I realize this may be a simple question, but I am having trouble proving proving this, mostly with selecting a suitable delta.

My attempt so far:

$x^2$ is continuous on $[0,1]$ if $\forall\epsilon>0,\exists\delta>0$ such that $$0<\vert x-a\vert<\delta\,x\in [0,1]\Rightarrow\vert x^2-a^2\vert<\epsilon$$ for all $a\in [0,1]$.

Since $a$ and $x$ are in $[0,1]$, the maximum value of $a+x$ is 2. Take $\delta=\min\{1,\frac{\epsilon}{2}\}$. Then $$0<\vert x-a\vert<\delta\Rightarrow\vert x^2-a^2\vert=\vert x-a\vert\vert x+a\vert<2\vert x-a\vert=2\delta\leq\epsilon$$

I think my logic makes sense but I am unsure about my step involving $\delta$. Is this a correct proof? If not, how would I improve it?

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    $\begingroup$ Looks good to me! Now I have a question for you: why take $\delta$ to be the minimum of $1$ and $\epsilon/2$? Why not just let it be $\epsilon/2$? $\endgroup$ – layman Oct 23 '15 at 4:57
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    $\begingroup$ In the end you need to get a strict inequality according to the definition So you might want to make $\delta = \epsilon /4 $ $\endgroup$ – happymath Oct 23 '15 at 4:58
  • $\begingroup$ This proof is perfect and is exactly how a continuity proof is supposed to go! My only thought is as $x^2$ is continuous on unbound R, can you do this without using the max value of a + x? $\endgroup$ – fleablood Oct 23 '15 at 5:04
  • $\begingroup$ Oooh, the "less than 2*...." should be a less than or equal. Very minor but as it need to be a strict inequality.... $\endgroup$ – fleablood Oct 23 '15 at 5:11

The only improvement I can see here (as mentioned by happymath in the comments), is that the implication requires strict inequality. So, choosing something like $\delta=\frac{\epsilon}{3}$ yields a better result. This gives the implication:

$$ 2\delta=\frac{2\epsilon}{3}<\epsilon$$

This is the required condition. Of course, as you probably know, $\delta=\frac{\epsilon}{3}$ is a somewhat arbitrary choice. You can choose anything less than $\frac{\epsilon}{2}$ (i.e. $\delta=\frac{\epsilon}{2.0000000001}$, or $\delta=\frac{\epsilon}{123456789}$ ). Or, more generally, as proposed in the comments, consider $\delta<\min(1,\frac{\epsilon}{2})$.

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  • $\begingroup$ Or he could simply say, "let $\delta <\min(1 , \epsilon/2)$" $\endgroup$ – fleablood Oct 23 '15 at 5:07
  • $\begingroup$ It doesn't hurt that $\delta$ is chosen as the minimum of $1$ and $\epsilon/2$, but how does it help in this case? Why not just take $\delta = \epsilon/2$? $\endgroup$ – layman Oct 23 '15 at 5:13
  • $\begingroup$ @fleablood Also valid. I'm just trying to underscore the somewhat arbitrary nature of the choice here. $\endgroup$ – Alekos Robotis Oct 23 '15 at 5:14
  • $\begingroup$ It is arbitrary. But as it's arbitrary it needn't actually be specified. As you point out is be be anything less (very true and a good point; we're basically saying the same thing) and as anything less can always be found, we only have to say "let it be less". $\endgroup$ – fleablood Oct 23 '15 at 5:23
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    $\begingroup$ I took $\delta=\min\{1,\epsilon/2\}$ because when I was initially working out the problem, I set $\delta=1$ and noted that $1<x-a<1\Rightarrow -(x+a)<x^2-a^2<x+a\leq2$, and then I modified my definition of $\delta$ from that. But I didnt have a firm idea of what I was doing $\endgroup$ – Matt G Oct 23 '15 at 5:28

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