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)}$ $