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!
 A: Let $c = f(1).$ Then it's easy to see that $c > 0, \;$ since $\; 0 \leq f(0) < f(1).$
Since $\{x : \; f(x) \geq c\} \; \neq \; \emptyset \;$ (the number $1$ belongs to this set), it follows that $\{x : \; f(x) \geq c\} \;$ is a nonempty set that is bounded below (by $0),$ and hence by the least upper bound property of the real numbers we have that $\;\inf \{x : \; f(x) \geq c\}$ exists.
A: 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. 
A: 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 $c>0$.
A: The domain of the set is $[0, \infty)$ Thus $0 \le x$ for all $x \in [0,\infty)$.  There is nothing lower.  So $0$ is a lower bound of all subsets of $[0, \infty)$ and every subset of $[0, \infty)$ is bounded below.
Showing that $K \subset [0, \infty)$ is bounded below, is simply not an issue.  The only thing we have to show prove $\inf K$ existis is that $K$ is non-empty.  That is all.
To show that the must exist some $c > 0$ so that $\inf  \{x| f(x) \ge c\}$ we must simply prove there is some $c > 0$ so that $ \{x| f(x) \ge c\}$ is not empty.  Simply pick any $c= f(a) > 0$. As $f$ is strictly increasing then for $a > 0$ we know $f(a) > f(0) \ge 0$.  And for all $b \ge a; f(b) \ge f(a) = c$.  So $\{x|f(x) \ge c\} = [a, \infty)$ and $\inf  \{x| f(x) \ge c\} = a$.
======= oops; you wrote for some $c$; not for all $c$.  Nevermind the rest of this post =====
Which... I don't think we can.  Let $f(x)$ be assymptotic to $d < c$ and $\{x| f(x) \ge c\}$ is empty and $\inf \{x| f(x) \ge c\}$ does not exist.[$*$]
....
Also you claim that $0 \not \in \{x| f(x) \ge c\}$.  There is no reason to assume that at all.  It is perfectly possible for $f(0) \ge c$.[$**$]
[$*$] Example $f(x) = 1 - \frac 1{x+1}; c= 2$ then $ \{x| f(x) \ge c\}= \{x| \frac 1{x+1} \ge 2; x \ge 0\} = \{x| 1 \ge 2x + 2; x\ge 0\}=\{x| x \le -\frac 12 and x \ge 0\}= \emptyset$.
[$**$] Example $f(x) = 256 + x; c = 59$ then $f(x) = 256 > 59$ so $0 \in \{x| f(x) \ge c\}=\{x| 256 + x \ge 59; x \ge 0\} = \{x|x \ge - 197 and x \ge 0\} = \{x| x \ge 0\} = [0 , \infty)$.
