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

[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:

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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 $\lim s_{n_{k}} = s$.

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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 $\lim t_n = s$?

 

[Office_Shredder]

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]

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, $\lim t_n =s$.

 

[FactChecker]

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.

 

[Hall]

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?

 

[FactChecker]

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]

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

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.

 

[FactChecker]

I think I took "some" for "finite' in this definition (given in wikipedia)

Yes, that was wrong. "some" can be infinite.

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.