## I don't understand uniform continuity :(

 Quote by Arian.D I don't understand why the last part is wrong :(. Where did 'i' come to play? This is what I've done step by step written down in propositional calculus: Suppose that $\forall x_n, \forall y_n:|x_n - y_n| \to 0 \implies |f(x_n)-f(y_n)| \to 0$ is false: 1) Negation: $\exists x_n, \exists y_n: |x_n - y_n|\to 0 \land \neg(|f(x_n)-f(y_n)| \to 0)$ 2) conjunction elimination: $\exists x_n, \exists y_n: \neg(|f(x_n)-f(y_n)| \to 0)$ 3) $\neg(|f(x_n)-f(y_n)| \to 0) \equiv \forall\delta>0, \exists\epsilon>0 : |x_n - y_n|<\delta \implies |f(x_n)-f(y_n)|\geq \epsilon$ 4) $\exists x_n, \exists y_n, \forall \delta>0, \exists \epsilon>0: |x_n-y_n|<\delta \implies |f(x_n)-f(y_n)| \geq \epsilon$ 5) $\exists x_n, \exists y_n, \forall \delta>0, \exists \epsilon>0: |x_n-y_n|<\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon>0, \exists \delta>0, \forall x_n, \forall y_n : |x_n - y_n|<\delta \implies |f(x_n)-f(y_n)|<\epsilon)$ This is what my argument was in symbolic logic, I usually suck at propositional calculus but when I write things in that way I find it easier to argue with people.
For a constant n it is impossible that |x_n - y_n| < delta for all delta(unless x_n = y_n).

I'm not sure where you screwed up but it's not right.

 Quote by Skrew For a constant n it is impossible that |x_n - y_n| < delta for all delta(unless x_n = y_n). I'm not sure where you screwed up but it's not right.
hmmm, sounds you're right. Maybe I should've put up a quantifier like for any n in natural numbers....
well, except that part, do I have other mistakes? How would you rephrase my proof to make it right?

 Quote by Arian.D hmmm, sounds you're right. Maybe I should've put up a quantifier like for any n in natural numbers.... well, except that part, do I have other mistakes? How would you rephrase my proof to make it right?
I think your first statement does not correctly state what you wanted it to state.

In particular, your for alls need to be in terms of sequences, not elements of the sequence.

 Quote by Arian.D Very great sentences. Well, the last part is important, |x| is not differentiable everywhere (it's not differentiable at x=0) but yet it is uniformly continuous over R. Right? I used your intuitive method.
Yes, exactly.

If you can "see" without any need for explanation or symbols that 1/x and ex are continuous but not uniformly continuous, that's what I'm trying to explain.

Because as x gets larger, ex gets a LOT LOT LOT larger. And as x gets close to zero, 1/x gets a LOT LOT LOT larger. Both 1/x and ex get stretched.
Once you see that you can intuitively see that they must fail to be uniformly continuous.

Then the epsilons and deltas will be easer.

The things people were saying about differentiability, boundedness, and the compactness of the domain, are conditions that ensure that a given function is uniformly continuous. But they are not the definition or the meaning of uniform continuity. The meaning is in the stretchiness.

 Quote by Skrew I think your first statement does not correctly state what you wanted it to state.
Yea. me too. But I don't know how to reword my proof. So if we exclude the details, have I got the strategy right in my proof, or my proof is totally wrong?

 Quote by SteveL27 Yes, exactly. If you can "see" without any need for explanation or symbols that 1/x and ex are continuous but not uniformly continuous, that's what I'm trying to explain. Because as x gets larger, ex gets a LOT LOT LOT larger. And as x gets close to zero, 1/x gets a LOT LOT LOT larger. Both 1/x and ex get stretched. Once you see that you can intuitively see that they must fail to be uniformly continuous. Once you can see that, the epsilons and deltas will be easer.

Well, yea, I've got the intuition now I think. ex could be written as ex=sinh(x)+cosh(x), because ex is not uniformly continuous, then by what I proved earlier, sinh(x) 'OR' cosh(x) should be not uniformly continuous as well. Using the intuitive method, one could see that both sinh(x) and cosh(x) are in fact not uniformly continuous and it sounds reasonable because both have ex in their equations that make them get steeper and steeper as we go to infinity (and also because e-x fails to cancel out the fast growth of ex in sinh(x)). Am I right? If yes, then I've got the intuition right.

 Quote by Arian.D Yea. me too. But I don't know how to reword my proof. So if we exclude the details, have I got the strategy right in my proof, or my proof is totally wrong?
The problem starts with the forall statements, the forall needs to be applied to the sequences, not the elements of the sequences.

Also a bounded function can fail to be uniformly continuous.

For example sin(1/x) on (0, 1).

 Quote by Skrew The problem starts with the forall statements, the forall needs to be applied to the sequences, not the elements of the sequences.
Ahh, that's right. I guess my proof is flawed even more than that, because I didn't use the fact that |xn-yn| goes to zero.
if $|x_n - y_n| \to 0$ then $\exists M\in\mathbb{N}>0,\forall\epsilon>0,\forall n\in\mathbb{N}: n\geq M \implies |f(x_n)-f(y_n)| < \epsilon$
maybe this is where that i could come to play, because this time when I negate this there must exist n=i that has the property you said. Am I right?

 Quote by Arian.D Ahh, that's right. I guess my proof is flawed even more than that, because I didn't use the fact that |xn-yn| goes to zero. if $|x_n - y_n| \to 0$ then $\exists M\in\mathbb{N}>0,\forall\epsilon>0,\forall n\in\mathbb{N}>M: n\geq M \implies |f(x_n)-f(y_n)| < \epsilon$ maybe this is where that i could come to play, because this time when I negate this there must exist n=i that has the property you said. Am I right?
yess

 Quote by Skrew yess
So what do I need to do now to have my proof completed?
Because after this step I think other steps are justified, or not?

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