Given least upper bound $\alpha$ for $\{\ f(x) : x \in [a,b] \ \}$, $\forall \epsilon > 0 \ \exists x$ s.t. $\alpha - f(x) < \epsilon$ I can't figure out how all of this follows. Taken from Ch.8 of Spivak's Calculus.

If $\alpha$ is the least upper bound of  $\{\ f(x) : x \in [a,b] \ \}$ then, $$\forall \epsilon > 0 \ \exists x\in [a,b] \ \ \ \ \ \ \ \alpha - f(x) < \epsilon$$ 
  This, in turn, means that 
  $$ \frac{1}{\epsilon} < \frac{1}{\alpha - f(x)}$$

 A: By definition of a least upper bound $\alpha$, any smaller value than $a$ is not an upper bound. So take $\epsilon>0$. $\alpha-\epsilon$ is not an upper bound of $S=\{\ f(x) : x \in [a,b] \ \}$. Which means that it exists $x \in [a,b]$ such that $f(x)>\alpha -\epsilon$ or $\alpha -\epsilon <f(x)$ as desired.
A: The definition of $\alpha $ is the least upperbound of $\{f(x)| x\in [a,b]\}$.

1) $\alpha$ is an upper bound of $\{f(x)| x\in [a,b]\}$

That means for all $x \in [a,b]$, $\alpha \ge f(x)$.

2) If $y < \alpha$ then $y$ is not an upper bound of $\{f(x)| x\in [a,b]\}$

Alternatively that means.

2') If $y < \alpha$ then there is an $x\in [a,b]$ so that $y < f(x)$.

Now for every $\epsilon > 0$ then $\alpha - \epsilon < \alpha$ so by 2') it follows that:
For every $\epsilon > 0$ the $\alpha - \epsilon < \alpha$ and so there is an $x\in[a,b]$ so that $\alpha - \epsilon < f(x) \le \alpha$ which in turn means:
$\alpha - f(x) < \epsilon$.
Now $\alpha \ge f(x)$ so if $\alpha - f(x) \ge 0$.  and if $f(x)\ne 0$ then 
$\alpha - f(x) < \epsilon  \implies \frac 1{\alpha - f(x)} > \frac 1{\epsilon}$.
However your book made an error.  If $f(x) = \alpha$ this is not true.
... 
A counter example would be $f(x) =x$ if $x < b$ but $f(x)= b+1$ if $x \ge b$.
Then $\{f(x)|x \in [a,b]\} = [a,b)\cup \{b+1\}$ and $\alpha = b+1$. 
For any $\epsilon: 0 < \epsilon < 1$ we have $b\in [a,b]$ and $b-\epsilon < f(b) \le b+1$ but $b$ is the only possible $x \in [a,b]$ where $b < b-\epsilon < f(x) \le b+1$ (because if $x < b$ then $f(x) = x < b$).
So it is true that there is an $x \in [a,b]$ where $(b+1) - f(x) < \epsilon$ but there isn't any $x \in [a,b]$ where $\frac{1}{(b+1) - f(x)} > \frac 1{\epsilon}$ because $\frac{1}{(b+1) - f(x)}  = \frac{1}{(b+1) - f(b)} =\frac 10$ is not defined.
