If a sequence converges, then all subsequences of it have same limit

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• Hall
In summary: It's sort of a simple proof, probably to get you used to proving things. But it is an important fact. Simple proofs can seem like a "language game", but you should just tolerate that. There will be much more difficult proofs in the...
Hall
Let's say we're given a sequence ##(s_n)## such that ##\lim s_n = s##. We have to prove that all subsequences of it converges to the same limit ##s##. Here is the standard proof:

Given ##\epsilon \gt 0## there exists an ##N## such that
$$k \gt N \implies |s_k - s| \lt \varepsilon$$
Consider any subsequence ##(s_{n_{k}})##. Since ##n_k \geq k##, therefore ## k \gt N \implies n_k \gt N##. Hence,
$$k \gt N \implies |s_{n_{k}} - s| \lt \varepsilon$$
So, we can write ## \lim s_{n_{k}} = s##.

But it seems to me that this proof is more of a consequence of the language than logic, I know the idea of this proof is logical: if all the terms after a certain point gets nearer and nearer to a point, then no matter how many finite terms we delete, the higher terms will still be near to that same point (because subsequence also contains infinite terms, and eventually the higher terms will get to ##s## only); but this rigorous proof seems merely to be language game.

If we write the subsequences as ##t_n## (and not as ##s_{n_k}##), then how we would proceed to show ##\lim t_n = s##?

We know for each ##n##, ##t_n=s_k## with ##k\geq n##. So if you pick ##N## such that ##s_k## is close to s for ##k\geq N##, then ##t_n## must be close as well if ##n\geq N##

Hall
Office_Shredder said:
We know for each ##n##, ##t_n=s_k## with ##k\geq n##. So if you pick ##N## such that ##s_k## is close to s for ##k\geq N##, then ##t_n## must be close as well if ##n\geq N##
I had the same thing in mind. Here is how I planned to do it:

Given ##\lim s_n = s##, means that for any given ##\varepsilon \gt 0##, there exits ##N## such that ##n \gt N \implies |s_n - s| \lt \varepsilon##.

Definition of subsequence: If in a sequence we delete off a finite number of terms, then the remaining terms, after being ordered, are said to be forming a subsequence.

In the sequence ##(s_n)## we can do the cancellation as follows:
$$\bcancel{s_1}, s_2, s_3, \bcancel{s_4}, \cdots \bcancel{s_N}, \bcancel{s_{N+1}}, s_{N+2}, \bcancel{s_{N+3}}, \cdots$$
##t_1 = s_2##, ##t_2 = s_3##, ##\cdots##, ##t_N = s_{N+2}##, ##\cdots##. All the terms ##t_n##, for ##n \gt N##, corresponds to terms ##s_n## for ##n \gt N## and which are within epsilon of ##s##. Therefore, for the sequence ##(t_n)##, ##n \gt N \implies |t_n -s| \lt \varepsilon##. Hence, ##\lim t_n =s##.

Hall said:
If we write the subsequences as ##t_n## (and not as ##s_{n_k}##), then how we would proceed to show ##\lim t_n = s##?
You must tie it back to the original ##s_n## sequence or you have nothing. Why would you want to remove the reference to the fact that it is a subsequence? That is the most important fact.

FactChecker said:
You must tie it back to the original ##s_n## sequence or you have nothing. Why would you want to remove the reference to the fact that it is a subsequence? That is the most important fact.
Can you please check the facts of the proof I have presented in post #3?

Hall said:
Can you please check the facts of the proof I have presented in post #3?
Your definition of subsequence is wrong. You can remove an infinite number and still have an infinite number remaining for a subsequence. Other than that, you immediately tie the ##t_i## sequence back to the ##s_j## sequence, so I don't think that it has any significant difference from the original proof.

I don't mean to criticize. It is good for you to think about such things.

Hall
FactChecker said:
Your definition of subsequence is wrong. You can remove an infinite number and still have an infinite number remaining for a subsequence.
I think I took "some" for "finite' in this definition (given in wikipedia)
In mathematics, a subsequence of a given sequence is a sequence that can be derived from the given sequence by deleting some or no elements without changing the order of the remaining elements
FactChecker said:
Other than that, you immediately tie the ti sequence back to the sj sequence, so I don't think that it has any significant difference from the original proof.
Yes. My problem with the standard proof was that it seemed more like a language game, we exploited that ##k## quite commercially.

Hall said:
I think I took "some" for "finite' in this definition (given in wikipedia)
Yes, that was wrong. "some" can be infinite.
Hall said:
Yes. My problem with the standard proof was that it seemed more like a language game, we exploited that ##k## quite commercially.
It's sort of a simple proof, probably to get you used to proving things. But it is an important fact. Simple proofs can seem like a "language game", but you should just tolerate that. There will be much more difficult proofs in the future.

Hall

1. What does it mean for a sequence to converge?

Convergence in a sequence means that as the terms in the sequence increase, they eventually get closer and closer to a specific value, called the limit. In other words, the sequence approaches a specific value as the number of terms increases.

2. How is a subsequence different from a sequence?

A subsequence is a sequence that is obtained by removing some terms from the original sequence. The remaining terms must still be in the same order as they were in the original sequence, but they do not have to be consecutive.

3. Why is it important for all subsequences of a convergent sequence to have the same limit?

If all subsequences of a convergent sequence have the same limit, it means that the sequence is consistently approaching a specific value. This allows us to make accurate predictions and calculations based on the sequence, knowing that it will eventually reach the same limit regardless of which subsequence we choose.

4. Can a sequence have more than one limit?

No, a sequence can only have one limit. If a sequence has more than one limit, it is considered to be divergent rather than convergent.

5. How can we prove that all subsequences of a convergent sequence have the same limit?

To prove that all subsequences of a convergent sequence have the same limit, we can use the definition of convergence and the properties of subsequences. We can also use proof by contradiction, assuming that there are two different limits and showing that it leads to a contradiction.

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