# Continuity on $[a,b]$ implies uniform continuity on $[a,b]$

I don't understand the step underlined in green.

I understand that for any $n$ , $|f(x_n)-f(y_n)|\geq \varepsilon$ where $x_n, y_n$ satisfy the conditions given regarding a function being not uniformly continuous. However if say $x_n=\frac{1}{2n}, y_n=\frac{1}{3n}$ then $|x_n-y_n|=\frac{1}{6n}$ and $|f(\frac{1}{2n})-f(\frac{1}{3n})|\geq \varepsilon$

so if we take $n=1$ , $|f(\frac{1}{2})-f(\frac{1}{3})|\geq \varepsilon$

but what if $|x_{k_n}-y_{k_n}|\geq\frac{1}{n}?$

• I don't understand what the issue is. By construction the sequences $x_n,y_n$ satisfy $|f(x_n)-f(y_n)|\ge\varepsilon$, $|x_n-y_n|<1/n$. (PS: I hate it when proofs wordlessly assume the axiom of (countable) choice in a proof that doesn't need it.) Jan 28, 2015 at 13:47
• Incidentally, I find it is much more natural to prove Heine-Borel and then prove Heine-Cantor (this result) from that. The proof there is quite simple: pointwise continuity furnishes a (generally uncountable) open cover of the interval, made up of the intervals $(x-\delta,x+\delta)$ whose existence is guaranteed by continuity at $x$. Heine-Borel gives a finite subcover, and now this finite subcover has a minimal value of $\delta$ that has the needed property.
– Ian
Jan 28, 2015 at 14:09
• @Mario Carneiro Say $x_{1_n}=x_1$ but $y_{1_n}=y_2$ from what I have assumed $|x_{1_n}-y_{1_n}|$ is not necessarily $< \frac{1}{n}$? Jan 28, 2015 at 14:54
• Ah, you've misread the subscripts. $x_{k_n}$ is $x_{(k_n)}$, not $(x_k)_n$. So $k_n$ is a sequence, and if $k_n=1$ then you get $x_{k_n}=x_1$, $y_{k_n}=y_1$. (You should take a clue from the TeX code you used to type that in!) Jan 28, 2015 at 14:57
• @ Mario Carneiro But $x_{k_n}$ is a subsequence of $x_n$ Jan 28, 2015 at 15:21

Your problem stems from the somewhat unfortunate circumstance that the author denotes a subsequence of the sequence $(x_n)_{n\geq1}$ by $(x_{k_n})$ instead of $(x_{n_k})_{k\geq1}$.

He begins with two sequences $(x_n)_{n\geq1}$, $(y_n)_{n\geq1}$ behaving badly insofar as $|f(x_n)-f(y_n)|\geq\epsilon_0$ for all $n$, even though $|x_n-y_n|\to0$ as $n\to \infty$. In order to be able to invoke the continuity of $f$ we'd like to have all $x_n$, $y_n$ near some point $\xi\in[a,b]$. Bolzano's theorem guarantees that there is some $\xi\in[a,b]$ such that "infinitely many" $x_n$, $y_n$ are in the immediate neighborhood of $\xi$. To be exact: There is a selection function $$\sigma:\quad{\mathbb N}\to{\mathbb N}, \qquad k\mapsto n_k$$ such that the "good" $x_n$, namely the selected $x_{n_k}$ $\>(k\geq1)$, actually converge to $\xi$. It is then easily seen that $$\lim_{k\to\infty} x_{n_k}=\lim_{k\to\infty} y_{n_k}=\xi\ .$$ Since $f$ is continuous at $\xi$ this implies $$\lim_{k\to\infty} \left(f\bigl(x_{n_k}\bigr)-f\bigl(y_{n_k}\bigr)\right)=f(\xi)-f(\xi)=0\ .$$ On the other hand, we have $|f(x_n)-f(y_n|\geq\epsilon_0$ for all $n$, good or bad, so that we arrive at a contradiction.

Complete Proof:

To obtain a contradiction assume that $f$ is not uniformly continuous on $[a,b]$. Then there exists $\epsilon _{0}>0$ such that for any $\delta >0$, there exist $x_{\delta},y_{\delta}\in [a,b]$ such that $\vert x_{\delta}-y_{\delta}\vert <\delta$ and $\vert f(x_{\delta})-f(y_{\delta})\vert \geq \epsilon _{0}$.

Therefore for any $n\in \mathbb{N}$, there exists $x_{n},y_{n}\in [a,b]$ such that $\vert x_{n}-y_{n}\vert <1/n$ and $\vert f(x_{n})-f(y_{n})\vert \geq \epsilon _{0}$. (A)

Since for any $n\in \mathbb{N}$, $x_{n}\in [a,b]$ we have $(x_{n})$ is a bounded sequence. Then by Bolzano-Weieistrass theorem, $(x_{n})$ has a convergent sub sequence $(x_{n_{k}})$. Then there exists $u\in \mathbb{R}$ such that $(x_{n_{k}})$ converges to $u$.($u\in [a,b]$ since $[a,b]$ is complete)

Now we need to show $(y_{n_{k}})$ also converges to $u$.

Let $\epsilon >0$ be given. Then there exists $K_{\epsilon}\in \mathbb{N}$ such that for each $k>K_{\epsilon}$, $\vert x_{n_{k}}-u\vert <\dfrac{\epsilon}{2}$. Observe that there exists $K_{0}\in \mathbb{N}$ such that for each $k>K_{0}$, $n_{k}>\dfrac{2}{\epsilon}$. Choose $K=\max \{K_{\epsilon},K_{0}\}$. Now let $k>K$. Then $$\vert y_{n_{k}}-u\vert =\vert y_{n_{k}}-x_{n_{k}}+x_{n_{k}}-u\vert \leq \vert y_{n_{k}}-x_{n_{k}}\vert + \vert x_{n_{k}}-u\vert <\dfrac{1}{n_{k}}+\dfrac{\epsilon}{2} < \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon .$$ Therefore $(y_{n_{k}})$ converges to $u$.

Since $f$ is a continuous function on $[a,b]$ we have that both sequences $f(x_{n_{k}})$ and $f(y_{n_{k}})$ converge to $f(u)$. Then obviously the sequence $(f(x_{n_{k}})-f(y_{n_{k}}))$ converges to $0$.

Hence there exists $K_{\epsilon _{0}}\in \mathbb{N}$ such that for each $k>K_{\epsilon _{0}}$, $$\vert f(x_{n_{k}})-f(y_{n_{k}})-0\vert = \vert f(x_{n_{k}})-f(y_{n_{k}})\vert < \epsilon _{0}.$$ Now substitute ${n_{k}}$ for $n$ in the statement (A). Then we have that $$\vert f(x_{n_{k}})-f(y_{n_{k}})\vert \geq \epsilon _{0}.$$ But this is a contradiction and hence $f$ is uniformly continuous on $[a,b]$. $\square$

Note : Since $f$ is a continuous function on $[a,b]$, the only thing we can say is there exists $\epsilon _{0}>0$ such that for any $n\in \mathbb{N}$, there exist $x_{n},y_{n}\in [a,b]$ such that $\vert x_{n}-y_{n}\vert <1/n$ and $\vert f(x_{n})-f(y_{n})\vert \geq \epsilon _{0}$. Therefore you can't put $x_{n}=\dfrac{1}{2n}$ and $y_{n}=\dfrac{1}{3n}$ and like this. Actually it is not needed for the proof.