Can a function be uniformly continuous without being continuous? Help needed!

[sergey_le]

Homework Statement

Prove that if for any two sequences (xn), (yn) Which are sustained lim(yn-xn)=0 Happening
limf(yn)-f(xn)=0, Then f is uniformly continuous at R (hint: Use proof of Heine–Cantor theorem)

Relevant Equations

Heine–Cantor theorem

The first thing I thought about doing was to prove that f is continuous using the Heine–Cantor theorem proof.
But I do not know at all whether it is possible to prove with the data that I have continuous.
I would love to get help.
Thanks

 

[member 587159]

What is Heine-Cantor's theorem?

 

[sergey_le]

What is Heine-Cantor's theorem?

I meant a tease sentence that says if f is continuous in a closed section then f uniformly continuous .

 

[member 587159]

I meant a tease sentence that says if f is continuous in a closed section then f uniformly continuous .

This is wrong. You should replace "closed" with "compact". Because $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, while $\mathbb{R}$ is closed in itself.

But anyway, even if you correct that statement, I see no way to use it. I also don't understand what the word sustained does in your problem statement. The following is the correct English wording. Correct me if I'm missing something (you can relax the domains and codomains, the statement remains true in arbitrary metric spaces).

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for all sequences $(x_n)_n, (y_n)_n$ with $\lim_n (x_n-y_n)=0$ we have $\lim_n(f(x_n)-f(y_n))= 0$. Then $f$ is uniformly continuous.

Here is how I would proceed. I answered a similar post this morning of @Math Amateur , so this may be interesting for him as well.

We use the contrapositive of the statement. That is, if $f$ is NOT uniformly continuous, we prove that there are sequences $(x_n)_n, (y_n)_n$ such that $\lim_n (x_n-y_n)=0$ and $\lim_n (f(x_n)-f(y_n))\neq 0$.

Because $f$ is not uniformly continuous, we have (make sure you understand this!):
$$\exists \epsilon > 0: \forall \delta > 0: \exists x,y \in \mathbb{R}: |x-y|< \delta \land |f(x)-f(y)| \geq \epsilon$$

Fix an $\epsilon > 0$ as above. Consider the sequence $\delta_n:= 1/n, n \geq 1$. Then we can find corresponding sequences $(x_n)_n, (y_n)_n$ such that $|x_n-y_n| < \delta_n$ and $|f(x_n)-f(y_n)| \geq \epsilon$ for all $n \geq 1$.

Letting $n \to \infty$, we see that $\lim_n (x_n-y_n)=0$ while $\lim_n (f(x_n)-f(y_n)) \neq 0$ and we are done!

 

[sergey_le]

This is wrong. You should replace "closed" with "compact". Because $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, while $\mathbb{R}$ is closed in itself.

But anyway, even if you correct that statement, I see no way to use it. I also don't understand what the word sustained does in your problem statement. The following is the correct English wording. Correct me if I'm missing something (you can relax the domains and codomains, the statement remains true in arbitrary metric spaces).

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for all sequences $(x_n)_n, (y_n)_n$ with $\lim_n (x_n-y_n)=0$ we have $\lim_n(f(x_n)-f(y_n))= 0$. Then $f$ is uniformly continuous.

Here is how I would proceed. I answered a similar post this morning of @Math Amateur , so this may be interesting for him as well.

We use the contrapositive of the statement. That is, if $f$ is NOT uniformly continuous, we prove that there are sequences $(x_n)_n, (y_n)_n$ such that $\lim_n (x_n-y_n)=0$ and $\lim_n (f(x_n)-f(y_n))\neq 0$.

Because $f$ is not uniformly continuous, we have (make sure you understand this!):
$$\exists \epsilon > 0: \forall \delta > 0: \exists x,y \in \mathbb{R}: |x-y|< \delta \land |f(x)-f(y)| \geq \epsilon$$

Fix an $\epsilon > 0$ as above. Consider the sequence $\delta_n:= 1/n, n \geq 1$. Then we can find corresponding sequences $(x_n)_n, (y_n)_n$ such that $|x_n-y_n| < \delta_n$ and $|f(x_n)-f(y_n)| \geq \epsilon$ for all $n \geq 1$.

Letting $n \to \infty$, we see that $\lim_n (x_n-y_n)=0$ while $\lim_n (f(x_n)-f(y_n)) \neq 0$ and we are done!

Yes you understood my question right.
My English is not good and I translate my English as closely as I can.
It will take me some time to learn how to correctly write the correct terms in English.
So I guess you want me to ask the guy who tagged how to solve it?
By the way thank you very much for your patience :)

 

[member 587159]

Yes you understood my question right.
My English is not good and I translate my English as closely as I can.
It will take me some time to learn how to correctly write the correct terms in English.
So I guess you want me to ask the guy who tagged how to solve it?
By the way thank you very much for your patience :)

No, I gave you a proof. I suggest you take some time to understand all stems involved.

I just tagged this person because I answered a question of him that uses the same proof technique and thus the answer to this question may benefit him as well.

 

[sergey_le]

This is wrong. You should replace "closed" with "compact". Because $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, while $\mathbb{R}$ is closed in itself.

But anyway, even if you correct that statement, I see no way to use it. I also don't understand what the word sustained does in your problem statement. The following is the correct English wording. Correct me if I'm missing something (you can relax the domains and codomains, the statement remains true in arbitrary metric spaces).

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for all sequences $(x_n)_n, (y_n)_n$ with $\lim_n (x_n-y_n)=0$ we have $\lim_n(f(x_n)-f(y_n))= 0$. Then $f$ is uniformly continuous.

Here is how I would proceed. I answered a similar post this morning of @Math Amateur , so this may be interesting for him as well.

We use the contrapositive of the statement. That is, if $f$ is NOT uniformly continuous, we prove that there are sequences $(x_n)_n, (y_n)_n$ such that $\lim_n (x_n-y_n)=0$ and $\lim_n (f(x_n)-f(y_n))\neq 0$.

Because $f$ is not uniformly continuous, we have (make sure you understand this!):
$$\exists \epsilon > 0: \forall \delta > 0: \exists x,y \in \mathbb{R}: |x-y|< \delta \land |f(x)-f(y)| \geq \epsilon$$

Fix an $\epsilon > 0$ as above. Consider the sequence $\delta_n:= 1/n, n \geq 1$. Then we can find corresponding sequences $(x_n)_n, (y_n)_n$ such that $|x_n-y_n| < \delta_n$ and $|f(x_n)-f(y_n)| \geq \epsilon$ for all $n \geq 1$.

Letting $n \to \infty$, we see that $\lim_n (x_n-y_n)=0$ while $\lim_n (f(x_n)-f(y_n)) \neq 0$ and we are done!

Thank you.
And thank you for correcting my English.