I don't understand the solution to problem 18. (b) from chapter 8 of Spivak's Calculus:
18. A number $x$ is called an almost upper bound for $A$ if there are only finitely many numbers $y$ in $A$ with $y \geq x$. An almost lower bound is defined similarly.
(b) Suppose that $A$ is a bounded infinite set. Prove that the set $B$ of all almost upper bounds of $A$ is nonempty, and bounded below.
Solution Every upper bound for $A$ is surely an almost upper bound, so $B \neq \emptyset$. No lower bound for $A$ can possibly be an almost lower bound (since $A$ is infinite), so $B$ is bounded below by any lower bound for $A$.
I understand the first sentence. An upper bound $b$ for $A$ is an element such that $b \geq x$ for all $x$ in $A$, which satisfies the definition of an almost upper bound because there are only finitely many numbers $y$ in $A$ with $y \geq b$, namely zero (EDIT: I should have written one or zero here?).
But why can't a lower bound $c$ for $A$ be an almost lower bound and how does it follow that $B$ is bounded below by any lower bound for $A$?
If $c$ is a lower bound for $A$ then isn't it also an almost lower bound for $A$ by an argument similar to the one used for showing that every upper bound is also an almost upper bound?
Intuitively I would suspect that the infimum of $B$ is the supremum of $A$, which makes every lower bound for $A$ a lower bound for $B$.
It seems important to understand Spivak's solution clearly because in problem (c) he defines the limit superior of $A$ in terms of the infimum of $B$:
(c) It follows from part (b) that $\inf B$ exists; this number is called the limit superior of $A$, and denoted by $\limsup A$. ...