Let $f : [a,b] \rightarrow \mathbb{R}$ be continuous and $f(x) > 0$ for all $x \in [a,b]$. Show that there exists $\delta > 0$ such that $f(x) \geq \delta$ for all $x \in [a,b]$.
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Are you allowed to use that a continuous function $f$ attains minimum value on a closed interval? – Paramanand Singh Oct 29 '18 at 13:18
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If no such $\delta$ exists then there is a sequence $(x_n)$ in $[a,b]$ such that $0 < f(x_n) < 1/n$ and thus $f(x_n) \to 0$.
The sequence $(x_n)$ is bounded and by Bolzano-Weierstrass has a convergent subsequence $(x_{n_k})$. Since $[a,b]$ is closed there is a point $c \in [a,b]$ such that $x_{n_k} \to c$.
Can you finish by showing that $0 = \lim_{k \to \infty}f(x_{n_k}) = f(c)$ -- a contradiction?
RRL
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