I don't understand uniform continuity

[Arian.D]

The last part is wrong.

The statement "|x_n - y_n| ->0 does not imply |f(x_n) -f(y_n)| -> 0"

Says that for every delta > 0 there exists a i such that |x_i - y_i| < delta but |f(x_i) - f(y_i)| >= constant.

In particular your constant epsilon does not change when disproveing uniform continuity, only the delta.

Also e^x is uniformly continuous on any bounded interval.

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&gt;0, \exists\epsilon&gt;0 : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|\geq \epsilon
4) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon
5) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon&gt;0, \exists \delta&gt;0, \forall x_n, \forall y_n : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|&lt;\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.

 

[Arian.D]

There's an intuitive way to see that e^x is not uniformly continuous on the reals.

Visualize the graph of e^x. As you go to the right along the positive x-axis, e^x gets steeper and steeper. The graph "stretches out," so that if you go far enough to the right, you can make a very small change in x be a huge change in e^x.

Do you see that?

It's the exact same problem that one has with f(x) = 1/x. With that function, as you get closer to zero on the positive x-axis, you can make f(x) vary hugely. It's the same issue.

A function that's continuous is one that doesn't have a tear or rip in its graph. (Speaking intuitively, as I am in this post).

A function that's uniformly continuous is one that doesn't stretch too much.

So a function like 1/x or e^x is a function that is continuous, BUT it stretches out a lot. If we pulled any harder we'd tear it.

Think of the graph of a function as a piece of stretch Turkish taffy. Or silly putty if you like.

If a little pull makes a little stretch: it's uniformly continuous.

If a little pull can sometimes make a big stretch: its continuous but NOT uniformly continuous.

If a little pull tears it into two pieces: It's not continuous.

So the condition of continuity-but-not-uniform-continuity (for which there's no good word AFAIK, but there should be) is a middle state between being uniformly continuous and discontinuous.

Do you see that? Because if you get what I wrote, then you get uniform continuity. If you don't see the visualization, pushing the symbols around will not help with intuition.

Uniform continuity means that a small change in input will ALWAYS give you a small change in output. Continuity without uniform continuity means that a small change in input may give you a HUGE change in output; as with 1/x near zero, and e^x as you move to the right on the positive x-axis; but that you still won't tear the graph.

Discontinuity means that you can make a tiny change in the input and you'll rip the graph into two pieces.

By the way I would like to say that the discussion of compact sets is obscuring the basic intuition; and I think the OP should dwell on the examples of 1/x and e^x before trying to sling proofs about continuous functions on compact sets. My two cents on that one.

WOW. This is a very great post indeed. It helped a lot. I think you're somehow using the theorem which micromass posted that states 'if f is differentiable and its derivative is bounded then f is uniformly continuous'. Having accepted that as true, if the derivative is bounded, we don't expect the function to dramatically change for a small input because the rate of change for any input is bounded and can't get higher than some value. so all you said makes sense to me, it's still not fully intuitive for me, I mean I can't say if a function is uniformly continuous just by looking at the graph while I could definitely say whether a function is continuous or not by looking at its graph (Yes, I know I can't if the domain is R, because we could only graph a part of the function, not the whole of it), but still...

Your post helped a lot, I'll try to create some intuition behind uniform continuity by your post, it'll help me a lot in the future.

 

[SteveL27]

WOW. This is a very great post indeed. It helped a lot. I think you're somehow using the theorem which micromass posted that states 'if f is differentiable and its derivative is bounded then f is uniformly continuous'

Very glad this helped.

But I am NOT repeat NOT caring at all about theorems. Especially about differentiability, which has nothing to do with this.

We are looking at the intuition. If I needed to teach uniform continuity to a six year old I could do it. You're stretching a big glop of taffy, or silly putty. You can stretch it very gently; or you can stretch it very roughly but without tearing it; or you can rip it apart. Uniform continuity is pulling on a piece of taffy very gently.

Non-uniform continuity is pulling on it roughly without tearing it. And discontinuity is pulling on it so hard that you tear it into two pieces.

This is not about theorems. Only the intuition. Once you get the intuition you can bang out the proofs automatically. That's the secret of doing well in real analysis. If you try to sling the symbols before you understand the concepts, you're doomed. That's because it's about ten times faster to knock out an epsilon proof when you already have a clear visualization; than it is to struggle to put together a proof with only the symbols, lacking the understanding.

And this thread has become very symbol-driven when it's clear that you are struggling with the concepts.

Differentiability is irrelevant here. Most continuous functions aren't even differentiable, so the notions of continuity and uniform continuity should be understood on their own terms.

Please forget about differentiability; and please forget about compactness. Those two topics have greatly complicated this thread, when clarity is what's needed.

Think about taffy.

 

[Arian.D]

Very glad this helped.

But I am NOT repeat NOT caring at all about theorems.

We are looking at the intuition. If I needed to teach uniform continuity to a six year old I could do it. You can stretch it gently; or you can stretch it very roughly; or you can rip it apart. Uniform continuity is pulling on a piece of taffy very gently.

This is not about theorems. Only the intuition. Once you get the intuition you can bang out the proofs automatically. That's the whole point of real analysis. If you try to sling the symbols before you understand the concepts, you're doomed.

And this thread has become very symbol-driven when it's clear that you are struggling with the concepts.

Differentiability is irrelevant here. Most continuous functions aren't even differentiable, so the notions of continuity and uniform continuity should be understood on their own terms.

Please forget about differentiability; and please forget about compactness. Think about taffy.

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.

 

[Skrew]

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&gt;0, \exists\epsilon&gt;0 : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|\geq \epsilon
4) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon
5) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon&gt;0, \exists \delta&gt;0, \forall x_n, \forall y_n : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|&lt;\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.

 

[Arian.D]

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?

 

[Skrew]

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.

 

[SteveL27]

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 e^x are continuous but not uniformly continuous, that's what I'm trying to explain.

Because as x gets larger, e^x 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 e^x 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.

 

[Arian.D]

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?

Yes, exactly.

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

Because as x gets larger, e^x 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 e^x 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. e^x could be written as e^x=sinh(x)+cosh(x), because e^x 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 e^x 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 e^x in sinh(x)). Am I right? If yes, then I've got the intuition right.

 

[Skrew]

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).

 

[Arian.D]

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 |x_n-y_n| goes to zero.
if |x_n - y_n| \to 0 then \exists M\in\mathbb{N}&gt;0,\forall\epsilon&gt;0,\forall n\in\mathbb{N}: n\geq M \implies |f(x_n)-f(y_n)| &lt; \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?

 

[Skrew]

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}&gt;0,\forall\epsilon&gt;0,\forall n\in\mathbb{N}&gt;M: n\geq M \implies |f(x_n)-f(y_n)| &lt; \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

 

[Arian.D]

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?