Understanding Uniform Continuity: A Guide to Formalizing Proofs

[sergey_le]

Homework Statement

Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0

Relevant Equations

Uniform continuity

There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely

 

[member 587159]

Homework Statement:: Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0
Homework Equations:: Uniform continuity

There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely

The correct term is sequences. Not series.

I think you got the right idea but it contains some mistakes.

Let $\epsilon>0$. Choose $\delta>0$ as in the definition of uniform continuity.

Choose $N$ such that $|x_n−y_n|<\delta$ if $n\geq N$.

Can you conclude?

 

[PeroK]

Homework Statement:: Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0
Homework Equations:: Uniform continuity

There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely

You must be much clearer about how you get $N, \delta, \epsilon$.

Hint: try working backwards from the way you did it:

Let $\epsilon > 0$

How do you get $|f(y_n) - f(x_n)| < \epsilon$?

Answer: if $x_n, y_n$ are close enough together. This gives you $\delta$ (depends on $\epsilon$).

How do you get $x_n, y_n$ close enough together?

Answer: by taking $n$ large enough. This gives you $N$, which depends on $\delta$.

Can you formalise that strategy?

 

[sergey_le]

You must be much clearer about how you get $N, \delta, \epsilon$.

Hint: try working backwards from the way you did it:

Let $\epsilon > 0$

How do you get $|f(y_n) - f(x_n)| < \epsilon$?

Answer: if $x_n, y_n$ are close enough together. This gives you $\delta$ (depends on $\epsilon$).

How do you get $x_n, y_n$ close enough together?

Answer: by taking $n$ large enough. This gives you $N$, which depends on $\delta$.

Can you formalise that strategy?

I understand what you mean I tried for an hour now to translate my formulation into English.
But my English is not good and it turns out really not good.
I'm not sure how I write the passage from that if |xn−yn|<δ so |f(yn)−f(xn)|<ϵ
I'd love if you could help me with that.

 

[sergey_le]

The correct term is sequences. Not series.

I think you got the right idea but it contains some mistakes.

Let $\epsilon>0$. Choose $\delta>0$ as in the definition of uniform continuity.

Choose $N$ such that $|x_n−y_n|<\delta$ if $n\geq N$.

Can you conclude?

The correct term is sequences. Not series . Thanks I will use it
Can you please help me with a formal wording: /

 

[member 587159]

The correct term is sequences. Not series . Thanks I will use it
Can you please help me with a formal wording: /

Let $\epsilon > 0$. Because $f$ is uniformly continuous, we may choose $\delta > 0$ such that for all $x$ with $|x-y|< \delta$ we have $|f(x)-f(y)|< \epsilon$. Because $\lim_n (x_n-y_n)= 0$, there is $N$ such that $|x_n-y_n|< \delta$ whenever $n \geq N$. Thus, combining the two previous statements, we get $|f(x_n)-f(y_n)|< \epsilon$ whenever $n \geq N$. Thus we have proven that for any $\epsilon>0$, there is $N$ such that $|f(x_n)-f(y_n)|< \epsilon$ whenever $n \geq N$. This is exactly the definition of $\lim_n (f(x_n)-f(y_n)) = 0$ and we are done. $\quad \square$

 

[sergey_le]

Let $\epsilon > 0$. Because $f$ is uniformly continuous, we may choose $\delta > 0$ such that for all $x$ with $|x-y|< \delta$ we have $|f(x)-f(y)|< \epsilon$. Because $\lim_n (x_n-y_n)= 0$, there is $N$ such that $|x_n-y_n|< \delta$ whenever $n \geq N$. Thus, combining the two previous statements, we get $|f(x_n)-f(y_n)|< \epsilon$ whenever $n \geq N$. Thus we have proven that for any $\epsilon>0$, there is $N$ such that $|f(x_n)-f(y_n)|< \epsilon$ whenever $n \geq N$. This is exactly the definition of $\lim_n (f(x_n)-f(y_n)) = 0$ and we are done. $\quad \square$

Thanks so much for your understanding.
This stuff is new to me and I need to get used to a good formulation