Mark44 said:@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 said:I think s/he is trying to understand precisely where the proof s/he proposed is wrong.
I explained exactly where you went wrong in post #26, most of which I'll post again here.Happiness said: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.
You didn't reply to my challenge then. Can you do so now?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 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 said: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.
Mark44 said: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.
That's pretty much what I said.Happiness said: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##."
Which is what I've been saying for many posts.Happiness 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.
Mark44 said:That's pretty much what I said.
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.
You don't get to pick ε -- I do, and I chose it to be 0.1.Happiness said:Say, ##\epsilon'=2##, then the required ##\delta'## can be 0.6
Mark44 said:You don't get to pick ε -- I do, and I chose it to be 0.1.
Tell me the δ that works.
Mark44 said:Just to be completely clear, whatever δ you pick, every point in the interval (1 - δ, 1 + δ) has to map to the interval (0.9, 1.1) on the y-axis, using the function of Lavinia's counterexample.
No, a = 1, which is what I wrote in post #26 and reposted in #33. The question is not whether the function is continuous at 0, but whether we can extend that continuity a little away from 0.Happiness said:Say, ##\epsilon'=0.1##, then the required ##\delta'## can be ##\sqrt{0.05}-0.1\approx0.123607##, and the point in the ##h##-neighbourhood of ##a## (which is 0) can be 0.1.
##a=0##. ##a## is the point from which we extend continuity. And I ##seemingly## extended continuity to the point ##x=0.1##. But this is false, as explained in post #28.Mark44 said:No, a = 1, which is what I wrote in post #26 and reposted in #33. The question is not whether the function is continuous at 0, but whether we can extend that continuity a little away from 0.
You are misunderstanding. I don't care about showing that the function is continuous at x = 0. It is. The question is whether we can say that the function is continuous a little ways away from 0.Happiness said:No, this is not required. The claim just states that there exist an h>0. It doesn't say that h must be 1.
Disclosure: the last interval was originally (x- δ, x + δ). I changed it to (1 - δ, 1 + δ).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 (1 - δ, 1 + δ).
Mark44 said:You are misunderstanding. I don't care about showing that the function is continuous at x = 0. It is. The question is whether we can say that the function is continuous a little ways away from 0.
Here's the challenge again (third time's the charm).
Disclosure: the last interval was originally (x- δ, x + δ). I changed it to (1 - δ, 1 + δ).
What I asked you to do three times was to prove the function was continuous at x = 1. You did not prove that the function was continuous at x = 1 or even at x = 0.1.Happiness said:I just proved that the function is continuous at ##x=0.1## and you care about showing that the function is continuous at ##x=0.1##. (But the proof is wrong. See post #28.)
Mark44 said:What I asked you to do three times was to prove the function was continuous at x = 1. You did not prove that the function was continuous at x = 1 or even at x = 0.1.
Since you have changed the goal posts on me, let's see if you can convince me that f is continuous at x = 0.1.
Here a = 0.1 and f(a) = 0.01
I choose ε = 0.005.
What is δ here? Whatever you choose for δ has to be such that for all x ∈ (.1 - δ, .1 + δ), then f(x) ∈ (.005, .015)
There is no paradox. Your claim is simply not true.Happiness said:I'm not trying to prove my claim. If you read the first post carefully, you will appreciate the paradox it brings out.
The algorithm makes no sense. There is no general "one size fits all" algorithm that can be used.Happiness said:The algorithm to find such a ##\delta'## for any ##\epsilon'## is given in the first post.
Happiness said:We are not trying to disprove the claim. That has been done. We are trying to appreciate the paradox, and figure out why the claim is wrong.