Q:

If $f:[0,1) \to \mathbb{R}$ is continuous and $f(x) > 0$ for all $x$, then there exists $c > 0$ such that $f(x)\geq c$ for all $x$.

If $f:[0,1) \to \mathbb{R}$ 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.

A:

Note that if $y \in [0,1[$, then there is some $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x) - f(y)| < f(y)/2$, implying that $f(x) > f(y)/2$; hence $f(x) > 2^{-1}\inf f[0,1[$ for all $y \in [0,1[$. 
A problem is that $\inf f[0,1[$ can be $=0$, as shown by the example that Thomas provided; hence you know the statement under consideration is not true. But it may be interesting to note that, if we impose a further condition on $f$, say $\inf f[0,1[ > 0$, then the conclusion follows.

