I need to prove that a certain function is continuous (with an epsilon-delta proof). I want to restrict my $\delta$ to be $\delta \leq \frac{1}{2}$ (bigger than $\frac{1}{2}$ complicates things). If I were to choose $\delta = \epsilon$ I believe my proof wouldn't be valid for all $\epsilon$ since I'm restricting $\delta$ and thereby $\epsilon$ to be smaller than or equal to $\frac{1}{2}$.

How can I solve this?

Can I just say "choose $\delta \leq \epsilon$" or do I really need an equal sign in between those two?

If so, could I solve this by saying "choose $\delta = \min\{\frac{1}{2},\epsilon\}$? In this case, do I have to write out two proofs? One for $\delta = \frac{1}{2}$ and one for $\delta = \epsilon$?


1 Answer 1


You can choose $\delta$ however you want. The definition begins "For all $\varepsilon > 0$, there exists a $\delta > 0$ such that ..."

This means you need to let $\varepsilon$ be arbitrary. But once you fix your $\varepsilon$, all you need to do is find one specific $\delta$ that works for it. And $\delta$ is allowed to depend on $\varepsilon$. So both $\delta = \varepsilon$ and $\delta \le \varepsilon$ are allowed.

If you really need your $\delta$ to be no larger than $1/2$, then the $\min\{1/2, \varepsilon\}$ idea you suggested will work fine. And if you carefully write your proof then I think you shouldn't need to do two separate cases. To be sure I'd need to see your proof or at least what you're trying to prove.

  • $\begingroup$ The problem is if I fix $\epsilon > \frac{1}{2}$ I can't have $\delta = \epsilon$ since I don't want $\delta > \frac{1}{2}$. But choosing $\delta \leq \epsilon$ and smaller than $\frac{1}{2}$ solves that. Thanks! $\endgroup$
    – JuliusC
    Sep 19, 2016 at 0:57
  • $\begingroup$ @JuliusC Although we say "for all $\varepsilon > 0$" the truth is we really only care about really teeny tiny positive values of $\varepsilon$. In other words we only care about positive values of $\varepsilon$ that are arbitrarily close to zero. So technically it's ok to assume $\varepsilon < 1/2$ in general, but make sure it's ok in your class before you do it there. $\endgroup$
    – user307169
    Sep 19, 2016 at 2:45

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