0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

$\delta_k$ 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 $0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.