Are continuous functions on sequentially compact sets u-continuous?

  • #1
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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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  • #2
Office_Shredder
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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##.
 
  • #3
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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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  • #4
Office_Shredder
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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)##
 
  • #5
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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##.
 
  • #6
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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.
 
  • #7
Stephen Tashi
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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##.
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##.


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}##.
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##.

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?


 
  • #8
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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}##.
 
  • #9
Stephen Tashi
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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?
 
  • #10
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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.
 
  • #11
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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.
 
  • #12
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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.
 
  • #13
Svein
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I would use the finite intersection property of compact sets.
 
  • #14
Stephen Tashi
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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\}##.
 
  • #15
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Let ##\delta'>0## and ##\epsilon>0##.
This should give you pause. There's no relation between ##\epsilon## and ##\delta'##?

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.
 
  • #16
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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.
 
  • #17
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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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  • #18
Office_Shredder
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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.
 
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  • #19
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Thanks for all the feedback, everyone.
 

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