I'm trying to show that if we have a continuous function $f:[a,b]\to\mathbb{R}$ with $f(x)>0$ on it's entire domain, then there exists an $\epsilon>0$ such that $f(x)\geq\epsilon$ for all $x\in[a,b]$.
I've figured that f being continuous on a closed interval means it's uniformly continuous, and this must have something to do with it, because I can think of functions such as $e^x$ which if defined from $\mathbb{R}\to\mathbb{R}$ do not satisfy this condition (since it's not uniformly continuous).
I'm struggling with the rigour and how to write this up in particular - any ideas please?
Suppose there is no such $\epsilon$. Then for every $n \geq 1$ we can find $x_n \in [a,b]$ such that $f(x_n) < 1/n$. The sequence $x_n$ is bounded and hence by Bolzano-Weierstrass has a convergent subsequence $y_n$, whose limit $y$ lies in $[a,b]$ since $[a,b]$ is closed. Then by continuity of $f$:
$$ f(y) = f(\lim _{n\to \infty}y_n) = \lim_{n\to \infty} f(y_n) = 0 \,,$$ which is a contradiction.