Is "bounded" required for the definition below

Question

If $\ (a_n) \ $is an infinite sequence of real numbers and we let $\ (b_n) \ = sup${$\ a_n,a_{n+1},a_{n+2}...$}$\ \ $, and $\ (c_n)\ = inf${$\ a_n,a_{n+1},a_{n+2}...$}$\ $, don't we have to also state that $\ (a_n) \ $ is bounded (and by this I mean the set {$\ a_n \ $})? If the set {$\ a_n \ $} is bounded, then we know a supremum exists for {$\ a_n:n\in\mathbb{N}^{+} \ $}, thus satisfying $\ b_1 \ $. $\ b_2 \ $ would be the supremum of a smaller set, specifically {$\ a_{n+1}:n\in\mathbb{N}^{+} \ $}. $\ b_3 \ $ would be the supremum of a smaller set, specifically {$\ a_{n+2}:n\in\mathbb{N}^{+} \ $}. This continues, and thus we could say the sequence $\ (b_n) \ $ is monotone decreasing (since $\ b_n \ \ge b_{n+1}$). If we can say the set {$\ a_n \ $} is bounded, we could say that $\ b_{1} \ $is an upper bound for $\ (b_n) \ $ and that $\ c_{1} \ $ is a lower bound for $\ (b_n) \ $ and thus, if $\ (b_n) \ $ is bounded (and monotone decreasing), then $\lim_{n\to \infty}$$\ (b_n) \ $ = inf{$\ b_n \ $}, or the limit superior of $\ (a_n) \ $ = inf{$\ b_n \ $}. If we don't state that $\ (a_n) \ $ is a bounded sequence, doesn't all of this just fall apart? I don't understand how you could even choose $\ b_{1} \ $ or $\ c_{1} \ $if $\ (a_n) \ $ wasn't bounded since {$\ a_n \ $} (the set defining the terms of the sequence) would have no supremum or infimum.

Answer 1

You are right that you need the assumption that $(a_n)$ is bounded in order to conclude that each of $b_n$ and $c_n$ are real numbers.

On the other hand, sometimes one goes through the same logical steps when $(a_n)$ is not bounded, by allowing oneself to use the symbols $-\infty$ and $+\infty$. For instance, if $a_n$ is not bounded above then each $b_n$ is $+\infty$. Similarly if $(a_n)$ is not bounded below then each $c_n$ is $-\infty$. Of course these symbols $-\infty$ and $+\infty$ do not represent real numbers, so if one is in a setting where the sups and infs are supposed to be real numbers then, as you say, the whole thing indeed falls apart if $(a_n)$ is not bounded.