When proving that a function which is continuous on a closed interval $[a,b]$ is uniformly continuous, every proof is somewhat involved. Why does the following argument not suffice (or does it)? Since for all $c\in [a,b]$, there exists $\delta_k>0$ such that for all $\epsilon>0$, we have $|f(x)-f(c)|<\epsilon$ whenever $|x-c|<\delta_k$, then we can find the minimum element of the set of all such $\delta_k$ (which we term $\delta$) and thus for all $x,y$, we must have $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$ which is the definition of uniform continuity. I must be missing something...

  • $\begingroup$ That minimum element may not exist. For example, there is no minimum of the set $\{\frac 1n , n \in \mathbb N\}$. Sure, there is an infimum, but that is zero, and $\delta$ must be positive. $\endgroup$ – астон вілла тереса лисбон May 14 at 4:12
  • $\begingroup$ Besides, if this argument worked, then it would work on any domain, so uniform continuity would become equivalent to continuity. $\endgroup$ – астон вілла тереса лисбон May 14 at 4:15
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    $\begingroup$ The reason every proof is "involved" is because when $A\subset \Bbb R,$ the result is valid for all continuous $f:A\to \Bbb R$ IFF $A$ is compact. So we muse use the compactness of $[a,b]$ in the proof. $\endgroup$ – DanielWainfleet May 14 at 5:35
  • $\begingroup$ As pointed by other one can't guarantee a minimum of possibly infinite number of $\delta$. You should see common proof based on Heine Borel which allows us to get a finite set of $\delta$'s which are sufficient and therefore we can take a minimum. The fact that the domain of function is closed and bounded is important here. $\endgroup$ – Paramanand Singh May 14 at 11:31

$\delta_k$ is dependent on $c$. There are uncountably many choices for $c$ in $[a,b]$. How do you know that an uncountable collection of $\delta_k(c)$ attains its minimum ("we can find the minimum element of the set of all such $\delta_k$")? It may merely have an infimum or the limit inferior may fail to exist... ((Unfair example removed.)) or that limit might be $0$.

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