# Question about intervals in theorem 8-1 of Spivak's Calculus

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?

• 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 $j.$ Jun 1, 2022 at 10:13

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 $$\delta_1=\delta_0/2$$.
• You stopped the proof in this moment. Please analyse the remaining part of the proof but with $\delta_1$ instead of $\delta_0$. Jun 1, 2022 at 1:43