I am doing my best to understand the proof given to me in my class notes. It is attached below:
Proof
We prove this by contradiction. Suppose that $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\epsilon > 0$ such that for all $n \in \mathbb{N}$ there exist $x_n, y_n$ such that $|x_n – y_n| < 1/n, |f(x_n) – f(y_n)|> \epsilon$. By the Bolzano-Weierstrass Theorem, there exist $x_{n_k}, y_{n_k}$ which converge to $x, y $ respectively. Hence, using the first inequality: $x=y$. Therefore $f(x) = f(y)$, a contradiction.
Questions:
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Firstly, in order for the $x=y$ to work, we need $\lim_{k\to\infty}(x_{n_k} -y_{n_k})= 0 $. Why is this condition true? Does the fact that $x_{n_k}$ and $y_{n_k}$ are subsequences of $x_n$ and $y_n$ imply this?
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I took a look at a similar proof on this link: http://www.math.ku.edu/~lerner/m500f09/Uniform%20continuity.pdf which seems to completely discredit the proof in my course notes, as the very first question below is: "Why, in the proof of the theorem, can’t we just take a convergent subsequence of $x_n$
and a convergent subsequence of $y_n$ and proceed directly to the conclusion?
"
P.S. I am new to Analysis and a detailed explanation would be appreciated.
Best Answer
We have $|x_{n_k}-y_{n_k}|<\frac1{n_k}$ and as $n_k\to \infty$ (i.e., because we consider subsequences), we have $\frac1{n_k}\to 0$.
Yes, just taking a subsequence of $(x_n)$ and an unrelated subsequence of $(y_n)$ might not work, as in that case the limits $x$ and $y$ might also be unrelated. Just think of $x_n=(-1)^n+\frac1{n}$ and $y_n=(-1)^n+\frac1{2n}$. If you pick all even indices for the subsequence of $(x_n)$ and the odd indices for $(y_n)$, you get $x_{n_k}=(-1)^{n_k}+\frac1{2n_k}=1+\frac1{2k}\to 1$ and $y_{m_k}=(-1)^{m_k}+\frac1{2m_k}=-1+\frac1{4k+2}\to -1$. Actually, it suffices to pick a convergent subsequence $x_{n_k}$ of $(x_{n_k})$ with $x_{n_k}\to x$. Then automatically (per the argument under 1) the corresponding subsequence $y_{n_k}$ converges to $x$ as well.