Continuity at a point implies continuity in the neighborhood

[Happiness]

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.

Say, $\epsilon'=2$, then the required $\delta'$ can be 0.9, and the point in the $h$-neighbourhood of $a$ (which is 0) can be 0.1.

$|x-0.1|<0.9 \implies |f(x)-0.01|<2$

and it holds for all the $x$ in the range (-0.8, 1).

 

[Mark44]

Say, $\epsilon'=2$, then the required $\delta'$ can be 0.6

You don't get to pick ε -- I do, and I chose it to be 0.1.
Tell me the δ that works.

 

[Mark44]

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.

 

[Happiness]

You don't get to pick ε -- I do, and I chose it to be 0.1.
Tell me the δ that works.

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.

$|x-0.1|<\sqrt{0.05}-0.1 \implies |f(x)-0.01|<0.1$

and it holds for all the $x$ in the range $(-\sqrt{0.05}+0.2, \sqrt{0.05})$.

 

[Happiness]

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, this is not required. The claim just states that there exist an h>0. It doesn't say that h must be 1.

 

[Mark44]

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.

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]

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.

$a=0$. $a$ 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]

No, this is not required. The claim just states that there exist an h>0. It doesn't say that h must be 1.

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).

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).

Disclosure: the last interval was originally (x- δ, x + δ). I changed it to (1 - δ, 1 + δ).

 

[Happiness]

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).

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]

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.)

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)

 

[Happiness]

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)

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 to find such a $\delta'$ for any $\epsilon'$ is given in the first post.

We are not trying to disprove the claim. That has been done. We are trying to appreciate the paradox, and figure out why the argument in the claim is invalid/unsound.

 

[Mark44]

I'm not trying to prove my claim. If you read the first post carefully, you will appreciate the paradox it brings out.

There is no paradox. Your claim is simply not true.

The algorithm to find such a $\delta'$ for any $\epsilon'$ is given in the first post.

The algorithm makes no sense. There is no general "one size fits all" algorithm that can be used.

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.