Proving every continuous function on closed interval is uniformly continuous by smallest delta

Question

When proving the statement above, I completely understand the textbook proof with convergent subsequences and contradiction, but I wonder: when function is continuous by classic epsilon-delta continuity,for all possible points in the closed interval, there is a delta that satisfy epsilon-delta definition, so if we take the smallest delta, will that exact delta satisfy uniform continuity?

Answer 1

If $f$ is a continuous function from $[a,b]$ into $\Bbb R$ and if, for each $c\in[a,b]$ and each $\varepsilon>0$, you take $\delta_{\varepsilon,c}$ such that$$(\forall x\in[a,b]):|x-c|<\delta_{\varepsilon,c}\implies\left|f(x)-f(c)\right|<\varepsilon,$$then there is no reason why, for a fixed $\varepsilon$, the set $\{\delta_{\varepsilon,c}\mid c\in[a,b]\}$ has a minimum. If it has and if $\delta$ is that minimum, then, yes,$$(\forall x,y\in[a,b]):|x-y|<\delta\implies|f(x)-f(y)|<\varepsilon.$$

Here's an example. Let $f\colon[0,1]\longrightarrow\Bbb R$ be the null function. For each $c\in[0,1]$ and each $\varepsilon>0$, take$$\delta_{\varepsilon,c}=\begin{cases}1&\text{ if }c=0\\c&\text{ if }c>0.\end{cases}$$Then it is clearly true that$$(\forall x\in[0,1]):|x-c|<\delta_{\varepsilon,c}\implies\left|f(x)-f(c)\right|<\varepsilon.$$But, for each $\varepsilon>0$, the set $\{\delta_{\varepsilon,c}\mid c\in[0,1]\}$ is the set $(0,1]$, which has no minimum (and its infimum is $0$).

Answer 2

Your thought is a good one but as Jose's answer indicates, doesn't quite work.

I'd like to add, however, that you can indeed avoid talking about convergent subsequences and contradiction, and do something in the spirit of your suggestion.

Namely, given $\epsilon>0$, choose a $\delta_x$ for each $x$ in your interval $I$, satisfying the definition of continuity for $\epsilon/2$. Define balls $B_x:=B(x,\delta_x)$ of radius $\delta_x$ around $x$, and notice the diameter of $f(B_x)$ is at most $\epsilon$ for each $x$.

Moreover, these balls form a cover of $I$, so by a covering lemma of Lebesgue, there is something called a Lebesgue number $\delta>0$ such that any set of diameter $<\delta$ in $I$ is contained in one of the balls $B_x$. In particular, for $|x_1-x_2|<\delta$, $x_1,x_2\in B_x$ for some $x$, so $f(x_1),f(x_2)\in f(B_x)$. Since $f(B_x)$ has diameter at most $\epsilon$, we obtain $|f(x_1)-f(x_2)|\leq \epsilon$.