exisitance of inf Let $f:[0, \infty] \rightarrow [0, \infty]$ be a strictly increasing (therefore one-to-one) but not onto function, with $f(0)=0$. For some $c >0$, I want to prove that $\inf\{ x \mid f(x) \ge c\}$ always exists. How do I prove this? We know that 0 is not in the set $\{ x \mid f(x) \ge c\}$ but how can we show that 0 is lower bound of the set? Thanks very much!

As you note, $0$ is not in the set $\{x | f(x) \geq c\}$. This means that $0$ as a lower bound (not necessarily the greatest lower bound, but that's not a problem.
An important property of the real numbers is the "completeness" property. This means that if a set is bounded below (or above), then there exists an inf (or sup) for the set. Since your set is bounded below, an inf will always exist.
Edit: To show that $0$ is not in the set, we note that $f$ strictly increasing implies that $f(x) > 0$ for any $x > 0$. But because the domain of $f$ is $[0,\infty)$, $f(0) \geq 0$. This means that $f(c) > 0$ for any $c>0$.
