# Idea of a subsequence

1. Nov 5, 2009

### kingwinner

Idea of a "subsequence"

I don't fully understand the idea of a "subsequence"...

1) Say we have an infinite sequence {an} = {1,2,3,4,...}
Then if we only take {1,2,3} with FINITE number of elements, is it a subsequence of an?

2) If an is an infinite sequence, is an a subsequence of an? (i.e. when we talk about subsequence of an, does it include itself?)

2) Fact: Let an be a sequence. If every subsequence of an converges to a, then the sequence an converges to a.
WHY is this true?

Thanks for clarifying!

2. Nov 5, 2009

### fourier jr

Re: Idea of a "subsequence"

re: 2) what else could a_n converge to? denote a subsequence by a_f(n) where f(n) is strictly increasing. then use the definition of convergent sequence & the fact that n<m => f(n) < f(m)
I guess if f is strictly increasing then a_n wouldn't be a subsequence of itself... that seems too strict though; I don't see why a_n shouldn't be a subsequence of itself

3. Nov 5, 2009

### l'Hôpital

Re: Idea of a "subsequence"

Actually, it's a tautology. a_n is a subsequence of itself, so by definition, if all subsequences converge to a, so must a_n.

4. Nov 5, 2009

### fourier jr

Re: Idea of a "subsequence"

again, I don't see why a_n shouldn't be a subsequence of itself, but my copy of baby rudin says if n_1, n_2, ... are the indices of a subsequence, then n_1 < n_2 < ... which would make it impossible

5. Nov 5, 2009

### ordirules

Re: Idea of a "subsequence"

Well, first of all, keep in mind that a sequence keeps order, unlike a set. A set is like a bag you just throw numbers in and a sequence is like a pile where you stack numbers on, stacking it differently makes it a different sequence.

Typically, Mathematicians use parentheses() for sequences and brackets{} for sets. I would usually use () to denote a sequence because everything I've read uses them. (Books usually tell you about their notations in the first few pages somewhere)

A sub-sequence is just picking any number from the sequence, the only catch is they have to be in the right order.

For example, consider sequence:
(1,5,3,2,4,23)

(1,3,23) is a subsequence, but (3,1,23) is not because 3 is after 1 in the sequence.

However, if the sequence was:
(1,5,3,1,2,4,23)
(3,1,23) would now be a subsequence.

So that's the only rule in a sequence: preserve order.

Okay, now for the questions (try to answer them yourself before reading now :-)):

1) Yes, you can choose a finite number of elements from an infinite sequence to make a subsequence. (Just preserve the order)

2) Yes, a sequence can be a subsequence of itself. (You just pick all the numbers of the sequence)

3) Convergence only is used for infinite sequences because they never end. Every finite sequence converges technically (you can maybe imagine it as the finite sequence plus an infinite number of zeros after, some mathematicians would disagree with me though).
Now, a sequence converges if the next numbers are eventually getting smaller and smaller until they reach zero (at infinity).

For example, the infinite sequence: (1,1,1,1,.... ) does not converge, because you will see 1's even at infinity.

However, this sequence (1,1/2,1/4,1/8,1/16.....) does converge (I'm halving the number each time) If you go far off to infinity, the numbers will be smaller than the size of the smallest atom, so you can assume it to be pretty much zero for practical purposes.

Now, if you choose a subsequence of the sequence I just wrote you have two cases:
a) You chose a finite subsequence. This sequence will converge
b) You chose an infinite sequence. This will converge because you know at infinity these numbers get eventually very small... For example, if I decided to choose every other number (1, 1/4, 1/16,...) of the sequence, I would still have it converging.

I hope this kind of helps. Another thing to think of is that if a sequence does eventually become zero at infinity, then how would the next number of such a sequence relate to the number before? Would it be bigger or smaller?

6. Nov 5, 2009

### l'Hôpital

Re: Idea of a "subsequence"

Untrue.

(1,1,1,1,1...) does converge, to one.

A sequence a_n converges to L if for any given e>0, there exists an n_0 such that l a_n - L l < e for all n > n_0

Fourier:

What if n_1 = 1, n_2 = 2, n_3 = 3...? The subsequence would be the sequence itself.

If you n_i = i, it is still increasing, but it is just the sequence.

7. Nov 5, 2009

### ordirules

Re: Idea of a "subsequence"

Woops! You're definitely right!
Yes, a sequence has to converge to a number. (I was trying to remember what people defined sequences as)

Anyway, let me rephrase my last question:
If a sequence does eventually converge to a value, what is the relation between the differences of adjacent sequence numbers, do they start getting smaller or larger?
ex:
(1,1/2,1/4,1,8 etc...)
if 1/2/1-4 larger or smaller than 1/4-1/8?

Thanks for pointing out the rule Hopital :-)

8. Nov 5, 2009

### fourier jr

Re: Idea of a "subsequence"

never mind, i confused myself

9. Nov 5, 2009

### l'Hôpital

Re: Idea of a "subsequence"

It also convenient to point out that this is called the Cauchy Criterion and these kind of sequences are Cauchy Sequences. In R^n, Sequences are only cauchy iff they are convergent. However, this does not necessarily mean cauchy sequences are always convergent. If you take the sequence xn = 1/n in the space [1,0), then it will get close to 0, but it will never actually become zero, thus it will not converge.

10. Nov 6, 2009

### HallsofIvy

Re: Idea of a "subsequence"

Why would that make it impossible for {a_n} to be a subsequence of itself? You are just taking n_1= 1, n_2= 2, n_3= 3, etc. so "n_1< n_2< ..." becomes "1< 2< ..." which is certainly true.

A "subsequence" of {a_n} is just a subset of {a_n}- and a set is always a subset of itself.

11. Nov 6, 2009

### l'Hôpital

Re: Idea of a "subsequence"

I never said it was impossible. In fact, quite the opposite. I suggested a subsequence was really the sequence in question. That was my intention.

Sorry if it was unclear.

12. Nov 6, 2009

### ordirules

Re: Idea of a "subsequence"

Thanks for saying that, but now I'm a little confused... The number seems to approach zero when you go to infinity (although the sum of the sequence clearly doesn't). Why is it said that it does not approach zero? What criterion says so?

Sorry for using this post for my question. Is there a way to send messages through physics forums?

13. Nov 6, 2009

### l'Hôpital

Re: Idea of a "subsequence"

Not the sum of the sequence but the sequence itself. Yes, it goes to zero, but zero is not in its domain, therefore it can't converge to that which is not in its domain. Usually, xn is within the set of real numbers, of which zero is a part of, therefore I can converge there, but since zero is not in this set, then the sequence can't converge to it.

14. Nov 6, 2009

### rochfor1

Re: Idea of a "subsequence"

Definitely not. $$( ( -1 )^i )$$ is a perfectly good sequence that doesn't converge.

15. Nov 10, 2009

### kingwinner

Re: Idea of a "subsequence"

Definition:
A subsequence of a given sequence is understood as 'a sequence that can be derived from a given sequence by deleting some elements without changing the order of the remaining elements'.

1) Consider the sequence {a_n} = {1,2,3,4,...}
Then according to the definition above, if I delete all the terms from 4 onwards, I would have {1,2,3}, and the order is preserved, so {1,2,3} is a subsequence of a_n, right??

3) Fact 2: Let a_n be a sequence. If the sequence a_n converges to a, then EVERY subsequence of a_n also converges to a.

Why does it make sense???

But I think I can give a counterexample...
Take {b_n}={1,3,1,3,1,1,...,1,2,2,...2,...}
Then b_n converges to 2.
But now if we delete ALL the "2" terms (note: order is preserved, so it is a subsequence I think), I don't think the resulting subsequence can possibly converge to 2...??

Any help is appreciated!

16. Nov 10, 2009

### sutupidmath

Re: Idea of a "subsequence"

To kindwinner: I had the same dilema. my real analysis book also defines the subsequence in the following manner:
$$let \{a_n\}_{n=1}^{\infty}$$ be a sequence, and let $$\{n_k\}_{k=1}^{\infty}$$ be a strictly increasing sequence of integers, then we say that $$\{a_n_k\}_{k=1}^{\infty}$$ is a subsequence of a_n.

In a sence this def. excludes the possibility of having a finite number of terms of a_n to form a subsequence, but nevertheless, if we allow a finite number of terms of an to form a subsequence (which indeed is the case), then when it comes to the proposition: that every subsequence of a convergent sequence converges, and that even more it converges to the same limit. I think that they implicitly mean here every infinite subsequence, instead of simply every subsequence. Because, as you pointed out, another counter example would be say take a_n=1/n, it clearly is a convergent sequence with limit being zero. but any finite subsequence would not converge to zero. So again, i think they mean infinite subsequence.

17. Nov 10, 2009

### HallsofIvy

Re: Idea of a "subsequence"

Yes, your example is a "subsequence" but it is not an infinite subsequence. The correct statement of "Fact 2" is "If the infinite sequence a_n converges to a, then EVERY infinite subsequence of a_n also converges to a."

We don't really need to specify "infinite" sequence or "infinite" subsequence because "converge" is not even defined for finite sequences.

18. Nov 14, 2009

### kingwinner

Re: Idea of a "subsequence"

Thanks #16 and #17, this really clarifies the concept!

19. Nov 15, 2009

### HallsofIvy

Re: Idea of a "subsequence"

Yes, you are right. Your textbook should have specified infinite subsequences.

20. Nov 17, 2009

### kingwinner

Re: Idea of a "subsequence"

For the following discussion, sequence means infinite sequence, and subsequence means infinite subsequence.

Theorem:
The sequence a_n converges to a
<=> EVERY subsequence of a_n converges to a
<=> every subsequence has a subsequence of it which converges to a

Why is the last part true?

21. Nov 17, 2009

### l'Hôpital

Re: Idea of a "subsequence"

It's trivial, really.

a_n is a subsequence of itself.

22. Nov 18, 2009

### kingwinner

Re: Idea of a "subsequence"

Yes, I understand this part:
The sequence a_n converges to a
<=> EVERY subsequence of a_n converges to a

But I don't understand this part:
EVERY subsequence of a_n converges to a
<=> every subsequence has a subsequence of it which converges to a

23. Nov 18, 2009

### HallsofIvy

Re: Idea of a "subsequence"

I'm puzzled only about the use of "has a subsequence". In fact, if $\{a_n\}$ converges to a, the every subsequence of every subsequence converges to a!
This is just an extension of "if $\{a_n\}$ converges to a then every subsequence converges to a" to one more level. If $\{a_n\}$ converges to a, then every subsequence converges to a. Since that subsequence converges to a, it follows, by that same theorem, that evey subsequence, of that subsequence, must also converge to a.

24. Nov 18, 2009

### l'Hôpital

Re: Idea of a "subsequence"

If a_n is a subsequence of itself, then we can say a_n is a subsequence of subsequence of itself.

25. Nov 20, 2009

### kingwinner

Re: Idea of a "subsequence"

Yes, I think you are right. If {a_n} converges to a, the every subsequence of every subsequence converges to a, so in particular, this implies every subsequence has a subsequence of it which converges to a.

So the following is definitely true:
EVERY subsequence of a_n converges to a
=> every subsequence has a subsequence of it which converges to a

Now it remains to show that:
every subsequence of a_n has a subsequence of it which converges to a
=> EVERY subsequence of a_n converges to a