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Give f is continuous on the interval [a,b]

Show that given any epsilon >0, we can subdivide the interval [a,b] into smaller pieces, each of size smaller than some delta >0 such that, on each small subinterval the difference between the max and min of the function f is smaller than epsilon/(b-a).

So in this question, we subdivide the interval into infinite subinterval and show that the difference between the max and min of the function f is so extremely small? How can we approach this question?

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There's a lemma in Topology called the Lebesgue Number Lemma and it is:

Let $\mathcal{A}$ be an open covering of a compact space $(X, d)$. Then there is $\delta \gt 0$ such that for each subset $A$ of $X$ having diameter $\lt \delta$, there exists an element of $\mathcal{A}$ containing it.

There's a proof in Munkres'. Let me know if you need the book.

Let's construct a cover of $X = [a,b]$ such that for any subdivision of $X$ into consecutive intervals of length $\delta$, means that the interval is contained in an element of the cover. And we can construct such a cover with the additional property that $f(B) \subset B_{\epsilon}$, for each $B$ in the cover and for any $\epsilon \gt 0$ that we choose.

Proof. Let $\epsilon' \gt 0$, cover $Y = f([a,b])$ by open balls of radius $\epsilon = \frac{\epsilon'}{2(b-a)}$. The inverse images of these open balls form a cover of $X$. Then let $\delta$ be the Lebesgue number of the covering. Then choose any $\delta' \lt \delta$. Then a subdivision of $[a,b]$ into intervals of length $\delta'$ means that for each closed interval $I$ we have $f(I) \subset B_{\epsilon}(y)$ for some $y$. Since $I$ is closed it is compact and so $f$ being continuous takes on a minimum and a maximum on $I$. Because of our clever choice of $\epsilon$ we see that $f_{max} - f_{min} \lt \frac{\epsilon'}{b-a}$. QED

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