# Are continuous functions on sequentially compact sets u-continuous?

• B
• Eclair_de_XII
In summary: Also, "by construction" is a phrase that you use when you want to highlight that you have used some inductive construction to create a sequence. That is not what you have done here. You are constructing something here, but it's not a sequence. You have constructed a proof.
Eclair_de_XII
TL;DR Summary
(1) ##f## is continuous
(2) ##f## is defined on a sequentially compact interval ##K##

Prove that ##f## is uniformly continuous.
Suppose ##f## is not uniformly-continuous. Then there is ##\epsilon>0## such that for any ##\delta>0##, there is ##x,y\in K## such that if ##|x-y|<\delta##, ##|f(x)-f(y)|\geq \epsilon##.

Choose ##\delta=1##. Then there is a pair of real numbers which we will denote as ##x_1,y_1## such that if ##|x_1-y_1|<\delta##, uniform-continuity fails to hold.

Now choose ##\delta=\frac{1}{2}## and find a pair of real numbers ##x_2,y_2## that also fail this condition. Choose a sequence of real numbers in this fashion and denote them as: ##\{x_n\},\{y_n\}##

These are sequences contained in ##K##. Therefore, for ##\{x_n\}##, there exists a subsequence ##x_{n_k}\rightarrow z\in K##.

Moreover, if ##\epsilon>0##, choose ##N_\epsilon\in\mathbb{N}## that is at least ##\frac{1}{\epsilon}##. Then if ##k\geq N_\epsilon##, ##n_k\geq n_{N_\epsilon}##. Note that ##n_{N_\epsilon}\geq \frac{1}{\epsilon}##. Hence, by construction:

##|x_{n_k}-y_{n_k}|<\frac{1}{n_k}\leq \epsilon##

By the Triangle Inequality, ##y_{n_k}\rightarrow z##.
By the Triangle Inequality, ##|f(y_{n_k})-f(x_{n_k})|<\epsilon##, which contradicts the initial claim that every pair of images of the points ##x_i,y_i## constructed thus are not Cauchy. Hence, ##f## must be uniformly-continuous.

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Eclair_de_XII said:
Suppose ##f## is not uniformly-continuous. Then there is ##\epsilon>0## such that for any ##\delta>0##, there is ##x,y\in K## such that if ##|x-y|<\delta##, ##|f(x)-f(y)|\geq \epsilon##.

I think that you want to write is for any ##\delta>0##, there is ##x,y\in K## such that both ##`|x-y|<\delta## and ##|f(x)-f(y)| \geq \epsilon##.

I don't think this greatly affects what you write afterwards, but from what you wrote anyone can prove a function is not uniformly continuous by just picking x and y sufficiently far away.

I don't understand what triangle inequality you are using to prove that ##|f(x_{n_k}) - f(y_{n_k})| < \epsilon##.

Office_Shredder said:
I don't think this greatly affects what you write afterwards, but from what you wrote anyone can prove a function is not uniformly continuous by just picking x and y sufficiently far away.

I don't understand what triangle inequality you are using to prove that ##|f(x_{n_k})-f(y_{n_k})|<\epsilon##.
I see. Thank you for the tip.

I have trouble interpreting mathematical statements sometimes. Although, it's actually more often than not that this is the case. In fact, I feel like it is the one of the main difficulties that stymie my progress when trying to figure out math problems.

As for the other thing (note that I replaced all the ##x_{n_k}## with ##x'_k## for simplicity's sake):

(1) There is ##N_z## and ##N_\epsilon## such that if ##k\geq N_z,N_\epsilon##:

##x'_k\rightarrow z##
##|x'_k-y'_k|\rightarrow 0##.

Let ##\epsilon>0## and choose ##N=\sup\{N_z,N_\epsilon\}## such that if ##k\geq N##:

##|x'_k-y'_{k}|<\epsilon/2##
##|z-x'_k|<\epsilon/2##

The last inequality implies ##f(x'_k)\rightarrow f(z)## by continuity.

By the Triangle Inequality, ##|y'_k-z|<\epsilon##

Hence, ##f(y'_k)\rightarrow f(z)## by continuity, which is something I think I neglected to mention.

(2) Since ##f## is continuous:

##|f(y'_k)-f(z)|<\epsilon/2##
##|f(z)-f(x'_k)|<\epsilon/2##

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In your post, ##\epsilon## is a measure of how far apart ##x_k## and ##y_k## are. Most versions of continuity do not use the same ##\epsilon## to also measure how far apart ##f(x_k)## and ##f(y_k)##

Should I replace the measure of how far apart ##y_k## and ##x_k## are with a different letter, like ##\delta##, then?

Let ##\epsilon>0##. Also, this ##\delta## can be basically any positive number, including the ##\delta## such that if ##x_k,y_k\in K## have the property ##|x_k-y_k|<\delta##, then ##|f(x_k)-f(y_k)|<\epsilon##.

Eclair_de_XII said:
Should I replace the measure of how far apart ##y_k## and ##x_k## are with a different letter, like ##\delta##, then?

Let ##\epsilon>0##. Also, this ##\delta## can be basically any positive number, including the ##\delta## such that if ##x_k,y_k\in K## have the property ##|x_k-y_k|<\delta##, then ##|f(x_k)-f(y_k)|<\epsilon##.
In your original post, you used the same letter ##\epsilon## for two different parts of the proof, where these numbers are not related. You should choose something different for the second ##\epsilon##.

##x_k, y_k## are two specific points that you have found. You have, therefore, an existential qualifier, not a universal if-then.

Eclair_de_XII said:
Suppose ##f## is not uniformly-continuous. Then there is ##\epsilon>0## such that for any ##\delta>0##, there is ##x,y\in K## such that if ##|x-y|<\delta##, ##|f(x)-f(y)|\geq \epsilon##.
The correct phrase is "such that ##|x - y| < \delta## and ##|f(x) - f(y)| \geq \epsilon##" since a statement of the form "if ##|x - y| < \delta ## then..." is True when ##|x - y | \geq \delta##.
Eclair_de_XII said:
Choose a sequence of real numbers in this fashion and denote them as: ##\{x_n\},\{y_n\}##
You want to chose two sequences (plural) , not "a" sequence (singular).
By "In this fashion" you mean that ##(x_i,y_i)## is a pair of real numbers chosen such that for ##\delta_k = \frac{1}{k}## , ##|x_k - y_k| \leq \delta_k## and ##|f(x_k) - f(y_k) | \geq \epsilon##.
Eclair_de_XII said:
These are sequences contained in ##K##. Therefore, for ##\{x_n\}##, there exists a subsequence ##x_{n_k}\rightarrow z\in K##.

Eclair_de_XII said:
Moreover, if ##\epsilon>0##, choose ##N_\epsilon\in\mathbb{N}## that is at least ##\frac{1}{\epsilon}##.
We are assuming ##\epsilon > 0## so you should not say "if ##\epsilon > 0##". It's better to say: Since ##\epsilon > 0## there exists a positive integer ##N_e## such that ##N_e > 1 /\epsilon##.

Eclair_de_XII said:
Then if ##k\geq N_\epsilon##, ##n_k\geq n_{N_\epsilon}##. Note that ##n_{N_\epsilon}\geq \frac{1}{\epsilon}##. Hence, by construction:
You haven't defined "##n_k##" and "##n_{N_\epsilon}##". Assuming ##k## is an index, then ##x_k## and ##y_k## are terms of sequences. How are the ##n_i## defined?

Stephen Tashi said:
Assuming ##k## is an index, then ##x_k## and ##y_k## are terms of sequences. How are the ##n_i## defined?
Well, in the opening post, I wrote an algorithm stating how to get each ##x_n##. By sequential compactness, there existed a subsequence ##x_{n_k}## that converged to some number ##z##. Then later in post #3, I relabeled this subsequence ##x'_k## and sort of neglected to maintain that notation. In any case, they are the indices of the subsequence ##x_{n_k}##.

Eclair_de_XII said:
Well, in the opening post, I wrote an algorithm stating how to get each ##x_n##. By sequential compactness, there existed a subsequence ##x_{n_k}## that converged to some number ##z##. Then later in post #3, I relabeled this subsequence ##x'_k## and sort of neglected to maintain that notation. In any case, they are the indices of the subsequence ##x_{n_k}##.

You still haven't defined ##n_k## and ##n_N## except to say they are indices of sequences ##\{x'\}, \{y'\}##. Do ##n_k## and ##n_N## have special properties? Or are they arbitrary indices?

Let's say we have a sequence ##a_n## defined by ##n## for ##n\in\mathbb{N}##.

Define a subsequence ##a_{n_k}## by choosing all the odd integers. The first odd integer is ##1##, so ##a_{n_1}=1##. The second odd integer is ##3## so ##a_{n_2}=3##.

In general, ##\{a_{n_k}\}## represents the subsequence of ##\{a_n\}## obtained by selecting certain terms of the latter sequence, starting from the first element. The sub-sub-script represents the term's index in the new subsequence.

Sorry. This notation was sort of the norm in my real analysis class, so I did not consider the possibility that it would not be universally understood by the mathematicians who did not take the class with me.

I think Stephen probably misread something in the original post, I think your sequences are well defined. The only thing you missed in your attempts was a correct epsilon-delta proof of the fact that ##|f(x_{n_k})-f(y_{n_k})|## gets small enough - I think you understand why intuitively for the right reason, but you haven't correctly proven it in this thread.

PeroK
Office_Shredder said:
I think you understand why intuitively for the right reason, but you haven't correctly proven it in this thread.
Let me give it a go, then.

Let ##f## be a continuous function defined on a sequentially compact set ##K##. Suppose ##f## is not uniformly continuous. Then there is a positive ##\epsilon'## such that for all positive ##\delta##, there are points ##x_\delta,y_\delta\in K## such that even though ##|x_\delta-y_\delta|<\delta##, ##|f(x_\delta)-f(y_\delta)|\geq \epsilon'##.

There are points ##x_1,y_1\in K## such that ##|x_1-y_1|<1## and ##|f(x_1)-f(y_1)|\geq \epsilon'##. Moreover, there are points ##x_2,y_2## with the property that ##|x_2-y_2|<1/2## and ##|f(x_2)-f(y_2)|\geq \epsilon'##. Construct two sequences of points ##\{x_n\},\{y_n\}\subset K## in this fashion.

Since ##\{x_n\}## is contained within a sequentially compact set, it follows by definition that there is a subsequence ##\{x_{n_k}\}## that converges to some ##z\in K##.

Let ##\delta'>0## and ##\epsilon>0##.

Then there is a natural number ##N_z## with the property that whenever ##n_k\geq N_z##, ##|x_{n_k}-z|<\delta'/2##.

Since ##\{y_n\}## is contained within a sequentially compact set, there is a subsequence ##\{y_{n_k}\}## that converges to some number in ##K##. We show that this number is ##z## by first proving that ##|y_{n_k}-x_{n_k}|\rightarrow 0##, then using the Triangle Inequality.

Choose ##N'>2/\delta'##. Then whenever ##n_k\geq N'##:

##|x_{n_k}-y_{n_k}|<1/n_k\leq 1/N'<\delta'/2##

Now choose ##N=\max\{N',N_z\}##. Then whenever ##n_k\geq N##:

##|x_{n_k}-y_{n_k}|<\delta'/2##
##|x_{n_k}-z|<\delta'/2##

\begin{align*}
\delta'&>&|x_{n_k}-y_{n_k}|+|z-x_{n_k}|\\
&\geq&|(x_{n_k}-y_{n_k})+(z-x_{n_k})|\\
&=&|-y_{n_k}+z|
\end{align*}

Since ##f## is continuous, it follows that:

##|f(y_{n_k})-z|<\epsilon/2##
##|-f(x_{n_k})+z|<\epsilon/2##

Hence:

\begin{align*}
\epsilon&>&|-f(x_{n_k})+z|+|f(y_{n_k})-z|\\
&\geq&|(-f(x_{n_k})+z)+(f(y_{n_k})-z)|\\
&=&|-f(x_{n_k})+f(y_{n_k})|
\end{align*}

contrary to the assumption that none of the images of ##x_i,y_i## through ##f## can be as close together as desired.

Hence, ##f## must be uniformly continuous.

I would use the finite intersection property of compact sets.

Eclair_de_XII said:
In general, ##\{a_{n_k}\}## represents the subsequence of ##\{a_n\}## obtained by selecting certain terms of the latter sequence, starting from the first element. The sub-sub-script represents the term's index in the new subsequence.
Perhaps what you are trying to say is that the term with index ##N_k## in subsequence ##\{x'\}## of sequence ##\{x\}## is defined to be the term that has index ##k## in the sequence ##\{x\}##.

Eclair_de_XII said:
Let ##\delta'>0## and ##\epsilon>0##.
This should give you pause. There's no relation between ##\epsilon## and ##\delta'##?

Eclair_de_XII said:
Since ##\{y_n\}## is contained within a sequentially compact set, there is a subsequence ##\{y_{n_k}\}## that converges to some number in ##K##. We show that this number is ##z## by first proving that ##|y_{n_k}-x_{n_k}|\rightarrow 0##, then using the Triangle Inequality.
I think you can just delete that sentence about sequential compactness. It hands you a random subsequence that might not converge to z. I agree ##y_{n_k}## converges to z and I agree with you proof of why, but it has nothing to do with the fact that the space is sequentially compact.

At the end, ##\epsilon## is literally an arbitrary number that has no relation to anything else you have done, so I don't think this is true. Also, you said you were going to show this is smaller than ##\epsilon'##, not ##\epsilon##.

You probably need to use the fact that f is continuous at z, which I suspect is what you were hinting at but didn't really do anything with.

PeroK
Eclair_de_XII said:
Let me give it a go, then.

Let ##f## be a continuous function defined on a sequentially compact set ##K##. Suppose ##f## is not uniformly continuous. Then there is a positive ##\epsilon'## such that for all positive ##\delta##, there are points ##x_\delta,y_\delta\in K## such that even though ##|x_\delta-y_\delta|<\delta##, ##|f(x_\delta)-f(y_\delta)|\geq \epsilon'##.

There are points ##x_1,y_1\in K## such that ##|x_1-y_1|<1## and ##|f(x_1)-f(y_1)|\geq \epsilon'##. Moreover, there are points ##x_2,y_2## with the property that ##|x_2-y_2|<1/2## and ##|f(x_2)-f(y_2)|\geq \epsilon'##. Construct two sequences of points ##\{x_n\},\{y_n\}\subset K## in this fashion.

Since ##\{x_n\}## is contained within a sequentially compact set, it follows by definition that there is a subsequence ##\{x_{n_k}\}## that converges to some ##z\in K##.

Let ##\delta'>0## and ##\epsilon>0##.

Then there is a natural number ##N_z## with the property that whenever ##n_k\geq N_z##, ##|x_{n_k}-z|<\delta'/2##.

Since ##\{y_n\}## is contained within a sequentially compact set, there is a subsequence ##\{y_{n_k}\}## that converges to some number in ##K##. We show that this number is ##z## by first proving that ##|y_{n_k}-x_{n_k}|\rightarrow 0##, then using the Triangle Inequality.

Choose ##N'>2/\delta'##. Then whenever ##n_k\geq N'##:

##|x_{n_k}-y_{n_k}|<1/n_k\leq 1/N'<\delta'/2##

Now choose ##N=\max\{N',N_z\}##. Then whenever ##n_k\geq N##:

##|x_{n_k}-y_{n_k}|<\delta'/2##
##|x_{n_k}-z|<\delta'/2##

\begin{align*}
\delta'&>&|x_{n_k}-y_{n_k}|+|z-x_{n_k}|\\
&\geq&|(x_{n_k}-y_{n_k})+(z-x_{n_k})|\\
&=&|-y_{n_k}+z|
\end{align*}

Since ##f## is continuous, it follows that:

##|f(y_{n_k})-z|<\epsilon/2##
##|-f(x_{n_k})+z|<\epsilon/2##

Hence:

\begin{align*}
\epsilon&>&|-f(x_{n_k})+z|+|f(y_{n_k})-z|\\
&\geq&|(-f(x_{n_k})+z)+(f(y_{n_k})-z)|\\
&=&|-f(x_{n_k})+f(y_{n_k})|
\end{align*}

contrary to the assumption that none of the images of ##x_i,y_i## through ##f## can be as close together as desired.

Hence, ##f## must be uniformly continuous.
I think you miss details and use sloppy notation . For example, you find a convergent subsequent of ##x_n##, then you find a convergent subsequence of ##y_n##, but you don't notice that maybe these two subsequences have no terms in common. For example, we could have:
$$x_n = 1, 2, 1, 2, 1, 2 \dots$$ and $$y_n = 3/2, 5/2, 5/4, 9/4, 9/8, 17/8 \dots$$ I.e. both sequences have two convergent subsequences. You might pick ##x_{n_k} = 1, 1, 1 \dots## and ##y_{n_k} = 5/2, 9/4, 17/8 \dots## and the two subsequences have no terms in common and converge to different points.

Then, you paper over this by assuming that ##n_k## is the same sequence in both cases.

Instead, what you should have noticed is that once you have picked ##x_{n_k}## you must use the same terms for ##y_{n_k}## and prove that ##y_{n_k}## converges to the same point as ##x_{n_k}##.

You had the same general problem in your first attempt, where you reused ##\epsilon## in two different contexts.

My advice would be to try to write down clearly how your proof works. Leave the rigour until you have a working strategy. For example, what I would do is something like:

1) Assume that ##f## is not uniformly continuous.

2) Generate two sequences ##x_n, y_n## that get arbitrarily close while their function values do not.

3) Use sequential compactness to find a subsequence ##x_{n_k}## that converges to some point ##z##.

4) Show that ##y_{n_k}## converges to the same point ##z##.

5) Use the continuity of ##f## at ##z## to show that ##f(x_{n_k})## and ##f(y_{n_k})## must converge to ##f(z)##.

6) Show that ##f(x_{n_k})## and ##f(y_{n_k})## must get arbitrarily close - which contradicts point 2).

7) Hence assumption 1) must be false and ##f## is uinformly continuous.

IMO, you need this clear strategy in your head before you start epsilon-delta'ing. Otherwise, you keep getting lost in the woods.

Let ##f## be a continuous function defined on a sequentially compact set ##K##. Suppose ##f## is not uniformly continuous. By definition, there is ##\epsilon>0## such that for any ##\delta>0##, there are points ##x,y\in K## such that ##|x-y|<\delta## but ##|f(x)-f(y)|\geq \epsilon##.

In particular, choose ##\delta=1## and find points ##x_1,y_1\in K## with the property that ##|x_1-y_1|<1## and ##|f(x_1)-f(y_1)|\geq \epsilon##. Now choose ##\delta=\frac{1}{2}##, then find points ##x_2,y_2\in K## with the property that ##|x_2-y_2|<\frac{1}{2}## and ##|f(x_2)-f(y_2)|\geq\epsilon##.

Continuing in this fashion, we obtain two subsequences ##\{x_n\},\{y_n\}## with the property that for any positive number ##\delta##, we can choose any integer ##N>\frac{1}{\delta}## in order to ensure that ##|x_N-y_N|<\frac{1}{N}<\delta##.

Since ##K## is sequentially compact, it follows that there must exist a subsequence of ##\{x_n\}##, which we shall denote as ##\{x_{n_k}\}##, that converges to some ##z\in K##. Now for each ##x_{n_k}##, choose ##y_{n_k}## such that ##|x_{n_k}-y_{n_k}|<\frac{1}{m}##, where ##m## is the index of ##x_{n_k},y_{n_k}## in their respective parent sequences.

This gives us a sequence ##\{y_{n_k}\}## with the property that ##|y_{n_k}-x_{n_k}|\rightarrow 0##. Moreover, since a given ##y_{n_k}## gets arbitrarily close to ##x_{n_k}##, which gets arbitrarily close to ##z##, it follows that this ##y_{n_k}## must get arbitrarily close to ##z##, as well.

Let ##\epsilon'>0##. Then there is a ##\delta'>0## such that for all ##x\in K## (in particular, the terms in the sequence constructed above) with the property that whenever ##|x-z|<\delta'##, it follows that ##|f(x)-f(z)|<\frac{\epsilon'}{2}##.

There is an integer ##N_1## such that if ##n_k\geq N_1: |z-y_{n_k}|<\delta'##
There is an integer ##N_2## such that if ##n_k\geq N_2: |x_{n_k}-z|<\delta'##

Choose ##N\equiv\sup\{N_1,N_2\}## such that if ##n_k\geq N##:

Hence,
\begin{align*}
|f(x_{n_k})-f(z)|<\frac{\epsilon'}{2}\\
|f(z)-f(y_{n_k})|<\frac{\epsilon'}{2}
\end{align*}

It follows that:

\begin{align*}
\epsilon'&>&|f(z)-f(y_{n_k})|+|f(x_{n_k})-f(z)|\\
&\geq&|[f(z)-f(y_{n_k})]+[f(x_{n_k})-f(z)]|\\
&=&|-f(y_{n_k})+f(x_{n_k})|
\end{align*}

contrary to the assumption that ##|f(x_n)-f(y_n)|\nrightarrow 0## for all ##x_n,y_n##.

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This all looks good, except you used ##\epsilon## at the very start then switched to an unrelated ##\epsilon'## at the bottom. You just need to not rename it.

Eclair_de_XII
Thanks for all the feedback, everyone.

## 1. What is a continuous function?

A continuous function is a mathematical function that preserves the relationship between input and output values. This means that small changes in the input result in small changes in the output, and the function does not have any sudden jumps or gaps in its graph.

## 2. What does it mean for a set to be sequentially compact?

A set is sequentially compact if every sequence within the set has a convergent subsequence. This means that the sequence has a limit point within the set.

## 3. What is the definition of u-continuous?

U-continuous is a type of continuity that applies to functions on sequentially compact sets. It means that for every sequence in the set, the limit of the function at the limit point of the sequence is equal to the value of the function at that point.

## 4. How is u-continuity different from other types of continuity?

U-continuity is a stronger form of continuity compared to other types, such as pointwise continuity. It requires the function to be continuous not just at individual points, but also at the limit points of sequences within the set.

## 5. What are some examples of u-continuous functions on sequentially compact sets?

Some examples of u-continuous functions on sequentially compact sets include the identity function, polynomial functions, and trigonometric functions. These functions have the property that the limit of the function at the limit point of any sequence within the set is equal to the value of the function at that point.

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