the mean value theorem:$$u \ge f(u)= f^\prime(\xi) u \ge f^\prime (\frac{1}{2}) u =\frac{4}{9} u$$where 0 \le \xi \le u \le ... 0 To prove that a given metric space is complete, you need to prove that every Cauchy Sequence has a limit. So take x_n a sequence of x_n\in X  which is also a Cauchy Sequence, by the given metric. From the Cauchy quality what do you know? How does this help you with the Cauchy interpretation of Limit*? Cauchy said that a sequence has a limit iff for ... 0 Hint: You need only prove triangle inequality.$$\ln(1+|x_1-x_2|)+\ln(1+|y_1-y_2|)=\ln((1+|x_1-x_2|)(1+|y_1-y_2|))$$use this and \ln x is an increasing function. 2 Without any further assumption on A and B, there is no conclusion about their intersection: In X = \def\R{\mathbf R}\R with the euclidian metric, A = [0,1] and B = (1,2) have d(A,B) = 0 and empty intersection, but A and B' = [1,2] have also d(A, B') = 0. Even if we suppose A and B to be closed, we can have d(A,B) = 0 with empty ... 1 It seems the following. c. Let d be a continuous metric on  X \times X . If f\in C(X,\Bbb R) in an unbounded function, then d’(x,y)=d(x,y)+|f(x)-f(y)| is an unbounded continuous metric on  X \times X . d. ii \Rightarrow iii. Assume that d is an unbounded continuous metric on  X \times X . Fix an arbitrary point x_0\in X and put ... 0 Suppose x is not in the set, call it E. E is closed so x is not a limit point. So there is an \epsilon such that B(\epsilon, x) has no point of E. We can find an m such that 0 < \frac{1}{m}< \epsilon. So B(\frac{1}{m}, x) has no point of E. So for all y in E, d(x, y) > \frac{1}{m}. So x is not in the \frac{1}{m} set of E. So x is ... 0 If you apply any non-linear homeomorphism of \Bbb R^2 to the standard metric, you'll get a homogeneous non-translation invariant metric. For example$$d(x_1,x_2,y_1,y_2)=\sqrt{(x_1-x_2)^2+(e^{y_1}-e^{y_2})^2}$$In general this takes the form d(x,y)=d'(f(x),f(y)), where f is a nonlinear homeomorphism and d' is the standard metric or any other ... 0 If I={0,1} with topology given by the empty set and the whole set is applied trivially on I with point topology then this application is not continuous. The preimage of the open set \{1\} is not opened. 2 Added: Answer to new question: No, just take I = \mathbb {Q}\cap [0,\infty) as you did earlier. Define f:I\to \mathbb {R} by setting f(x) = 1/x, x > 0,f(0)=0. Certainly not: Let I= \{0\} \cup \{1/n: n \in \mathbb {N}\}, with the topology it inherits from the usual one on \mathbb {R}. Define f(1/n) = 1, n = 1,2,\dots , f(0) = 0. Then f is ... 0 No, not necessarily, since it depends on the topology on I. If I has a trivial topology, where the only open sets are I and \emptyset, then it is easy to construct a counterexample. Let I = \mathbb{Q} be the set of rational numbers, but with the trivial topology I just defined, and let E be the set of real numbers, with the usual topology. ... 1 Ok now the proof is basically correct (when we are working in metric spaces)! Some remarks: As Umberto P. also noted in his answer in the related question you asked, I am not fond of the notation "N_{\varepsilon, \mathbb{R}} (\phi (x)) < \alpha". In fact in the other thread you write "\phi(Y) \le \alpha" for a set Y, which is still a bit more ... 2 Hint Show that$$\max_{i=1,...,n}\{|x_i-y_i|\} \leq \rho_p(x,y)\leq n^{1/p}\max_{i=1,...,n}\{|x_i-y_i|\}$$2 I'm not sure why you need an \epsilon-\delta proof when we aren't trying to prove continuity. We are just trying to prove \{x \mid \phi(x) \geq a \} is closed if \phi is continuous. It suffices to show the complement is open. But since \{x \mid \phi(x) \geq a \} = \phi^{-1}( [a, \infty) ), and f^{-1}(B^{c}) = (f^{-1}(B))^{c} for every function ... 2 Your proof isn't wrong, but (in my opinion) it uses proof by contradiction unnecessarily and I'm not sure about "\phi(Y) \le \alpha". Just argue directly. If (x_m) \subseteq A (what is the point of the overline?) is a convergent sequence with limit x, the continuity of \phi implies \phi(x_m) \to \phi(x). By an elementary order property of limits, ... 2 Hint: let \mathbb Z^n denote the integer lattice in \mathbb R^n. Which functions f : \mathbb Z^n \to \mathbb R are uniformly continuous? 3 Firstly, from \rho(x,z)\le \rho (x,y)+\rho(y,z) you can conclude that \frac{1}{\rho(x,z)}\ge \frac{1}{\rho (x,y)+\rho(y,z)}, but absolutely not the inequality you wrote. A hint to solve the exercise is to think of it more generally. Seek conditions on a function f\colon \mathbb R_+ \to \mathbb R_+ that will guarantee that if \rho is a metric on some ... 1 I'm assuming the following assumuption: Let f:X \to Y be a continuous function. Clearly for any B \subseteq Y$$\begin{align}\text{Int}(B) &\subseteq B \\ \implies f^{-1}(\text{Int}(B)) &\subseteq f^{-1}(B) \\ \implies \text{Int}(f^{-1}(\text{Int}(B))) &\subseteq \text{Int}(f^{-1}(B))\end{align}$$Now recall that \text{Int}(B) is open so ... 0 If we're using a continuous map: We denote by Int(B) = \cup_{U \subset A}U where each U in the union is open. This is the usual definition of "interior" in topology. For each U, U \subset E \implies f^{-1}(U) \subset f^{-1}(E). And f^{-1}(U) is open (f is continuous). Intf^{-1}(B) = \cup_{V \subset f^{-1}(B)}V  where all the V are open. ... 0 Hint. Show that the set of balls with rational (that is points in \mathbf Q^n) centers and rational radii forms a basis for the euclidian topology. To do so, you have to prove, that for each open U \subseteq \mathbf R^d, we have$$ U = \bigcup \left\{B_r(x) \mid r \in \mathbf Q, x \in \mathbf Q^d, B_r(x) \subseteq U \right\} $$0 Take any x\in S. Then x^2<2. In particular, due to density of rationals in \mathbb R, there exists some \dfrac{p^2}{q^2}\in\mathbb Q where p,q\in\mathbb Z with x^2< \dfrac{p^2}{q^2}<2. So take r=d(x,\dfrac{p}{q}) then consider the ball B(x; \dfrac{r}{2}). Then for any y\in B(x;\dfrac{r}{2})\cap\mathbb Q we have ... 1 If (x_l)_{l \in \mathbb N} is a Cauchy sequence w.r.t. d_v, then for arbitrary \epsilon > 0 we may choose N \in \mathbb N such that$$ \forall n, k > N : d_v(x_k, x_n) = \lambda^{v(x_k - x_n)} < \epsilon. $$But$$ \lambda^{v(x_k - x_n)} < \epsilon \Leftrightarrow v(x_k - x_n) > \frac{\ln(\epsilon)}{\ln(\lambda)}.  As \$\epsilon \to ...

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