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Thanks!

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Thanks!

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Office_Shredder

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I think I do address this. I set ##\delta = \frac{\epsilon}{3|y|}## when ##\frac{\epsilon}{3|y|} < \frac{|y|}{2}## so that the restriction ##\delta < |y|## is maintained. In the proof I state

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When you write

[tex] \delta 3 |y| = \epsilon [/tex]

that should really be a less than or equal to sign, because if delta happened to be |y|/2 (in the event where epsilon IS big) then this is not going to be an equality. Also the proof would be better structured by removing the "One can easily see that.... with this in mind pick delta " and replace it with writing down what delta is going to be equal to, and then stating all the consequences - your proof does not require you to explain why you make a choice of definition before stating it (this is something that took me a long time to get over doing and I still do from time to time, and it is a bad habit). It's not wrong but it's not the cleanest way to structure a final proof.

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Let ##f:\mathbb{R} \to \mathbb{R}## be a function such that ## y \mapsto y^2##. To show continuity at every point of ##f##, it is sufficient to demonstrate:

##\forall u \;\; \forall \epsilon > 0 \;\; \exists \delta : \forall y \; \left ( |y-u| < \delta \implies |y^2 - u^2| \right ) ##

Choose any ##\epsilon > 0## and real ##u##.

Case 1: ##u=0##

Let ##\delta = \sqrt{\epsilon}##. It is clear that ##|y| < \sqrt{\epsilon} \implies |y^2| = \epsilon##.

Case 2: ##u \neq 0##

Choose any ##\delta > 0## such that ##\delta \leq \frac{\epsilon}{3|u|}## and ##\delta \leq \frac{|u|}{2}##. (Note: If ##|y-u| \geq \delta## then the continuity condition is met and the argument is complete; assume ##|y-u| < \delta## holds for the remainder of the proof) Since ##\delta \leq \frac{|u|}{2} < |u|##, it follows ##|y+u| < 3|u|##. Therefore, ##|y^2-u^2| = |y+u||y-u| < 3|u| \delta \leq \epsilon ## ■

I want to make sure of a few things:

- ##\delta## can be a function of both ##\epsilon## and ##u## (but not ##y##) for pointwise continuity since they both proceed ##\delta##'s quantifier?
- Is ##(|y-u| < \delta) \wedge (\delta < |u|) \implies |y+u| < 3|u|## obvious enough to state without proof?
- Obviously, once ##\delta## is chosen, ##|y-u|## could be ##\geq \delta## since both were chosen arbitrarily and without respect to ##\delta##. If this inequality does not hold, neither do the following arguments. Is the parenthetical "(Note: If ##|y-u| \geq \delta## then the continuity condition is met and the argument is complete; assume ##|y-u| < \delta## holds for the remainder of the proof)" sufficient to tackle this problem?

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mathwonk

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then to have y^2 - y0^2 < e, you just have to make both 2dy0 and d^2 less than e/2.

e.g. take d < min(1, e/2), and also d < e/(2y0).

oh now i see why you took y0 ≠ 0. but of course if y0 = 0, then 2dy0 is always < e/2 for any d, so you don't need to worry about that case.

- #7

Office_Shredder

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You need a < ε at the end here, but the statement looks good other than that.Let ##f:\mathbb{R} \to \mathbb{R}## be a function such that ## y \mapsto y^2##. To show continuity at every point of ##f##, it is sufficient to demonstrate:

##\forall u \;\; \forall \epsilon > 0 \;\; \exists \delta : \forall y \; \left ( |y-u| < \delta \implies |y^2 - u^2| \right ) ##

It should read [itex] |y^2| < \epsilon [/itex]Case 1: ##u=0##

Let ##\delta = \sqrt{\epsilon}##. It is clear that ##|y| < \sqrt{\epsilon} \implies |y^2| = \epsilon##.

- ##\delta## can be a function of both ##\epsilon## and ##u## (but not ##y##) for pointwise continuity since they both proceed ##\delta##'s quantifier?

Yes, this is correct.

[*]Is ##(|y-u| < \delta) \wedge (\delta < |u|) \implies |y+u| < 3|u|## obvious enough to state without proof?

[*] Obviously, once ##\delta## is chosen, ##|y-u|## could be ##\geq \delta## since both were chosen arbitrarily and without respect to ##\delta##. If this inequality does not hold, neither do the following arguments. Is the parenthetical "(Note: If ##|y-u| \geq \delta## then the continuity condition is met and the argument is complete; assume ##|y-u| < \delta## holds for the remainder of the proof)" sufficient to tackle this problem?

- #8

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mathwonk: I found your post quite insightful, even though I am going for a somewhat higher degree of rigor in my proof. Thanks!

I have created a similar thread here, which incorporates your feedback.

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