Equivalent metrics convergent sequences. (Convergent sequences in metric spaces.

Equivalent metrics convergent sequences Prove this. Ñ " and does not converge in . (Convergent sequences in metric spaces. Nov 4, 2022 · I am wondering whether the concept of two metrics defining the "same convergent sequences" (for instance, when we are interested in the metrics being topologically equivalent) also includes the respective limits of the sequences. $\endgroup$ Sep 29, 2016 · I can understand why two topologies having the same converging sequences doesn't make them equal. hence, those metrics are not equivalent and not uniformly equivalent. Metric Spaces The main concepts of real analysis on \(\real\) can be carried over to a general set \(M\) once a notion of distance \(d(x,y)\) has been defined for points \(x,y\in M\). Show that for any metric space (x, d), p(a,b) dla, b) 1+ d(a,b) defines an equivalent metric. So, if you find a sequence which is convergent, bu not eventually constant, in the other metric, you have proved that the two metrics are not equivalent. ) (x_n, x) \xrightarrow{} 0$, which shows that the metrics have the same convergent sequences, and hence the Nov 23, 2016 · I want to show that a sequence of points of $\mathbb{R}^n$ is convergent with respect to the Euclidean metric if and only if it is convergent with respect to the square metric. (b) Every convergent I can easily see that metrically equivalent metrics are topologically equivalent. So if the complement of a set is open with respect to one metric, the complement of the set will be open with respect to all equivalent metrics. when restricted to any specific convergent sequence, these metrics become strongly equivalent. Apr 24, 2022 · Completeness is such a crucial property that it is often imposed as an assumption on metric spaces that occur in applications. Show that p is a metric that is equivalent to d. Some authors say that two metrics d and p on a set M are equivalent if they generate the same open sets. In a general topological space, conver-gence must be formulated in terms of nets instead of countable sequences. Here's what I am doing - As $p$ and $d$ are equivalent metrics they generate the same open sets. (convergent) sequences of its elements. $\endgroup$ – Transcribed Image Text: Two metrics d, p on a set X are said to be equivalent if they define the same convergent sequences. ) Let (a n) n2N be a sequence of real numbers. Example 3. Any convergent sequence in any metric space is a Cauchy sequence. If d is any metric on M , show that the metrics , and , defined are all equivalent to d. We know that the set of all balls of a metric space is a base for the topology induced by the metric, so we must only prove that each ball of $(X,d_1)$ is an open set with respect to $\tau_2$ and viceversa. lecture 4 (01/24/24) Properties of open sets. 2. Even though a Cauchy sequence may not converge, here is a partial result that will be useful latter: if a Cauchy sequence has a convergent subsequence, then the sequence itself converges. A sequence {x n} in a metric space (X,ρ) converges to x ∈ X provided lim n→∞ ρ(x n,x) = 0. Since we are talking about sequences, closed should be best defined as "every sequence in the closed set converges to a point in the set. Let $\struct {X, d}$ be a metric space. , a sequence that converges in d_1 converges in d_2 and vice versa)? Yes. In particular, convergence in any norm is equivalent to coordinate-wise convergence, as you say. ) Let (X,d X) be a metric space, and let (a n) n2N be a sequence of Mar 13, 2018 · The usual metric on $\mathbb N$ is equivalent to the discrete metric, and any metric on a finite set is equivalent to the discrete metric Hot Network Questions Creating polygon from selected lines in QGIS May 18, 2020 · Definition. , for all U ⊆ X, U is open in the metric space (X, d) iff U is open in the metric space (X, d′ ). $\begingroup$ What does it mean that they are equivalent in this particular context? There are many equivalence relations. In particular, the notions of convergent and Cauchy sequences apply in any normed space. 43 (Directed Sets, Nets). Jan 25, 2022 · (Hence every metric is equivalent to a bounded metric. Upload Image. Sep 1, 2021 · Lemma. I. $\begingroup$ Being a Cauchy sequence or not is not a purely topological concept. My main reference in what regards metric spaces is Dieudonné's "Foundations of Modern Analysis" (chapter 3). For i E. 1 Convergence In metric spaces, convergence is de ned with respect to sequences indexed by the natural numbers (De nition A. 10: (Complete metric spaces) A metric space (X;d) is said to be complete if, and only if, every Cauchy sequence in (X;d) is in fact convergent in (X;d). Academic Discipline And Sub-Disciplines Mathematics, Functional Analysis. A subset C is I am wondering whether the concept of two metrics defining the "same convergent sequences" (for instance, when we are interested in the metrics being topologically Nov 4, 2017 · I'll prove one direction of the equivalence between the first two bullets, hopefully you will understand how to prove the other direction. The proof for general metric spaces is essentially the same as the proof for R and is left as an exercise. Nov 8, 2014 · I would like to study if these metric are equivalent, uniformly equivalent. $$ Let $\varepsilon >0$. Show that two metrics d, p on X are equivalent if and only if the convergent sequences in (X, d) are the same as the convergent sequences in (X, p). Here d (x n, x) is the metric (distance) in X. Hence, if the metrics are uniformly equivalent, the space is complete in one metric if and only if it is complete in the other. Then we say that the sequence converges to a ∞ ∈R, and write lim n→∞ a n = a ∞, if Definition 1. 2. ♦ Not every metric is induced from a norm; the metric in Exercise 1. That is, for each ε > 0 there is N ∈ N such that for every n ≥ N, ρ(x n,x) < ε. Mar 5, 2025 · Two metrics g_1 and g_2 defined on a space X are called equivalent if they induce the same metric topology on X. De nition A. The convergent sequences are the same in both norms. Jan 30, 2024 · This statement and what you are asking are equivalent in arbitrary metric spaces, here is the proof: Convergent sequences in metric space $(\mathbb{R}^n,d)$ 2. Convergence of sequences. Two equivalent definitions of convergent sequences? Ask Question Asked 11 years, 2 months ago. Then, d a and d b are equivalent if and only if they lead to identical set of convergent sequences; i. 1 Convergent sequences in metric spaces Definition 1. And a set is closed if and only if its complement is open. Let $ X $ be a space with two different metrics $ d_1,d_2$ such that the two topological spaces $ (X,d_1),(X,d_2) $ have the same convergent sequences. However, at the points in the subset the restricted metric is equal to the larger metric. But if the norms are equivalent, the topology they generate is the same. De nition 2. Every metric g on X Jun 13, 2016 · Since every Cauchy sequence has a convergent subsequence, Compactness and sequential compactness are equivalent in metric space but not always in others. I think that they are not equivalent because if we think of any rational sequence converges to irrational. sequence (x n) n≥1 ⊂ X is is convergent to x, if lim n→∞ d(x n,x) = 0. Remark 6. A sequence in a metric space $X$ is a function $x:\N\to X$. But it must make them similar in some sence to be specified, and if so in what possible ways ? Furthermore, I was wondering if two metrics having the same converging sequences makes them topologically equivalent ? Nov 21, 2023 · Equivalence of Definitions of Convergent Sequence in Metric Space; Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology for a proof that this definition is equivalent to that for convergence in the induced topology. A. A metric space (M;d) is said to be complete if and only if every Cauchy sequence of elements of Mconverges to an element of M. Changing a metric to a equivalent metric does not. Closed and bounded sets in R. But what I said is still true: the set of all sequences which are Cauchy in ALL the metrics for a completely metrizable topology is precisely the set of convergent sequences. Exercise 1. You need an uniform structure for that. Hint. 1. 1 Convergence Definition 14. For example, does not converge inÐ8Ñ Ð ß. Equivalent metrics are Oct 16, 2022 · $\begingroup$ While the notion of a convergent sequence does not depend on the choice of (equivalent) metrics, the notion of Cauchy sequence does. Define p on X x X by p(x, y) = min(1, d(x, y)), x, y ∈ X. 3. However, space , where is the (q 1, q 2) metric like obtained from (q 1, q 2)-quasi metric-like p, is T 2, and the convergent sequences in (X, p) converge only to a unique limit. " Oct 16, 2017 · $\begingroup$ For your second question, see my edit. 42. Exercise 2. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Feb 21, 2018 · There are several definitions of equivalent metrics and one of them says convergent sequences and their limits coincide for the two metrics. Changing a metric to a equivalent metric does not change the open sets $\begingroup$ Each convergent sequence is compact, because each open neigborhood of its limit contains all but finitely many elements of the sequence. Results about I can easily see that metrically equivalent metrics are topologically equivalent. convergent sequences to nonconvergent sequences. ) Prove that every subset of a metric space M can be written as the intersection of open sets. And by contrapositive, if you have two metrics on a space such that the space is complete in one but not in the other, the two metrics aren't uniformly equivalent, and a fortiori not strongly equivalent. For example, you might require that all sequences are equivalent (this is a very loose relation) or you might require that sequences are equivalent only if they are equal (this is a very strict relation). . Jan 16, 2023 · I understand subspaces and why you can have a set and a metric that make up a metric space, and say you take a subset of that set you can have a subspace by restricting the metric to that subset. That is, two arbitrary terms and of a convergent sequence become closer and closer to each other provided that the index of both are sufficiently large. We will work with functions between general metric spaces, so we need to start by de ning continuity of functions and conver-gence of sequences in metric spaces. )Let (X,d X) be a metric space, and let (a n) n∈N be a sequence Jun 19, 2024 · Convergent Sequence is Cauchy Sequence; Definition:Complete Metric Space: a metric space in which the converse holds, that is a Cauchy sequence is convergent. Define on X x X by p(x, y) = min(I, d(x, y)), where xy ∈ X. Problem 1 (Equivalence of metrics). e. $\endgroup$ – Result of Convergent sequence in Equivalent Metrics | L20 | TYBSc Maths | Completeness @ranjankhatu #Completemetricspace#completeness #convergent #sequence Feb 5, 2021 · $\begingroup$ You've defined "equivalent metrics" in terms of convergence of sequences, which for metric topologies amounts to getting the same topology. 4. This motivates the following de nition: De nition 1. R is a complete metric space, meaning that any Cauchy sequence of elements of R converges to an element of R. Ñ Ð0Ð8ÑÑ Ð ß. this is equivalent to de ning them in terms of a topology. Bounded sets. Two metrics on X are equivalent if they determine the same open subsets. May 13, 2023 · We say that two metrics d and d′ defined on the same set X are equivalent iff they have the same open sets, i. Prove that, given x ∈ X and as sequence (x n)∞ =1 ⊂ X, the above definition is equivalent to the fact that (x n) n∈N converges to x, as a net, with respect to the metric topology. w change the open sets and therefore does not change which sequences converge. 37 (Metric equivalence and convergent sequences) Let d a and d b be two different metrics on the same set X . ) Let (X,d X) be a metric space, and let (a n) n2N be a sequence of Mar 2, 2025 · It turns out that this is equivalent to a metric space having no limit points at all, as well as every convergent sequence being eventually constant (take it as an exercise to prove that these three notions are in fact equivalent, if it's not too trivial that is). 11 (Sequence convergence on different metrics) Consider the space of all continuous real valued functions defined on the interval [ 0 , 1 ] denoted by C [ 0 , 1 ] . Explain why the only convergent sequences in X are the ones which are eventually constant. Since the space is (sequentially) compact, this sequence has a convergent subsequence. X is equivalent to a discrete metric space. Oct 6, 2018 · $\begingroup$ @PaulFrost Yes, different metrics for a completely metrizable topology may have different sets of Cauchy sequences. We consider sequences of graphs and define various notions of convergence related to these sequences: ``left convergence'' defined in terms of the densities of homomorphisms from small graphs into the graphs of the sequence, and ``right convergence'' defined in terms of the densities of homomorphisms from the graphs of the sequence into small graphs; and convergence in a suitably defined metric. SO if a set is closed in one metric, it is closed with respect to all equivalent metrics. 12. If you have a different definition I can show you how to prove the equivalence. Nov 15, 2020 · $\begingroup$ If you use the sequentially compact definition (which is equivalent to compactness in a metric space), then the proof is quite easy: Take a Cauchy sequence in the compact metric space. Any metric space equipped with the discrete metric has this property. If X is a normed space, then d(f,g) = kf −gk defines a metric on X, called the induced metric. Show that two metrics d,p on X are equivalent if and only if the convergent sequences in (X,d) are the same as the convergent sequences in (X,p). Let's assume x ∈ E ∩ ∂E. A different (stronger) condition is often attached to "equivalent metries", but with your definition in place you are starting well. 3. Two metrics d and p on a set M are said to be equivalent if they generate the same convergent sequences; that is, d(xn , x ) ----} 0 if and only if p(xn , x ) ----} 0. $p$ and $d$ are such that $kd(x,y) \leq p(x,y) \leq td(x,y)$ for every $x, y \in X$, $k$ and $t$ are positive constants. $\endgroup$ – I am wondering whether the concept of two metrics defining the "same convergent sequences" (for instance, when we are interested in the metrics being topologically Sequences. See Exercise 3. " Now since the sequence $(x_n)$ does not contain any convergent subsequence, it must be unbounded by the Bolzano theorem. Thus Q with the usual metric is not complete, R with the usual Jan 8, 2022 · Cauchy Sequences. That is, $\struct {X, d_1}$ and $\struct {X, d_2}$ are two different metric spaces on Feb 14, 2021 · I am trying to figure out which definitions make this proof the easiest. $\endgroup$ – Alex Ravsky Commented Apr 23, 2013 at 2:22 Jun 6, 2020 · I am wondering whether the concept of two metrics defining the "same convergent sequences" (for instance, when we are interested in the metrics being topologically Show that it is possible to have a bounded sequence in a metric space which has no convergent subsequences. Let (X,‰) be a metric space and let A,A k be any nonempty closed subsets of X. 11. Uniformly equivalent metrics have the same Cauchy sequences. 4 Convergent sequences in metric spaces14. Let (X,d) is a metric space, and if Theorem 1. Answer to Let di, d2 be equivalent metrics on a set X. (Convergent sequences in R. Let $X$ be a set upon which there are two metrics $d_1$ and $d_2$. $(\Rightarrow)$ Assume that $(x_n)_n$ is Cauchy, that is $$\lim_{N\rightarrow \infty}\sup_{p\in\mathbb{N}}d(x_N,x_{N+p})=0. Let X be an infinite set with a discrete metric. , a sequence is convergent in ( X , d a ) if and only if it is also convergent in ( X , d b ) and it has same limit in Is it true that two metrics d_1 and d_2 on a set X are equivalent iff they have the same convergent sequences (i. Convergent sequences are bonded. Proof. 13. Nov 24, 2016 · Your intuition is correct, to an extent: in $\mathbb R^n$, all norms are equivalent (i. it would be convergent to 0 in d but to 1 in σ. , they define the same notion of convergence). Once you have that, it is easy to see We will also show that convergence from the left is equivalent to convergence in metric for a suitable notion of distance between two weighted graphs. 0. 7. The point to which the sequence converges is the limit of the sequence and we write {x n} → x to denote the convergence of {x n} to x. Math Mode Stack Exchange Network. For your first question, note that the definition of equivalence that I give (and I believe you are also using) is silent about sequences that do not converge. Let (X,‰) be a metric space. Real Number Line is Complete Metric Space; Complex Plane is Complete Metric Space; Thus in $\R$ and $\C$ a Cauchy sequence and a convergent sequence are equivalent concepts. We will also show that convergence from the left is equivalent to convergence in metric for a suitable notion of distance between two weighted graphs. (Hence equivalence is a metric and not a topological property. If fp 14. 2). My thoughts were the following. For any nonempty closed subsets A,A k µ X, it is said that the sequence {A k} is Wijsman convergent to A if lim k!1 d(x,A k)˘d(x,A) for each x2 X and it is denoted by W-limA k ˘ A. Let a relation $\sim$ be defined on Two metrics on X are equivalent if they determine the same open subsets. Define p on X X X by p(x,y) = min(1,d(x,y)), x,y e X. Two metrics and on X are strongly or bilipschitz equivalent or uniformly equivalent if and only if there exist positive constants and such that, for every ,, (,) (,) (,). Nov 6, 2024 · I read that: If $d_1,d_2 $ are metrics on $X \neq \emptyset $ then $d_1,d_2 $ are equivalent metrics $\iff$ they have the same convergent sequences. Modified 11 years, The sequence $(x_n)$ in the metric space $(X,d The main purpose of this handout is to give two properties of functions that are equivalent to continuity, one in terms of sequences and one in terms of open sets. Well my approach is this: "=>" Let E be an open set in X. In fact if you go out far enough in the sequence these metrics agree. Finally, we will show that convergence from the left is equivalent to the property that the graphs in the sequence have asymptotically the same Szemerédi partitions. This is the case iff, for every point x_0 of X, every ball with center at x_0 defined with respect to g_1: B(x_0,r_1;g_1)={x in X|g_1(x_0,x)<r_1} (1) contains a ball with center x_0 with respect to g_2: B(x_0,r_2;g_2)={x in X|g_2(x_0,x)<r_2}, (2) and conversely. Theorem 3. There, precompact stands for the same as totally bounded. Show that two metrics d and p on X are equivalent if and only if the convergent sequences in (X, d) are the same as the convergent sequences in (X, p). Since equivalent metrics admit the same class of open sets, they admit same convergent sequences. 1 (Convergent and Cauchy Sequences): Hence, if the metrics are uniformly equivalent, the space is complete in one metric if and only if it is complete in the other. lecture 5 (01/26/24) Convergent sequences are bounded. (Convergent sequences inR. Mar 26, 2018 · Two metrics: d 1 and d 2 on a metric space X are called equivalent when there exists M >= 1 such that M-1 d 1 (x,y) <= d 2 (x,y) <= Md 2 (x,y) That definition corresponds to strongly equivalent metrics. Apr 12, 2023 · Definition. These two homeomorphic metric spaces have exactly the same convergent sequences but they do not have the same Cauchy sequences. May 30, 2016 · Keywords: Partial metric space; equivalent Cauchy sequences; 0-equivalent 0-Cauchy sequences. 8 ([16]). spaces. Oct 3, 2020 · The definition of equivalent metrics is the following in my context: Two metrics are equivalent if convergence of any sequence in one metric implies the convergence of the other, and vise versa. Note. 2 corresponding to convergence in measure is an example. Show that two metrics d,p on X are equivalent if and only if the convergent sequences (X,d) are the same as the convergent sequences in (X,p). Question: 3. Show that d is equivalent to the discrete metric if and only if every point in X is an isolated point Homework Equations Two metrics [itex] d, d' [/itex] are said to be equivalent if they have to same convergent sequences with the same limit points It's well known that for metric spaces the following is true. On the other hand, when considered as a sequence of elements of R (the set of real numbers), the previously mentioned sequence converges to sqrt(2) in R. Note that every convergent sequence in (X, p) is a Cauchy sequence. But the change might Cauchy sequencescreate or destroy May 6, 2021 · As a result, a right (left) convergent sequence might converge to more than one limit. I have proved $(\Rightarrow). Definition 2. ) Let (a n) n∈N be a sequence of real numbers. Results about convergent sequences in the context of metric spaces can be found here. I'm trying to prove that if the convergent sequences of $(X,d)$ and $(X,\rho)$ are the same, then the metrics $d$ and $\rho$ are equivalent. Dec 1, 2020 · Note that equivalent metric spaces have the same convergent sequences but need not have the same Cauchy sequences—equivalence therefore does not conserve completeness. $\endgroup$ – 1 Convergent sequences in metric spaces Definition 1. (See Section 2). 5. If a sequence converges to a limit , its terms must ultimately become close to its limit and hence close to each other. Question: Two metrics d, p on a set X are said to be equivalent if they define the same convergent sequences. Once you have that, it is easy to see So the set will be open in all equivalent metrics. (Convergent sequences in metric spaces. "Two metrics on X are equivalent iff they induce the same convergence. Then x ∊ E and $\begingroup$ This is a great example for: "A metric that is strongly equivalent (definition 2) to a complete metric is also complete; the same is not true of equivalent metrics (definition 1) because homeomorphisms do not preserve completeness". Apr 3, 2015 · I am wondering whether the concept of two metrics defining the "same convergent sequences" (for instance, when we are interested in the metrics being topologically the usual metric, as well as in all discrete metric spaces. Two metrics on a set Xare called (topologically) equivalent, if they generate the same topology on X. (Recall that we have defined equivalence to mean that d and p generate the same convergent sequences. This is the definition I am using. $ Also I can prove Two metrics $d,p$ on $X$ are equivalent if they determine the same open subsets. Then choose some sequence in X all of whose points are distinct. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Let [itex] (X,d) [/itex] be a metric space. But two equivalent Sep 7, 2019 · We suppose that two metrics must be equivalent and we want to prove that the topologies induced by that metrics are equal. (Hence every metric is 1 Convergent sequences in metric spaces Definition 1. Show that two metrics $d,p$ on $X$ are equivalent if and only if the convergent sequences in $(X,d)$ are the same as the convergent sequences in $(X,p)$. Then X\E is closed in X. In contrast to the sufficient condition for topological equivalence listed above, strong equivalence requires that there is a single set of constants that holds for every pair of points in , rather than potentially different Mar 26, 2018 · The example in post #15 is very nice, and makes a good point, but notice it is somewhat irrelevant to the question of convergence, since all convergent sequnces are bounded. ) Let (X,d X) be a metric space, and let (a n) n2N be a sequence of May 1, 2016 · $\begingroup$ I accepted the edit. . Therefore, your overall argument does not fly. A sequence {x n} in a metric space X is said to converge if there is a point x ∈ X which for any ε < 0 there exists an integer n ε such that n ≥ n ε implies that d (x n, x) < ε. Oct 16, 2022 · $\begingroup$ While the notion of a convergent sequence does not depend on the choice of (equivalent) metrics, the notion of Cauchy sequence does. I am trying to show two equivalent metrics $p$ and $d$ on a set $X$ have the same convergent sequences. Show that p is a metric that is equivalent to d Aug 2, 2022 · Stack Exchange Network. ) Nov 2, 2016 · A convergent sequence is a Cauchy sequence that has a limit point in the space. Prove: (i) Two metrics dand d0 on a set X are topologically equivalent, if and only if the convergent sequences in (X;d) are the same as the convergent sequences in (X;d0). Let $\CC \sqbrk X$ denote the set of all Cauchy sequences of elements from $X$. (a) Prove that equivalent metrics have the same closed sets, the same compact sets, and the same convergent sequences. Jan 18, 2024 · lecture 3 (01/22/24) Metric spaces, Cauchy-Schwarz, triangle inequality of the Euclidean metric, open balls in metric spaces. And finding such a sequence isn't to hard either - have exactly the same convergent sequences but they do not have the same Cauchy sequences . also Show that the usual metric on is equivalent to the A sequence which converges in the metric space (X, d 1) may not converge in the metric space (X, d 2) unless the metrics are equivalent. Show that for any metric space (x, d), d(a, b) 1+ d(a, b) p(a, b) : defines an equivalent metric. But, this definition really does make no sense to me. It is said that the Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Jul 31, 2017 · $\begingroup$ Is this definition of convergence just completely equivalent to saying that: convergence in a normed space is convergence in the metric space induced by the norm; or am I misunderstanding something? $\endgroup$ Convergent Sequences Subsequences Cauchy Sequences Convergent Sequence De nition A sequence fp ngin a metric space (X;d) is said to converge if there is a point p 2X with the following property: (8 >0)(9N)(8n >N)d(p n;p) < In this case we also say that fp ngconverges to p or that p is the limit of fp ngand we write p n!p or lim n!1 p n = p. Then we say that the sequence converges to a1 2 R, and write lim n!1 a n = a1,if Definition 1. In the usual notation for functions the value of the function $x$ at the integer $n$ is written Question: 4. baff zrltfh qnaj xmbl mtqx faxm wlodbmu grxe yxnjv vszdg mnnyd tuisza hbdhb uyonqq hamo