Prove that the set of all almost upper bounds is nonempty, and bounded below. 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 $\limsup A$. ...

 A: You are asking: "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?"
Yes, it is, but the problem is possibly here. When the Author says: "No lower bound for $A$ can possibly be an almost lower bound (since $A$ is infinite), [...]", he might like to say "No lower bound for $A$ can possibly be an almost upper bound for $\boldsymbol{A}$ (since $A$ is infinite)". That's because there are infinitely many elements of $A$ that are greater than whatever lower bound of $A$ you may choose and not finitely many as are needed by an element (in this case incidentally a lower bound of $A$) to be an almost upper bound of $A$. It follows that every almost upper bound of $A$ is strictly grater than any lower bound of $A$, that is the set of almost upper bounds of $A$ is bounded below.
A: Infimum of $B$ is not the supremum of $A$ as you might think. Limit superior is almost always presented in a complicated fashion in many textbooks and most attempts to present this in a simple manner don't work. Plus there is an overhead of temporary terminology like "almost upper bound".
You should try to understand the following important properties of limit superior which all presentations try to derive using their approaches:


*

*It is defined only for infinite sets and has finite value only if the set is bounded. Let $A$ be such a set and $M=\limsup A$. 

*If $M'>M$ then only a finite number of members of set $A$ are greater than or equal to $M'$. In terms of Spivak, $M'$ is an almost upper bound of $A$. 

*If $m'<M$ then infinitely many members of $A$ exceed $m'$. 



Spivak has tried to collect all numbers like $M'$ into another set $B$ which is bounded below. This is the part where you have an issue. First note that if $x\in B$ then any number greater than $x$ is also in $B$. Thus if $B$ is not bounded below then $B=\mathbb{R} $. And this is absurd if $A$ is a bounded infinite set with some lower bound $c$. In this case you can see that infinitely many members of $A$ are greater than or equal to $c$ and hence $c\notin B$ which contradicts $B=\mathbb {R} $. It is thus clear that $B$ is bounded below.

Another common approach which you may want to compare is the following. First we define accumulation point (which unlike "almost upper bound" is not temporary):

Let $A$ be a non-empty set of real numbers. A number $p$ is said to be an accumulation point or a limit point of $A$ if for every positive number $\epsilon$ there is an element $x\in A$ such that $x\neq p$ and $x\in(p-\epsilon, p+\epsilon) $.

If $A$ is an infinite bounded set then there is a theorem by the name of Bolzano-Weierstrass which guarantees an accumulation point of $A$. The set of all accumulation points of $A$ is called the derived set of $A$ and denote by $A'$. It can be proved that this set $A'$ has a maximum and a minimum. The maximum element of $A'$ is called limit superior of $A$ and minimum element of $A'$ is called limit inferior of $A$. 
A: Let $c=\inf A$ and let $d\in (-\infty,c]$.  Then $d$ is a lower bound for $A.$ So $B_d= A$ \ $\{d\}$  is an infinite subset of $A$ and every member of $B_d$ is greater than $d$, so $d$ is not an almost-upper bound for $A .$ Hence any almost-upper bound for $A$ must belong to $(c,\infty).$ So $c$ is a lower bound for the set of almost-upper bounds for $A.$
Examples: (1). Let $A=\{1/n: n\in \mathbb N\}.$ Then $\inf A=0,$ and $x$ is an almost-upper bound for $A$ iff $x>0.$
(2). Let $A=\{1-1/n:n\in \mathbb N\}.$ Then $\inf A=0$, but $x$ is an almost-upper bound for $A$ iff $x\geq 1.$
A: It's easy to see that an upper bound is an  almost upper bound, so that $B$ is not bounded.
Suppose $x \in B$, and let $y$ be a lower bound of $A$, existing as $A$ is bounded. Then, there exist only finitely many elements in $A$ such that $x$ is smaller than all these elements, while $y$ is smaller than all the elements.
Since $A$ is infinite, $x$ is definitely not smaller than all elements of $A$. So let it be greater than some $p \in A$, say. But then, since $y$ is a lower bound of $A$, $y \leq p$. Combining these statements, $x \geq y$ follows.
The importance of infiniteness of $A$ is emphasized, for example if $A = \{ 5\}$, then every real number is an almost upper and almost lower bound for $A$, since $A$ is finite, so that any real number can possibly fail only finitely many comparisons, which is neglected by the above definition.
The set of almost upper bounds is then bounded below. That bound will probably be called the lowest almost upper bound, similar to how the $\limsup$ and $\liminf$ definitions operate.
A: $A$ is bounded, and any upperbound $b$ for $A$ (which exists) is in $B$ as well, showing $B \neq \emptyset$.
If $L = \inf A$, then pick any $b \in B$. There are finitely many $a \in A$ such that $b < a$ and infinitely many (the remaining ones) $a \in A$ such that $L \le a \le b$ ($L$ is a lower bound for $A$). Hence $L$ is a lower bound for $B$, as claimed. 
