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So, here's the problem:

Let $\{a_n\}_{n \in \mathbb{N}}$ be a sequence of real numbers such that all the terms of the sequence belong to the interval $[4,9)$. Then, prove or disprove the assertion that there exists a convergent subsequence $\{b_n\}$ such that $\lim_{n \to \infty}b_n \geq 4$.


Proof Attempt:

I claim that there exist no convergent subsequences that have limit strictly less than 4. Since this is a bounded sequence of real numbers, it will have a convergent subsequence and that subsequence must have limit greater than or equal to 4.

To prove this, suppose that all convergent subsequences must have limit strictly less than 4. We will pick one of them and say that the limit is $c$. Then, consider an $\epsilon$ neighbourhood of $c$ such that $c+\epsilon < 4$.

We can certainly define this because, for instance, $\epsilon = \frac{4-c}{2}$. Then, this neighbourhood of $c$ must contain infinitely many terms of the subsequence. In other words, there are terms of the original sequence that are outside of the given interval. This is a contradiction.

It follows that such a convergent subsequence cannot exist.

Does the proof above work? If it doesn't, then why? How can I fix it?

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  • $\begingroup$ How do you know that the sequence has any convergent subsequences at all? (It does, of course, but I don't know which tools are available to you to know that.) $\endgroup$ – Robert Shore Jun 25 '20 at 10:11
  • $\begingroup$ Any bounded sequence has a convergent subsequence. That’s the statement of the Bolzano-Weierstrass Theorem. In this case, our given sequence is bounded above by 9 and below by 4. $\endgroup$ – Abhi Jun 25 '20 at 10:12
  • $\begingroup$ That's fine. You should just start your proof by noting that Bolzano-Weierstrass guarantees the existence of a convergent subsequence so all that's left to determine is whether the limit of that subsequence can be less than 4. $\endgroup$ – Robert Shore Jun 25 '20 at 10:14
  • $\begingroup$ Yeap, will take note of that. But I mean, if I am writing these sorts of proofs, then would I need to explicitly mention the name of the theorem every single time? Could I not just make the assertion without ever explicitly invoking the name? $\endgroup$ – Abhi Jun 25 '20 at 10:20
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    $\begingroup$ Wow I haven't even begun undergraduate.......I'm going to try and follow that practise of including details in every step of my argument :D Thank you so much for your assistance and input. $\endgroup$ – Abhi Jun 25 '20 at 20:18
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I claim that there exist no convergent subsequences that have limit strictly less than 4.

To prove this, suppose that all convergent subsequences must have limit strictly less than 4.


That is not how you would argue by way of contradiction. Because what you then have shown is that not all convergent subsequences have limit strictly less than $4$. However, one of them could.

Although, your proof then actually works with only an arbitrary such subsequence, so just modifying your statement to

To prove this, suppose that there exists a convergent subsequence with limit strictly less than 4.

will do the job.

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  • $\begingroup$ Ahhh okay so how I started the argument was the issue but everything else works, right? Thank you so much. $\endgroup$ – Abhi Jun 25 '20 at 8:17
  • $\begingroup$ Correct. If I were being a more pedantic corrector, I might also demand that you state that the choice of $\epsilon = \dfrac{4-c}{2}$ is indeed strictly greater than $0$ but your proof is correct. $\endgroup$ – Aryaman Maithani Jun 25 '20 at 8:20
  • $\begingroup$ Alright thank you so much, I will include an argument for why that's strictly greater than 0. I'm sort of still learning how to construct these proofs properly so I make silly mistakes like the ones you mentioned. Once again, thanks for your input $\endgroup$ – Abhi Jun 25 '20 at 8:22
  • $\begingroup$ As an aside, if you have seen that every bounded sequence has a convergent subsequence, you might've also seen that if $b_n \ge c$ for all $n$, then $\lim_n b_n \ge c$. You could've directly used that? (You have indeed just written a proof of that statement.) $\endgroup$ – Aryaman Maithani Jun 25 '20 at 8:22
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    $\begingroup$ Ah well, that certainly sounds easier. I'm sort of trying to prove most of these statements straight from the definition because I'm getting a bit more practise for my Analysis course that's going to begin soon. $\endgroup$ – Abhi Jun 25 '20 at 8:27

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