# Proof that a continuous function on a closed interval is uniformly continuous

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...

• 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. – астон вілла тереса лисбон May 14 at 4:12
• Besides, if this argument worked, then it would work on any domain, so uniform continuity would become equivalent to continuity. – астон вілла тереса лисбон May 14 at 4:15
• 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. – DanielWainfleet May 14 at 5:35
• 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. – 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 $$0$$.