I've taught how to use $\epsilon, \delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.

Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.

Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?

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    $\begingroup$ A linear function, perhaps? $\endgroup$ – paw88789 Mar 2 '19 at 23:10
  • $\begingroup$ @paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that? $\endgroup$ – Alec Mar 2 '19 at 23:18
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    $\begingroup$ @Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about? $\endgroup$ – Steven Gubkin Mar 3 '19 at 14:45
  • $\begingroup$ @StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases. $\endgroup$ – Alec Mar 3 '19 at 16:27

I think this cannot be understood without a contrasting example where it fails. So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = \frac{1}{x}$ over the open interval $(0,1)$. It is continuous over that interval, but not uniformly continuous. Fix an $\epsilon > 0$; then for any $\delta > 0$ one can arrange the difference in $f$-values to exceed $\epsilon$ by getting close enough to $x=0$.

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    $\begingroup$ Also show $1/x$ is uniformly continuous on $[1,\infty)$, or $[a,\infty)$ for $a>0$. $\endgroup$ – KCd Mar 7 '19 at 3:32

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