Prove that a converging sequence is bounded

  • Thread starter kehler
  • Start date
  • #1
kehler
104
0

Homework Statement


Suppose that the sequence {an}converges. Show that the sequence {an} is bounded.

The Attempt at a Solution


Since the sequence converges, for every delta>0, there must exist a number N such that for every n>=N,
|an - x|< delta. Therefore, for n>=N, -delta+x < an < delta + x.
So I've proven that for n>=N, the sequence is bounded between -delta+x and delta+x.

But I don't know how to prove that for n<N, an is also bounded. I know that there are only a finite number of elements before the sequence starts to converge. Is there a theorem stating that all finite sets are bounded?
 

Answers and Replies

  • #2
But I don't know how to prove that for n<N, an is also bounded. I know that there are only a finite number of elements before the sequence starts to converge. Is there a theorem stating that all finite sets are bounded?

I would consider looking at the maximum of a finite set.
 
  • #3
statdad
Homework Helper
1,495
36
To expand a little:
Pick an [tex] \varepsilon > 0 [/tex], and use convergence to conclude that

[tex]
|x_i - a | < \varepsilon
[/tex]

for all [tex] i \ge N [/tex].

You now have two sets of elements of the sequence: those with [tex] i \ge N [/tex] and those for smaller [tex] i [/tex]. You should be able to argue that both sets are bounded, which means ...
 
Last edited by a moderator:
  • #4
Dick
Science Advisor
Homework Helper
26,263
620
Yes. All finite sets are bounded. Prove it by induction. (If you really need a proof).
 

Suggested for: Prove that a converging sequence is bounded

Replies
4
Views
421
Replies
7
Views
342
Replies
4
Views
358
  • Last Post
Replies
18
Views
438
Replies
9
Views
486
Replies
2
Views
778
Replies
15
Views
458
Replies
8
Views
559
Replies
2
Views
197
Top