If $f:[0,\infty) \to \mathbb{R}$ is continuous and $f(x) > 0$ for all $x$, then there exists $c > 0$ such that $f(x)\geq c$ for all $x$.

Question

If $f:[0,\infty) \to \mathbb{R}$ is continuous and $f(x) > 0$ for all $x$, then there exists $c > 0$ such that $f(x)\geq c$ for all $x$.

I feel that this is a false statement given the order of the quantifiers (since this implies there exists one $c$ that works for all $x$), but I'm having difficulty coming up with a counterexample.

Edit:

Using Thomas' suggestion, would it suffice to show that since $f(x)\geq c$, $$\lim_{x\to\infty} f(x)=0>c\geq0$$,

and thus c=0?

Answer 1

Given any $c>0$ and assuming it is known that $e^{-x}$ satisfies $lim_{x\rightarrow \infty}e^{-x}=0$, there is by definition of the limit a real number $R>0$ sucht that $|e^{-x}| < \frac{c}{2}$ for every $x>R$. This shows, directly, that no real number $c>0$ with $f(x)\ge c$ can exist.