Proving Convergence of Sequences in a Metric Space and Closed Sets

  • Thread starter Thread starter ShengyaoLiang
  • Start date Start date
  • Tags Tags
    Analysis
Click For Summary

Homework Help Overview

The discussion revolves around proving properties of sequences in a metric space, specifically focusing on convergence and the characteristics of closed sets. The original poster presents two questions related to these topics, seeking assistance in understanding the proofs required.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the triangle inequality and the definitions of convergence in metric spaces. Some suggest logical structures to approach the proofs, while others express uncertainty about the problems.

Discussion Status

There is a mix of attempts to clarify the problems and provide hints without revealing complete solutions. Some participants have shared their reasoning and interpretations, while others question the original poster's approach and understanding of the material.

Contextual Notes

Participants note the challenges of the problems and the original poster's feelings of difficulty. There is an indication that the forum has guidelines regarding the nature of homework help, which may influence the discussion.

ShengyaoLiang
Messages
23
Reaction score
0
two question on analysis...

1)Prove : Let (An) amd (Bn) be sequences in a metric space S such that d (An , Bn) → 0. Then (An) converges if and only if (Bn) converges, and if they converge, they have the same limit.
2)Prove: Let C be a closed set and let (Xn) be a sequence in C converging to a. Then, a∈ C.

Could anyone help me out? Thanks!
i still have no idea on how to prove...
 
Physics news on Phys.org
i get the feeling that you are not doing these problems by yourself.

hint: triangle inequality.
 
。。。。。。。。。。。。。。。。。
 
anyway, after looking the class notes ,i still have no idea...


this is tooooo difficult for me.
 
triangle inequality:

︱a+b︱≤︱a︱+︱b︱ and ︱︱a︱-︱b︱︱≤︱a - b︱
 
i know how to prove : Let (Xn)be a sequence converging to a and to b. then a=b...

Is the prove above is similar as 1) ?
 
look at this

d(An,A) < d(An,Bn) +d(Bn,A)

what can u say about the left hand side if Bn converges to A.

The problem gives u that d(An,Bn) goes to zero and that Bn converges to A. It asks you to, using this information, show that Bn must converge.

draw a picture of examples.

Let A_n = 0 for all n and let Bn=1/n. look at what this theorem is saying.
 
What you should do with a problem like this is just write down the logical structure that you need to prove the result and look at the definitions of the entities involved. What I mean is this:

1) Show (An converges) implies (Bn converges). The symmetry in the assumptions then gives you the reverse implication. To show this, assume An converges, and then show that Bn converges. Use the definition of convergence, the axioms of a metric space, and the hypotheses of the problem. After that, assume they converge, to say a and b, and show that a=b.

2) With this one you may want to do the contrapositive. Show that if a is not in C, no sequence of elements of C may converge to it. Hint: C is closed, so it's compliment is open. What is the definition of an open set?
 
things i did:

1) if An converges to a, i can show that Bn converges to a , also.
similarly, if Bn converges to b, i can show An converges to b , also.
therefore, An converges iff Bn converges. and, obveriously, a=b
2) suppose Xn converges to a and let C = [m,n] then, m≤Xn≤n and i can show
m<a<n , by using hbhd. Thus, a in C

Is the idea correct? anyway~hehe i have hand in my assignment...
 
  • #10
You apparently also did not read the agreements you were supposed to when you registered with this forum. This is not the place for schoolwork problems! I am moving this to "homework, calculus and beyond".

(Look closely at the triangle inequality for these.)
 
  • #11
anyway...i am sorry...
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
3
Views
2K
Replies
3
Views
10K