Are continuous functions on sequentially compact sets u-continuous?

[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.

 

[Office_Shredder]

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 $|f(x_{n_k}) - f(y_{n_k})| < \epsilon$.

 

[Eclair_de_XII]

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:

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

 

[Office_Shredder]

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 $f(y_k)$

 

[Eclair_de_XII]

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 $|f(x_k)-f(y_k)|<\epsilon$.

 

[PeroK]

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.

 

[Stephen Tashi]

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?

 

[Eclair_de_XII]

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 $x_{n_k}$.

 

[Stephen Tashi]

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?

 

[Eclair_de_XII]

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.

 

[Office_Shredder]

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.

 

[Eclair_de_XII]

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.

 

[Svein]

I would use the finite intersection property of compact sets.

 

[Stephen Tashi]

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 $\{x\}$.

 

[Office_Shredder]

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.

 

[PeroK]

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.

 

[Eclair_de_XII]

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 $x_n,y_n$.

 

[Office_Shredder]

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.