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?
Best Answer
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
will do the job.