MHB Confirm Answers on Homework Sheet: Subsequence Convergence

Click For Summary
The discussion centers on the convergence properties of sequences and their subsequences. The original poster identifies four statements regarding sequences, providing answers and counterexamples for each. They conclude that statements (i), (ii), and (iv) are false, while (iii) is true based on the Bolzano-Weierstrass theorem. A correction is suggested regarding the divergence of a specific sequence, prompting further exploration of its behavior. The conversation emphasizes the importance of understanding the conditions under which subsequences converge.
Carla1985
Messages
91
Reaction score
0
Question from my homework sheet. Can someone confirm I've got these correct.

Let (an)n∈N be any sequence of real numbers. Which of the following statements are true?
Give precise references to the results in the Lecture Notes for those which are true. Construct counter examples for those that are false.
(i) Every sequence (an)n∈N has a convergent subsequence.
(ii) If (an)n∈N has a convergent subsequence, then (an)n∈N is convergent.
(iii) If (an)n∈N is bounded, then (an)n∈N has a convergent subsequence.
(iv) If (an)n∈N has a convergent subsequence, then (an)n∈N is bounded.

My answers:
i) False: {1,2,3,4,5,6...) has no convergent subsequence.
ii) False: {-1/2, 1/4, -1/8, 1/16, -1/32...} diverges but has subsequence {1/4, 1/16, 1/64...} which converges
iii) True (Bolzano Weierstrass theorem)
iv) False: {1, 1, 2, 1/2, 3, 1/3...} is unbounded but has subsequence {1, 1/2, 1/3, 1/4...} which converges

 
Physics news on Phys.org
Looks good to me.

[EDIT] See Evgeny.Makarov's post below for a correction.
 
Last edited:
Carla1985 said:

ii) False: {-1/2, 1/4, -1/8, 1/16, -1/32...} diverges but has subsequence {1/4, 1/16, 1/64...} which converges
Does -1/2, 1/4, -1/8, 1/16, -1/32, ... really diverge?
 
Evgeny.Makarov said:
Does -1/2, 1/4, -1/8, 1/16, -1/32, ... really diverge?


So maybe try flipping what the alternating terms are doing:
{-1/2, 1/2, -1/4, 1, -1/8, 2, -1/16, 4, ...}
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K