Proving Convergence of Sequences in a Metric Space and Closed Sets

  • Thread starter ShengyaoLiang
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In summary, two question on analysis were asked, but the person could not answer them. They suggest looking at the class notes or doing a Contrapositive.
  • #1
ShengyaoLiang
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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...
 
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  • #2
i get the feeling that you are not doing these problems by yourself.

hint: triangle inequality.
 
  • #3
。。。。。。。。。。。。。。。。。
 
  • #4
anyway, after looking the class notes ,i still have no idea...


this is tooooo difficult for me.
 
  • #5
triangle inequality:

︱a+b︱≤︱a︱+︱b︱ and ︱︱a︱-︱b︱︱≤︱a - b︱
 
  • #6
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) ?
 
  • #7
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.
 
  • #8
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?
 
  • #9
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...
 

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