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Theorem 8-1: If $f$ is continuous on $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.

Spivak's Proof: For $\varepsilon > 0$ let's say that $f$ is $\varepsilon$-good on $[a,b]$ if there is some $\delta > 0$ such that for all $y,z \in [a,b]$, whenever $|z - y| < \delta$, then $|f(z) - f(y)| < \varepsilon$. Consider any particular $\varepsilon > 0$ and define $A = \{ x : a\le x\le b, f \text{ is } \varepsilon-\text{good on } [a,x] \}$. Then $A \neq \emptyset$, since $a \in A$, and $A$ is bounded above by $b$, so $A$ has supremum $\alpha$. Suppose that $\alpha < b$. Since $f$ is continuous at $\alpha$, there is some $\delta_0 > 0$ such that, if $|y - \alpha| < \delta_0$, then $|f(y) - f(\alpha)| < \varepsilon/2$. Consequently, if $|y - \alpha| < \delta_0$ and $|z - \alpha| < \delta_0$, then $|f(z) - f(y)| < \varepsilon$. So surely $f$ is $\varepsilon$-good on $[\alpha - \delta_0, \alpha + \delta_0]$. (proof continues here)

My issue is with the last line. In particular, Spivak has bounds $\alpha - \delta_0 < y < \alpha + \delta_0$ and $\alpha - \delta_0 < z < \alpha + \delta_0$. Yet, in the definition of $f$ being $\varepsilon$-good, it has to be true for all $y,z \in [\alpha - \delta_0, \alpha + \delta_0]$. This to me seems like a contradiction, since $y$ nor $z$ are in the endpoints. What am I missing?

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  • $\begingroup$ An alternate method is by contradiction, using the sequential compactness of $[a,b]. $ Suppose $e>0$ and that $x_n,y_n\in [a,b]$ with $|x_n-y_n|<(b-a)/n$ and $|f(x_n)-f(y_n)|\ge e$ for each $n\in\Bbb N$. Take a subsequence $(x_{n_i})_{i\in\Bbb N}$ that converges to $x\in [a,b].$ Then $y_{n_i}\to x$ too. The continuity of $ f$ requires $\lim_{j\to\infty}f(x_{n_j})=f(x)= \lim_{j\to\infty}f(y_{n_j}) .$ . But this is not possible if $ |f(x_{n_j}) -f(y_{n_j})|\ge e$ for every $j.$ $\endgroup$ Jun 1, 2022 at 10:13

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It is a tiny mistake, easy to deal with. Just change the conclusion: $f$ is $\varepsilon$-good on $[\alpha-\delta_1,\alpha+\delta_1]$, where $\delta_1=\delta_0/2$.

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  • $\begingroup$ I don't see how that suffices for the proof. $\endgroup$ May 31, 2022 at 20:49
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    $\begingroup$ You stopped the proof in this moment. Please analyse the remaining part of the proof but with $\delta_1$ instead of $\delta_0$. $\endgroup$
    – Mateo
    Jun 1, 2022 at 1:43

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