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I still don’t get, why continuity on a set does not imply uniform continuity.

If a function $f : S \to R$ is continuous on $S$, it means it is continuous at every point in S and I know that in this case each of the deltas depends not only on the choice of epsilon, but also on the point in S, while with uniform continuity, it should only depend on the choice of epsilon.

But if there is such an delta for every point in $S$, wouldn’t the smallest element in this set of deltas fulfil the requirement for uniform continuity?

José Carlos Santos
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Iceman94
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  • When there is, it does. –  Oct 02 '19 at 14:17
  • If $S$ is infinite then there are infinitely many deltas here, so there may not be a smallest one. (The greatest lower bound would work just as well, but the greatest lower bound of all those deltas may be $0$, and $\delta>0$ is required...) – David C. Ullrich Oct 02 '19 at 14:31
  • Thanks to everyone for helping me out! – Iceman94 Oct 02 '19 at 14:34

1 Answers1

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Because if you have a set $X$ of numbers greater than $0$, there is not necessarily an element of $X$ which is smaller than all other elements of $X$. Take $X=(0,1)$, for instance.

For instance, if $f\colon\mathbb R\longrightarrow\mathbb R$ is defined by $f(x)=x^2$, then, for each $\varepsilon>0$ and each $x\in\mathbb R$, if you take$$\delta_{\varepsilon,x}=\min\left\{1,\frac\varepsilon{1+2\lvert x\rvert}\right\},$$then$$\lvert y-x\rvert<\delta_{\varepsilon,x}\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\varepsilon.$$But, for each fixed $\varepsilon$,$$\{\delta_{\varepsilon,x}\mid x\in\mathbb R\}=(0,\varepsilon]$$and this set has no smallest element.

José Carlos Santos
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