Find all almost lower bounds and almost upper bounds of $\{\frac 1n: n\in \Bbb N\}$ 
A number $x$ is called an almost upper bound for $A$ if there are only finitely many numbers $y$ in $A$ with $y\ge x$. An almost lower bound is defined similarly.
(a) Find all almost lower bounds and almost upper bounds of the set
  $A=\{\frac 1n: n\in \Bbb N\}$
(Source: Problem 18 in chapter 8 of Spivak's Calculus)

My answer: Denote almost upper bound for A by $Al.sup(A)$ and similarly with almost lower bound by $Al.inf(A)$. The answer is: $x=Al.sup(A)=1$ since it is the ONLY number that has a finite $y$ $\in$ A (only one value: 1). $Al.inf(A)=1$ for the same reason.
Spivak's answer book: All $\alpha$ $\gt$ $0$ for $Al.sup(A)$ and all $\alpha$ $\le$ $0$ for $Al.inf(A)$
My question: What is wrong with my answer? I do not understand anything at all. What exactly again is the $Al.sup(A)$ and $Al.inf(A)$ Spivak wants to talk about?
I thank you very much for having helped me.
 A: Just go ahead and verify Spivak's answer. Let $\alpha >0$. How many element of the form $1/n$ satisfy $1/n > \alpha $? The answer is only finitely many and thus $\alpha $ is an almost upper bound. Now take any $\alpha \le 0$. How many elements of the form $1/n$ satisfy $1/n>\alpha $? The answer is all of them, so not just finitely many and thus $\alpha $ is not an almost upper bound. This shows that $\alpha $ is an almost upper bound if, and only if, $\alpha >0$. You can argue similarly for the answer to the almost lower bound. 
A: For a concrete example, consider $\beta = 0.21 > 0$. Notice that $\beta$ is an almost upper bound for $A$ because there are only finitely many elements in $A$ that are greater than or equal to $\beta = 0.21$. Indeed, the only four elements of $A$ that prevent $\beta$ from being a true upper bound for $A$ are:


*

*$\frac{1}{1} = 1 \geq 0.21$

*$\frac{1}{2} = 0.5 \geq 0.21$

*$\frac{1}{3} = 0.\overline 3 \geq 0.21$

*$\frac{1}{4} = 0.25 \geq 0.21$


All other elements of $A$ are smaller than $0.21$.
