# I Understanding uniform continuity....

1. May 9, 2016

### Alpharup

Let us have a continous function f which is uniformly continuous on [a,b] and [b,c]....
Then Spivak says, f is uniformly continuous on [a,c]...
For prving this, he invokes the continuity of f on b.....
My questions here are:
1.For a given ε, we have a δ1 which works on whole of interval [a,b] and similarly, for this ε, δ2 works for all of [b,c].....Why can't we take δ=min(δ1,δ2) on this whole inteval [a,c], and consider that this δ will work for a given ε. Why does he consider continuity at b?
2. For this point b again, he considers these cases:
a. Either x<b<y or y<b<a....
b. ιx-bι<δ3 for ε/2 and ιy-bι<δ3 for ε/2
c. he takes ιx-yι< min(δ1,δ2,δ3), then he proves ιf(x)-f(y)ι<ε...
Why should he consider the following assumption that x and y only lie in (b-δ3,b+δ3)? What if for all x and y lying outside?

2. May 9, 2016

### chiro

Hey Alpharup.

You probably have to consider that there is some possible discontinuity at b and rectify whether the epsilon-delta proof is consistent for both intervals.

If you assume continuity on two intervals that are right next to each other then you have to make sure that you have continuity in the neighborhood of the point where the two intervals join - it can't just be assumed.

Imagine two functions on the two intervals - both interval mappings are continuous in their own right but you can have a condition where they don't retain continuity at the point where they join.

Because this is a closed interval you will have continuity {since you have [] instead of ()} but without this condition it could possibly not hold.

3. May 10, 2016

### Alpharup

Yes, I get it.....But we have assumed a perfectly continuous function in [a,c], didn't we? Also, isn't it quite obvious that a δ so small in [a,c] will definitely work for a givenε? Let us take δ=0.000001 ....Won't it work for the whole interval?

4. May 10, 2016

### Alpharup

Well, I think I have understood my previous question( why can't we take a small delta)...The reason for my misunderstanding was that I bluntly followed the definition of uniform continuity.
Please do say whether my reasoning is right:
Uniform continuity is a property like any other property. If it is valid throughout an interval say [a,c], it is valid at any open interval (m,n) and any closed interval [p,q] belonging to the bigger and enclosing interval [a,c]. In this case, we know that uniform continuity property holds in both [a,b] and [b,c]. We know the continuity of function at b. We have to prove it is uniformly continous at [a,c]. By the above property, we talk about some open interval (b-δ,b+δ) whwre δ is some constant. Here we have to prove that the open interval is uniformly continous. Is my thinking right?

5. May 10, 2016

### chiro

With continuity it has to exist from both sides for an example such as this.

If you can show that the definition is satisfied at the point b then you are basically done since you have already assumed the rest of the interval is continuous.

It's basically just showing it exists at that point and then using the merger of the two assumptions to show it exists across the interval as a whole.

6. May 11, 2016

### Alpharup

But, there is little confusion in the last step of the proof:
Let us take x and y such that x<b<y;
and let a and y be the points near to b satisfying |x-b|<δ3 and |y-b|<δ3.
Then what can be said about |x-y| ?

7. May 11, 2016

### Alpharup

I have attached the image of the proof.

In the last step he has taken |x-y|<δ. But how?

8. May 11, 2016

### tommyxu3

I think the lemma is the special case of the theorem:
If $f$ is conti. on a compact set $X,$ then $f$ is uniformly conit. on it.
And the condition of uniform continuity and uniform "Cauthy" is equivalent.

And I think in your book the step he didn't take but discussed the case.

9. May 11, 2016

### Alpharup

Well, I don't know anything about compactness

10. May 12, 2016

### Alpharup

Well, I have thought of a proof for the last part. Please evaluate the proof whether it is right.
f is contionuous at b and so
for every ε>0, there exists a δ3>0 such that for all x and y satisfying |x-b|<δ3 and |y-b|<δ3, it is implied that
|f(x)-f(b)|<(ε/2) and |f(y)-f(b)|<(ε/2)........(1)
if all x and y satisfy |x-b|<δ3 and |y-b|<δ3, then by triangle inequality, |x-y|<2*δ3
Also by triangle inequality |f(x)-f(b)|<(ε/2) and |f(y)-f(b)|<(ε/2). implies |f(x)-f(y)|<ε.
Thus we can write (1) as
for every ε>0, there exists a δ3>0 such that for all x and y satisfying |x-y|<2*δ3, implies |f(x)-f(y)|<ε.
or
|f(x)-f(y)|<ε

since δ=min(δ1,δ2,δ3)
δ<2*δ3.
So hence, if we take |x-y|<δ, it may imply |f(x)-f(y)|<ε.
Hence, we can conlude that if x<b<y or y<b<x, then
for every ε>0, there exists a δ>0 such that for all x and y satisfying |x-y|<δ, implies |f(x)-f(y)|<ε. Hence, we have proved.

11. May 13, 2016

### Stephen Tashi

That's a true statement, but it isn't a direct quotation of the definition of "f is continuous at b". (For example, the definition of "f is continuous at b" doesn't mention two points in the domain, x and y. )

12. May 13, 2016

### Alpharup

hmm....do you mean to say that the continuity definition mentions only one point, either X or y?

13. May 13, 2016

### Stephen Tashi

Yes.

14. May 13, 2016

### Alpharup

So, on what basis does Spivak assume the second point?

15. May 13, 2016

### lavinia

The function is assumed continuous on the entire interval [a,c].

16. May 13, 2016

### Alpharup

So, is he invoking continuity at y or b?

17. May 13, 2016

### lavinia

Continuity means continuity at a point in the domain of the function. So the assumption that f is continuous on [a,c] means that is it continuous at every point of [a,c] including b.

18. May 13, 2016

### Alpharup

So, how does he exploit this definition of continuity on the whole interval? How does he bring in another variable y?

19. May 13, 2016

### lavinia

It allows you to glue together the two neighborhoods of b - the one in [a,b} and the one in [b,c]. When you glue them together you get a neighborhood of b.

20. May 13, 2016

### Alpharup

Hmm...so X any y lie in this neighborhood of b ?

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