Does every subsequence of a_n converge to a?

[kingwinner]

Idea of a "subsequence"

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

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

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

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

Thanks for clarifying!

 

[fourier jr]

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

 

[l'Hôpital]

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.

 

[fourier jr]

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

 

[ordirules]

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?

 

[l'Hôpital]

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.

 

[ordirules]

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 :-)

 

[fourier jr]

never mind, i confused myself

 

[l'Hôpital]

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 :-)

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 x_n = 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.

 

[HallsofIvy]

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

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.

 

[l'Hôpital]

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.

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.

 

[ordirules]

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 x_n = 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.

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?

thanks for reading! :-)

 

[l'Hôpital]

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?

thanks for reading! :-)

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, x_n 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.

 

[rochfor1]

Yes, a sequence has to converge to a number.

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

 

[kingwinner]

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...??

Can someone please explain this?
Any help is appreciated!

 

[sutupidmath]

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 sense 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.

 

[HallsofIvy]

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...??

Can someone please explain this?
Any help is appreciated!

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.

 

[kingwinner]

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

 

[HallsofIvy]

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...??

Can someone please explain this?
Any help is appreciated!

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

 

[kingwinner]

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?

 

[l'Hôpital]

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?

It's trivial, really.

a_n is a subsequence of itself.

 

[kingwinner]

It's trivial, really.

a_n is a subsequence of itself.

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

 

[HallsofIvy]

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.

 

[l'Hôpital]

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

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

 

[kingwinner]

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.

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

But I don't see why this is always true. Does anyone have any idea about this?

Thanks for any help! :)

 

[kingwinner]

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

I think it would also work if we can show that:
ONE subsequence converges to a => every subsequence converges to a

If we can show that ONE subsequence converges to a, does this imply that EVERY subsequence converges to a?
If so, how to prove this?

Thanks!

 

[l'Hôpital]

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

I think it would also work if we can show that:
ONE subsequence converges to a => every subsequence converges to a

If we can show that ONE subsequence converges to a, does this imply that EVERY subsequence converges to a?
If so, how to prove this?

Thanks!

No.

Consider a_n = (-1)^n.

a_2n converges 1 for all values of, but a_n diverges. ALL subsequences must converge to the same value.