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

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  • $\begingroup$ 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.) $\endgroup$ Jan 28, 2015 at 13:47
  • $\begingroup$ 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. $\endgroup$
    – Ian
    Jan 28, 2015 at 14:09
  • $\begingroup$ @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 $< \frac{1}{n}$? $\endgroup$ Jan 28, 2015 at 14:54
  • $\begingroup$ 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!) $\endgroup$ Jan 28, 2015 at 14:57
  • $\begingroup$ @ Mario Carneiro But $x_{k_n}$ is a subsequence of $x_n$ $\endgroup$ Jan 28, 2015 at 15:21

2 Answers 2

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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.

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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.

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