Lemma: Extracting a convergent subsequence

[Hall]

Homework Statement

Given a sequence $(s_n)$
(i) If $t$ is in $R$, then there is a subsequence of $(s_n)$ converging to $t$
if and only if the set $\{n \in N : |s_n − t| \lt \varepsilon\}$ is infinite for all
ε > 0.

Relevant Equations

We shall see.

Let's us look at the first implication (I will post the reverse implication once this proof has been verified). We have to prove if there is a subsequence of $(s_n)$ converging to $t$, then there are infinitely many elements of $(s_n)$ lying within $\epsilon$ of $t$, for any $\varepsilon \gt 0$.

Let the subsequence be $(t_k)$ which converges to $t$. $\lim t_n = t$. So, for any $\epsilon \gt 0$ there is a $N$, such that
$$
k \gt N \implies |t_k - t| \lt \varepsilon$$
that means that there are infinite $t_k ~s$ satisfying $t- \varepsilon \lt t_k \lt t+ \varepsilon$, given that $k \gt N$.

As for every $t_k$ in $(t_k)$ corresponds to a $s_n$ in $(s_n)$, therefore, if there are infinite $t_k$ satisfying
$$
t- \varepsilon \lt t_k \lt t+\varepsilon$$
then there are infinite $s_n$ satisfying $t -\varepsilon \lt s_n \lt t+ \varepsilon$. Thus, the set $\{ n \in N : |s_n -t| \lt \varepsilon\}$ is infinite.

Is my proof correct?

 

[FactChecker]

Homework Statement:: Given a sequence $(s_n)$
(i) If $t$ is in $R$, then there is a subsequence of $(s_n)$ converging to $t$
if and only if the set $\{n \in N : |s_n − t| \lt \varepsilon\}$ is infinite for all
ε > 0.
Relevant Equations:: We shall see.

Let's us look at the first implication (I will post the reverse implication once this proof has been verified). We have to prove if there is a subsequence of $(s_n)$ converging to $t$, then there are infinitely many elements of $(s_n)$ lying within $\epsilon$ of $t$, for any $\varepsilon \gt 0$.

This is wrong in a subtle way. Look carefully at the statement of the problem. There might not be infinitely many distinct elements $s_n$, they might be the same $s_i$ repeated over and over. But there are an infinite number of indices, n. That is different.

 

[Hall]

Now, the backward implication:

Given that the set $\{n \in N : |s_n -t| \lt \varepsilon\}$ is infinite for all $\epsilon$, we have to prove that there exits a subsequence which converges to $t$.

Proof: Take $t_1 \in \left( t-1, t+1\right)$, and if $t_k \in \left( t- 1/k, t+ 1/k\right)$ then $t_{k+1} \in \left(t-1/(k+1), t+1/(k+1\right)$. The subsequence $(t_k)$, thus defined, gets closer and closer to $t$ and hence
we can write
$$
\lim t_n = t
$$

 

[FactChecker]

Now, the backward implication:

Given that the set $\{n \in N : |s_n -t| \lt \varepsilon\}$ is infinite for all $\epsilon$, we have to prove that there exits a subsequence which converges to $t$.

Proof: Take $t_1 \in \left( t-1, t+1\right)$, and if $t_k \in \left( t- 1/k, t+ 1/k\right)$ then $t_{k+1} \in \left(t-1/(k+1), t+1/(k+1\right)$. The subsequence $(t_k)$, thus defined, gets closer and closer to $t$ and hence
we can write
$$
\lim t_n = t
$$

You have to prove that these $t_k$s exist.
You have not addressed the issue that I mentioned in your first "proof". The sequence 1,1,1,1,1,1,1,... converges to 1, but there is only a single distinct element in the series. You should fix your first proof. You should also look carefully at your definition of a limit to see if it guarantees an infinite number of something within an $\epsilon$ distance or only one. In your proof, you should state what your definition of a limit guarantees.

You might think that I am being picky, but the whole purpose of you doing these proofs is for you to learn to be picky about the details.

 

[vela]

This is wrong in a subtle way. Look carefully at the statement of the problem. There might not be infinitely many distinct elements $s_n$, they might be the same $s_i$ repeated over and over. But there are an infinite number of indices, n. That is different.

I'm not seeing how your example constitutes a subsequence of $\{s_n\}$. In other words, $\{s_1, s_1, s_1, \dots\}$ is not a subsequence of $\{s_1, s_2, s_3, \dots\}$.

 

[FactChecker]

I'm not seeing how your example constitutes a subsequence of $\{s_n\}$. In other words, $\{s_1, s_1, s_1, \dots\}$ is not a subsequence of $\{s_1, s_2, s_3, \dots\}$.

If $s_i=s_1, \forall i$, then there is only one element but there are infinite indices. That is why the problem statement talks about the indices rather than the elements.

 

[vela]

Hmm, perhaps your point is too subtle for me. I get that @Hall didn't actually prove the set given in the problem is infinite, but I still don't see what the possible non-uniqueness of the elements of the sequence has to do with it.

 

[FactChecker]

Hmm, perhaps your point is too subtle for me. I get that @Hall didn't actually prove the set given in the problem is infinite, but I still don't see what the possible non-uniqueness of the elements of the sequence has to do with it.

With the problem as stated, there is no issue. It asks for an infinite number of indices. Unfortunately, the proof proposed in the post changed the statement to say there were an infinite number of elements. That is wrong. There might only be one element, repeated an infinite number of times. So the set only has one element. I think my point was missed.

 

[vela]

I guess you're saying that the sequence (1, 1, 1, 1,…) has only one element whereas I would consider it to have an infinite number of elements because the order matters.

 

[FactChecker]

I guess you're saying that the sequence (1, 1, 1, 1,…) has only one element whereas I would consider it to have an infinite number of elements because the order matters.

I would say that the set has only one element, no matter how many times it appears. That is why the statement of the problem talks about the number of indices, not the number of elements.

 

[Hall]

@FactChecker This will open a pdf file.

 

[FactChecker]

@FactChecker This will open a pdf file.

I think that it is essentially correct. I think that the indices can probably be handled in a simpler manner.