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This is question 5.3.18 from Bartle's Introduction to Real Analysis:

Let $I = [a, b]$ and $f: I \to \mathbb{R}$ s.t.
$\forall x \in I$ $\exists \delta_x, M_x > 0$ s.t. $\forall u \in V_{\delta_x}(x)$ $|f(u)| \le M_x$
Show $\exists M > 0$ s.t. $\forall x \in I$ $|f(x)| \le M$

I am very confused. Chapter 5.3 is about continuity but $f$ is not assumed to be continuous. Where should I start?

Alex Vong
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2 Answers2

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We have that

$\bigcup_{x \in [a,b]}V_{\delta_x}(x) \supseteq [a,b]$. Since $[a,b]$ is compact, there are $x_1,...,x_n \in [a,b]$ such that $\bigcup_{j=1}^nV_{\delta_{x_j}}(x_j) \supseteq [a,b]$.

Take $M=\max\{M_{x_1},...,M_{x_n}\}$

Fred
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  • Thanks for the short answer! But the concept of compactness is not taught in my introductory analysis course. Do you know of a solution that doesn't uses compactness? – Alex Vong Dec 13 '16 at 06:27
  • It seems compactness of closed bounded intervals can be proved using only nested interval theorem. So I think there is no problem now. – Alex Vong Dec 13 '16 at 06:32
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Without compactness: suppose that $f$ is unbounded. Let $I=[a,b]$

For each $n \in \mathbb N$ there is $x_n \in I$ such that

(*) $|f(x_n)| >n.$

Bolzano and Weierstrass say: there is a convergent subsequence $(x_{n_{k}})$

Without loss of generality we can assume that $(x_n)$ is convergent. Let $x_0$ be the limit of $(x_n)$. We have $x_0 \in I$

Then there is $N \in \mathbb N$ such that $x_n \in V_{\delta_{x_0}}(x_0)$ for $n>N$. Hence

$|f(x_n)| \le M_{x_0} $ for $n>N$. From (*) we get the contradiction

$n< M_{x_0}$ for $n>N$.

Fred
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