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!

If $\{ x \mid f(x) \ge c\} \neq \emptyset$ then its infimum exists, since it is a bounded-below subset of $R$.
Note that $x=0$ is a lower bound for the set. 
