I am wondering if the following statement is true, and if not, why not?
If $l-\epsilon < a_n < l + \epsilon$ for all $\epsilon > 0$, then $a_n = l$.
This is in the context of Cauchy Sequences. I had learned of the Squeeze Theorem in the past in calculus (although never a proof for it), and I presumed that the above statement would be true due to the squeeze theorem. But now with Cauchy Sequences, things seem to get a little more dicey, since we're dealing with epsilon neighborhoods. Can I not just treat $a_n$ as another variable like $l$? And if so, why not? Thanks in advance for the help!
Yes, the statement is true. More generally, the following two results hold:
In fact, (2) follows from (1), since $a > c -\epsilon$ if and only if $c < a + \epsilon$. So, it is enough to prove (1).
Proof. Suppose $a < b + \epsilon$ for all $\epsilon > 0$. We want to show that $a \le b$. For a contradiction, assume $a > b$, i.e. $m: =a- b > 0$. As $m > \frac{m}{2}$, $a > b + \frac{m}{2}$. But this contradicts the hypothesis, $a < b + \frac{m}{2}$. Therefore, $a \le b$ as desired. $\blacksquare$
From (1) and (2) it is clear that $$l - \epsilon < a_n < l + \epsilon\, \,\text{ for all }\epsilon > 0 \implies a_n = l$$