Continuity at a point implies continuity in the neighborhood

[Happiness]

I claim that if a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous at a point ##a##, then there exists a ##\delta>0## and ##|h|<\frac{\delta}{2}## such that ##f## is also continuous in the ##h##-neighbourhood of ##a##.

Please advice if my proof as follows is correct.

Continuity at ##a## means this: for every ##\epsilon>0##, there exists a ##\delta>0## such that ##|x-a|<\delta \implies|f(x)-f(a)|<\epsilon##. --- (*)

We want to show this: for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'##.

If we let ##\delta'## = ##\delta-|h|## and ##\epsilon=\frac{\epsilon'}{2}##, we will have ##|x-(a+h)|<\delta' \implies|x-a|<\delta\implies|f(x)-f(a)|<\epsilon\implies|f(x)-f(a+h)|<\epsilon'##

The details:

##|x-(a+h)|<\delta'##
##|x-(a+h)|<\delta-|h|##
##-\delta+|h|<x-a-h<\delta-|h|##
##-\delta\leq-\delta+|h|+h<x-a<\delta-|h|+h\leq\delta##
##|x-a|<\delta##

By (*),

##|f(x)-f(a)|<\epsilon##
##-\epsilon<f(x)-f(a)<\epsilon##
##-\epsilon<f(x)-f(a+h)+f(a+h)-f(a)<\epsilon##
##-\epsilon-f(a+h)+f(a)<f(x)-f(a+h)<\epsilon-f(a+h)+f(a)## --- (**)

By (*), when ##x=a+h##,

##\mid a+h-a\mid\,=\,\mid h\mid\,<\delta\implies\,\mid f(a+h)-f(a)\mid\,=\,\mid-f(a+h)+f(a)\mid\,<\epsilon##

Substituting this into (**), we get

##-2\epsilon<f(x)-f(a+h)<2\epsilon##
##\epsilon'<f(x)-f(a+h)<\epsilon'##

 

[lavinia]

Consider the function

$$f(x) = x^2$$ for x a rational number and
$$ f(x) = 0 $$ for x an irrational number.

This function is continuous at zero. But it is not continuous anywhere else.

 

[Happiness]

@lavinia

For that function, am I right to say that it is only continuous at ##x=0## and in its immediate neighbourhood but not anywhere else?

 

[Mark44]

@lavinia

For that function, am I right to say that it is only continuous at ##x=0## and in its immediate neighbourhood but not anywhere else?

No. The function Lavinia gave is continuous only at x = 0. "Immediate neighborhood" implies that for some δ > 0, the function is continuous at a point x within δ of 0.

 

[WWGD]

I claim that if a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous at a point ##a##, then there exists a ##\delta>0## and ##|h|<\delta## such that ##f## is also continuous in the ##h##-neighbourhood of ##a##.

Please advice if my proof as follows is correct.

Continuity at ##a## means this: for every ##\epsilon>0##, there exists a ##\delta>0## such that ##|x-a|<\delta \implies|f(x)-f(a)|<\epsilon##. --- (*)

We want to show this: for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'##.

If we let ##\delta'## = ##\delta-|h|## and ##\epsilon=\frac{\epsilon'}{2}##, we will have ##|x-(a+h)|<\delta' \implies|x-a|<\delta\implies|f(x)-f(a)|<\epsilon\implies|f(x)-f(a+h)|<\epsilon'##

The details:

##|x-(a+h)|<\delta'##
##|x-(a+h)|<\delta-|h|##
##-\delta+|h|<x-a-h<\delta-|h|##
##-\delta<-\delta+|h|+h<x-a<\delta-|h|+h<\delta## !~~~~~~~~~~~~~~~~~~~~~~~~~~~
##|x-a|<\delta##

By (*),

##|f(x)-f(a)|<\epsilon##
##-\epsilon<f(x)-f(a)<\epsilon##
##-\epsilon<f(x)-f(a+h)+f(a+h)-f(a)<\epsilon##
##-\epsilon-f(a+h)+f(a)<f(x)-f(a+h)<\epsilon-f(a+h)+f(a)## --- (**)

By (*), when ##x=a+h##,

##|a+h-a|=|h|<\delta\implies|f(a+h)-f(a)|=|-f(a+h)+f(a)|<\epsilon##

Substituting this into (**), we get

##-2\epsilon<f(x)-f(a+h)<2\epsilon##
##\epsilon'<f(x)-f(a+h)<\epsilon'##

Here is a problem I see. I added a !~~~~~~~~~~~~~~~~~~~~~ to your post to indicate the place. You seem to be canceling out -|h| with h , which does not hold if/when h is negative. In your layout, we only have ## |h| < \delta ##.

 

[Happiness]

Here is a problem I see. I added a !~~~~~~~~~~~~~~~~~~~~~ to your post to indicate the place. You seem to be canceling out -|h| with h , which does not hold if/when h is negative. In your layout, we only have ## |h| < \delta ##.

If ##h## is negative, then ##\delta-|h|+h<\delta-|h|<\delta##. So this is still valid.

But there is another problem. We need to ensure that ##0<\delta-|h|+h##. So we would require ##|h|<\frac{\delta}{2}##.

 

[Mark44]

@Happiness, Lavinia's example serves as a counterexample to your proof. The basic premise of your proof is flawed -- it can't be fixed.

 

[Happiness]

@Mark44, I'm trying to analyse Lavinia's example. But in the meantime, what's wrong with my proof?

 

[Mark44]

@Mark44, I'm trying to analyse Lavinia's example. But in the meantime, what's wrong with my proof?

There's not much to analyze in Lavinia's example.

I claim that if a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous at a point ##a##, then there exists a ##\delta>0## and ##|h|<\frac{\delta}{2}## such that ##f## is also continuous in the ##h##-neighbourhood of ##a##.

Lavinia's example negates your claim, so it's pointless to try to continue with it. For her example, f is continuous at only a single point, and not in any neighborhood of positive diameter around x = 0.

 

[Happiness]

There's not much to analyze in Lavinia's example.

Could you show me explicitly how Lavinia's example is not continuous at points other than 0? I'm not very good at this.

Lavinia's example negates your claim, so it's pointless to try to continue with it. For her example, f is continuous at only a single point, and not in any neighborhood of positive diameter around x = 0.

I'm not trying to save my claim. I'm trying to learn from my mistake. So it's important to find where exactly my mistake is.

 

[WWGD]

Basically, ## h -|h| =0 ##. Try for the negative and positive cases. This does not work in your ## \delta - \epsilon ## inequality; in your proof, it implies ##- \delta <- \delta ##. I don't know if it is _the_ problem, but it is a problem.

 

[WWGD]

If ##h## is negative, then ##\delta-|h|+h<\delta-|h|<\delta##. So this is still valid.

But there is another problem. We need to ensure that ##0<\delta-|h|+h##. So we would require ##|h|<\frac{\delta}{2}##.

Since ## -|h|+h =0## , your inequality becomes ## \delta < \delta-|h| < \delta ##.

 

[Happiness]

Since ## -|h|+h =0## , your inequality becomes ## \delta < \delta-|h| < \delta ##.

I don't understand your point. ##-|h|+h=0## only if ##h\geq0##.

##-\delta\leq-\delta+|h|+h<\delta-|h|+h\leq\delta## looks fine to me in both the case ##h<0## and the case ##h\geq0##.

 

[WWGD]

Ah, sorry, I misread. But you do get a problem when ## h > 0 ## , getting ## \delta < \delta -|h| < \delta ##

 

[Mark44]

There's not much to analyze in Lavinia's example.

Could you show me explicitly how Lavinia's example is not continuous at points other than 0? I'm not very good at this.

Lavinia's example negates your claim, so it's pointless to try to continue with it. For her example, f is continuous at only a single point, and not in any neighborhood of positive diameter around x = 0.

I'm not trying to save my claim. I'm trying to learn from my mistake. So it's important to find where exactly my mistake is.

I've drawn a crude picture that might help.
If f is continuous at x₀, then for a given ε > 0, there exists a δ > 0 so that if x is within (x₀ - δ, x₀ + δ), then f(x) will be in (f(x₀ - ε, f(x₀ + ε).
In my drawing, suppose some ε has been chosen. The band along the y-axis depicts the interval (f(x₀ - ε, f(x₀ + ε). What δ can we specify so that for any choice of x in (x₀ - δ, x₀ + δ), the interval I have marked along the x-axis, f(x) will be in the desired interval on the y-axis.

The answer is, we can't. For many of the x values in the interval along the x-axis, the function value is zero, which is well outside the band on the vertical axis.

 

[Mark44]

To go along with the image in my last post, let's take x₀ = 1, which is a rational number, so f(x₀) = 1. I choose ε = .04. So the interval along the y-axis is (.96, 1.04). Your job is to find a positive number δ for which every number in (1 - δ, 1 + δ) is mapped to the interval (.96, 1.04) on the y-axis.

Since I picked .04 for ε you might be tempted to take δ = .2 (.04 happens to be the square of .2). Now, does every number in the interval (.8, 1.2) map to the interval (.96, 1.04)? Some do and some don't. If x is a rational number between .8 and 1.2, then yes, its image is in the interval (.96, 1.04). However, if x is any irrational number between .8 and 1.2, then its image is 0, which is not between .96 and 1.04. In fact, there is no such δ that can be found, as long as x ≠ 0.

Hope that helps.

 

[Happiness]

To go along with the image in my last post, let's take x₀ = 1, which is a rational number, so f(x₀) = 1. I choose ε = .04. So the interval along the y-axis is (.96, 1.04). Your job is to find a positive number δ for which every number in (1 - δ, 1 + δ) is mapped to the interval (.96, 1.04) on the y-axis.

Since I picked .04 for ε you might be tempted to take δ = .2 (.04 happens to be the square of .2). Now, does every number in the interval (.8, 1.2) map to the interval (.96, 1.04)? Some do and some don't. If x is a rational number between .8 and 1.2, then yes, its image is in the interval (.96, 1.04). However, if x is any irrational number between .8 and 1.2, then its image is 0, which is not between .96 and 1.04. In fact, there is no such δ that can be found, as long as x ≠ 0.

Hope that helps.

Thanks a lot! Excellent explanation! Now I need to study my proof to see where I went wrong.

 

[Happiness]

Ah, sorry, I misread. But you do get a problem when ## h > 0 ## , getting ## \delta < \delta -|h| < \delta ##

Where did you get ## \delta < \delta -|h| < \delta ##?

It should be ##-\delta\leq-\delta+|h|+h<\delta-|h|+h\leq\delta##.

Suppose ##h=-\frac{\delta}{3}## (a negative number), then we have

##-\delta\leq-\delta+\frac{\delta}{3}-\frac{\delta}{3}<\delta-\frac{\delta}{3}-\frac{\delta}{3}\leq\delta##
##-\delta\leq-\delta<\frac{\delta}{3}\leq\delta##

which is valid.

 

[Mark44]

I claim that if a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous at a point ##a##, then there exists a ##\delta>0## and ##|h|<\frac{\delta}{2}## such that ##f## is also continuous in the ##h##-neighbourhood of ##a##.

Please advice if my proof as follows is correct.

Continuity at ##a## means this: for every ##\epsilon>0##, there exists a ##\delta>0## such that ##|x-a|<\delta \implies|f(x)-f(a)|<\epsilon##. --- (*)

We want to show this: for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'##.

So given an ε by someone else, you have to come up with the δ. I don't see any evidence of that in your work. In the line below, you seem to be taking the existence of such a δ for granted. The δ that you find depends on the ε that someone else gives you.

Think about this δ-ε as a sort of dialogue between you and someone who is not your friend. Since this person is not your friend, he's not motivated to do you any favors. He's likely to give you the smallest number he can think of, and challenge you to find a δ that works.

Non-friend: I choose ε = .001
You (after a bit of work): OK, then δ = .05 works

Non-friend: Hmmph! All right, how about ε = .0001?
You (a bit later): No problem. δ = .002 works

Etc.
After the non-friend realizes that no matter how smal a number he chooses for ε, you outsmart him and come up with a value for δ that works. He gives in, and cedes that the limit must be what you say it is.

If we let ##\delta'## = ##\delta-|h|## and ##\epsilon=\frac{\epsilon'}{2}##, we will have ##|x-(a+h)|<\delta' \implies|x-a|<\delta\implies|f(x)-f(a)|<\epsilon\implies|f(x)-f(a+h)|<\epsilon'##

 

[Happiness]

So given an ε by someone else, you have to come up with the δ. I don't see any evidence of that in your work. In the line below, you seem to be taking the existence of such a δ for granted. The δ that you find depends on the ε that someone else gives you.

After the non-friend realizes that no matter how smal a number he chooses for ε, you outsmart him and come up with a value for δ that works. He gives in, and cedes that the limit must be what you say it is.

The existence of ##\delta## is guaranteed by the continuity of ##f(x)## at ##x=a##. Isn't it?

And isn't it acceptable for ##\delta=\delta(\epsilon)##, that is, for ##\delta## to be a function of ##\epsilon##?

I've shown that ##\delta'(\epsilon')=\delta(\epsilon)-|h|=\delta(\frac{\epsilon'}{2})-|h|##.

Since ##h=h(\delta)##, we have ##\delta'(\epsilon')=\delta(\frac{\epsilon'}{2})-|h(\delta(\frac{\epsilon'}{2}))|##.

 

[Mark44]

The existence of ##\delta## is guaranteed by the continuity of ##f(x)## at ##x=a##. Isn't it?

Yes, but merely being continuous at a point doesn't guarantee that the function will be continuous anywhere else. That's what Lavinia's counterexample shows. Lavinia's example is continuous at x = 0, but nowhere else.

And isn't it acceptable for ##\delta=\delta(\epsilon)##, that is, for ##\delta## to be a function of ##\epsilon##?

I've shown that ##\delta'(\epsilon')=\delta(\epsilon)-|h|=\delta(\frac{\epsilon'}{2})-|h|##.

Since ##h=h(\delta)##, we have ##\delta'(\epsilon')=\delta(\frac{\epsilon'}{2})-|h(\delta(\frac{\epsilon'}{2}))|##.

Edit: Your assumption is that f is continuous at a, so yes, given any ε, there is a δ. The problem is that now you're trying to prove continuity at some nearby point, so now you have to come up with the δ that works for a given ε at that new point. For some functions (Lavinia's for example), this isn't possible.

 

[Happiness]

Yes, but merely being continuous at a point doesn't guarantee that the function will be continuous anywhere else. That's what Lavinia's counterexample shows. Lavinia's example is continuous at x = 0, but nowhere else.

I only assumed the function is continuous at ##x=a## and hence ##\delta## exists. I did not assume this: "the function is continuous at ##x=a+h## and hence ##\delta'## exists".

Instead, I've shown that since ##\delta## exists, ##\delta'## must exist, because ##\delta'=\delta-|h|##.

I suspect the mistake is related to ##h(\delta)##.

 

[Mark44]

I only assumed the function is continuous at ##x=a## and hence ##\delta## exists. I did not assume this: "the function is continuous at ##x=a+h## and hence ##\delta'## exists".

Right, you can't assume this. But this is what you have to prove, so at x = a + h, you have to find the δ.

I've shown that since ##\delta## exists, ##\delta'## must exist

No, because the δ that works at x = a (for a given ε) doesn't have to be the same one that works at x = a + h (for a given ε).

You are assuming that such a δ can be found (for the point x = a + h). What I am saying is that you can't make that assumption, and Lavinia's example should provide all the proof you need that what you are trying to prove is fatally flawed. It ain't true.

, because ##\delta'=\delta(\epsilon)-|h|##. I suspect the mistake is related to ##h(\delta)##.

 

[Happiness]

No, because the δ that works at x = a (for a given ε) doesn't have to be the same one that works at x = a + h (for a given ε).

Yes. They may not be the same, and that's why I've used different symbols (for delta): ##\delta## vs ##\delta'##.

You are assuming that such a δ can be found (for the point x = a + h). What I am saying is that you can't make that assumption, and Lavinia's example should provide all the proof you need that what you are trying to prove is fatally flawed. It ain't true.

I don't quite get your point. In fact, I've come up with a way to find such a ##\delta'## (delta prime) given any ##\epsilon'## (epsilon prime). Again, I've used different symbols (for epsilon), ##\epsilon## vs ##\epsilon'##.

 

[WWGD]

@Happiness: may be a good idea to use lavinia's function and look for f(a+h) and see why it fails.

EDIT: There is also the fact that you defined ## \delta' = \delta-|h|## . But you have not defined h itself, from what I read.

 

[Mark44]

Yes. They may not be the same, and that's why I've used different symbols (for delta): ##\delta## vs ##\delta'##.

You shown a δ' (= δ - |h|), but there is no guarantee that this value actually works.

The problem seems to be that you are thinking in terms of the symbols without understanding the underlying geometry.

Let's try this. Using Lavinia's function as f, let's say that a = 1. Never mind that a here is not within a "small" neighborhood of 0. If you insist I can take a to be as close to 0 as you like, but using a = 1 makes the calculations a lot simpler, so let's go with that.

I challenge you with ε = .1. What is your choice for δ so that if x is any number such that |x - 1| < δ, then |f(x) - 1| < .1? IOW, that f(x) is in the interval (0.9, 1.1). Keep in mind that your δ has to work for every x in the interval (x - δ, x + δ).

I don't quite get your point. In fact, I've come up with a way to find such a ##\delta'## (delta prime) given any ##\epsilon'## (epsilon prime). Again, I've used different symbols (for epsilon), ##\epsilon## vs ##\epsilon'##.

 

[Delta2]

Counterexample by lavinia seems correct to me, so we must look for a mistake in the proof.

I think the mistake relates to how ##h## is defined. It turns out that ##h=h(\delta)=h(\delta(\epsilon))=h(\delta(\frac{\epsilon'}{2}))## so ##h## depends on ##\epsilon'## which seems to be the problem because ##h## has to be constant, given, and ##\epsilon'## can be anything we want for which we have to find a ##\delta'(h,\epsilon')##. So what you doing in the proof is that ##h## is changing as ##\epsilon'## changes in order to keep the inequalities true, but ##h## has to be constant independent of ##\epsilon'##.

 

[Happiness]

Counterexample by lavinia seems correct to me, so we must look for a mistake in the proof.

I think the mistake relates to how ##h## is defined. It turns out that ##h=h(\delta)=h(\delta(\epsilon))=h(\delta(\frac{\epsilon'}{2}))## so ##h## depends on ##\epsilon'## which seems to be the problem because ##h## has to be constant, given, and ##\epsilon'## can be anything we want for which we have to find a ##\delta'(h,\epsilon')##. So what you doing in the proof is that ##h## is changing as ##\epsilon'## changes in order to keep the inequalities true, but ##h## has to be constant independent of ##\epsilon'##.

Yes, that is the mistake, I believe.

When we say a function ##f## is continuous in the ##h##-neighbourhood of a point ##a##, the statement (S1)

##"\,##for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'"##

is true for all ##\epsilon'>0##. But if ##h## depends on ##\epsilon'##, we may run into a problem (for ##h\neq0##).

Let's prove that there is a problem for the special case when ##\delta(\epsilon)## is strictly increasing, that is, a smaller ##\epsilon## implies (or requires) a smaller ##\delta##. The other cases are not proven.

As ##\epsilon'## decreases, ##\epsilon=\frac{\epsilon'}{2}## decreases, ##\delta(\epsilon)## decreases, and ##|h|<\frac{\delta}{2}## decreases.

Let's look at a specific example. Suppose ##f(x)=x##. Then, the required ##\delta## for ##f## is ##\delta(\epsilon)=\epsilon##.

Then, ##|h|<\frac{\delta}{2}=\frac{\epsilon}{2}=\frac{\epsilon'}{4}\implies\epsilon'>4|h|##.

So for a given ##h##, statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'\leq4|h|##.

In general, for other functions for which ##\delta(\epsilon)## is strictly increasing, we have ##|h|<\frac{\delta}{2}=g(\epsilon')##, where ##g(\epsilon')## is a strictly increasing function, since it is a composition of strictly increasing functions. Thus, ##g## has an inverse ##g^{-1}## that is also strictly increasing. We have ##\epsilon'>g^{-1}(|h|)##. So statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'\leq g^{-1}(|h|)##.

This proof only covers the case when ##\delta(\epsilon)## is strictly increasing. If ##\delta(\epsilon)## is not strictly increasing, then the claim may still be true. If it's not true, then the mistake in the claim has not been fully explained.

Statement S2: If ##\delta(\epsilon)## is not strictly increasing in the neighbourhood of a point ##a##, then ##f(x)## must be a constant ##c##. (I'm just claiming so; I don't know if it's true.)

Then, from the last sentence of the first post,

##-\epsilon'<c-f(a+h)<\epsilon'##

Let ##g^*(h) = c-f(a+h)##. Then,

##|g^*(h)|<\epsilon'##

Thus, statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'<|g^*(h)|##.

If statement S2 is indeed true, then we have covered all the cases.

 

[Mark44]

Yes, that is the mistake, I believe.

When we say a function ##f## is continuous in the ##h##-neighbourhood of a point ##a##, the statement (S1)

##"\,##for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'"##

is true for all ##\epsilon'>0##. But if ##h## depends on ##\epsilon'##, we may run into a problem (for ##h\neq0##).

Let's prove that there is a problem for the special case when ##\delta(\epsilon)## is strictly increasing, that is, a smaller ##\epsilon## implies (or requires) a smaller ##\delta##. The other cases are not proven.

As ##\epsilon'## decreases, ##\epsilon=\frac{\epsilon'}{2}## decreases, ##\delta(\epsilon)## decreases, and ##|h|<\frac{\delta}{2}## decreases.

Let's look at a specific example. Suppose ##f(x)=x##. Then, the required ##\delta## for ##f## is ##\delta(\epsilon)=\epsilon##.

Then, ##|h|<\frac{\delta}{2}=\frac{\epsilon}{2}=\frac{\epsilon'}{4}\implies\epsilon'>4|h|##.

So for a given ##h##, statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'\leq4|h|##.

In general, for other functions for which ##\delta(\epsilon)## is strictly increasing, we have ##|h|<\frac{\delta}{2}=g(\epsilon')##, where ##g(\epsilon')## is a strictly increasing function, since it is a composition of strictly increasing functions. Thus, ##g## has an inverse ##g^{-1}## that is also strictly increasing. We have ##\epsilon'>g^{-1}(|h|)##. So statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'\leq g^{-1}(|h|)##.

This proof only covers the case when ##\delta(\epsilon)## is strictly increasing. If ##\delta(\epsilon)## is not strictly increasing, then the claim may still be true. If it's not true, then the mistake in the claim has not been fully explained.

Statement S2: If ##\delta(\epsilon)## is not strictly increasing in the neighbourhood of a point ##a##, then ##f(x)## must be a constant ##c##. (I'm just claiming so; I don't know if it's true.)

Then, from the last sentence of the first post,

##-\epsilon'<c-f(a+h)<\epsilon'##

Let ##g^*(h) = c-f(a+h)##. Then,

##|g^*(h)|<\epsilon'##

Thus, statement S1 is not true for all ##\epsilon'>0##. Explicitly, it is not true when ##\epsilon'<|g^*(h)|##.

If statement S2 is indeed true, then we have covered all the cases.

@Happiness, what is your point here? In the OP you were trying to prove that if f is continuous at a, then f is continuous at every point in an h neighborhood around a. Since this is demonstrably not true for all functions, please tell me what it is you're trying to do here.

 

[WWGD]

I think s/he is trying to understand precisely where the proof s/he proposed is wrong.

 

[Happiness]

@Happiness, what is your point here? In the OP you were trying to prove that if f is continuous at a, then f is continuous at every point in an h neighborhood around a. Since this is demonstrably not true for all functions, please tell me what it is you're trying to do here.

I think s/he is trying to understand precisely where the proof s/he proposed is wrong.

Yes, I believe that if one cannot explain why a false or wrong idea/concept is false or wrong, then one has not really understood the idea/concept in its entirety.

And in addition, if the claim is true for a special class of functions, then it is of great mathematical interest, worthy of being studied.

 

[WWGD]

I suggested that you defined ## \delta' ## in function of h, but there are no results on what h is, i.e., for a given function, what is the value of ##\delta -|h|##, given ## \delta##? But I admit I did not go through the argument carefully -enough t find the specific point where there may be a mistake. But I agree, the aument should stand/fall on its own merits.

 

[Mark44]

Yes, I believe that if one cannot explain why a false or wrong idea/concept is false or wrong, then one has not really understood the idea/concept in its entirety.

I explained exactly where you went wrong in post #26, most of which I'll post again here.

You shown a δ' (= δ - |h|), but there is no guarantee that this value actually works.

The problem seems to be that you are thinking in terms of the symbols without understanding the underlying geometry.

Let's try this. Using Lavinia's function as f, let's say that a = 1. Never mind that a here is not within a "small" neighborhood of 0. If you insist I can take a to be as close to 0 as you like, but using a = 1 makes the calculations a lot simpler, so let's go with that.

I challenge you with ε = .1. What is your choice for δ so that if x is any number such that |x - 1| < δ, then |f(x) - 1| < .1? IOW, that f(x) is in the interval (0.9, 1.1). Keep in mind that your δ has to work for every x in the interval (x - δ, x + δ).

You didn't reply to my challenge then. Can you do so now?

And in addition, if the claim is true for a special class of functions, then it is of great mathematical interest, worthy of being studied.

I doubt it. Your claim in post #1 can be boiled down to this: "If a function f is continuous at a, and is continuous at every point in an h-neighborhood around a, then f is is continuous at every point in an h-neighborhood around a." So the "special class" of functions happens to be those that are continuous throughout some interval. In this case, the theorem you're trying to prove is trivially true.

 

[Happiness]

I doubt it. Your claim in post #1 can be boiled down to this: "If a function f is continuous at a, and is continuous at every point in an h-neighborhood around a, then f is is continuous at every point in an h-neighborhood around a." So the "special class" of functions happens to be those that are continuous throughout some interval. In this case, the theorem you're trying to prove is trivially true.

The claim is "If a function ##f## is continuous at ##a##, then there exist an ##h>0## such that ##f## is continuous at every point in an ##h##-neighbourhood around ##a##." Interestingly, post #28 proves that this is not even true for some functions that are continuous for all real numbers, eg., ##f(x)=x##. The implication is false.

 

[Mark44]

The claim is "If a function ##f## is continuous at ##a##, then there exist an ##h>0## such that ##f## is continuous at every point in an ##h##-neighbourhood around ##a##."

That's pretty much what I said.

Interestingly, post 28 proves that this is not even true for some functions that are continuous for all real numbers, eg., ##f(x)=x##. The implication is false.

Which is what I've been saying for many posts.

As I said before, it seems to me that a lot of what you wrote in this thread is just manipulation of symbols, without an understanding of the geometry that those symbols represent. For your own understanding, take up the challenge I gave in post #26 and again in post #34. Working with numbers and a graph of the function will give you some insight that algebraic manipulation of symbols can't.