$f:[a,b] \rightarrow R$ is continuous, then $f$ is uniformly continuous

Question

$f:[a,b] \rightarrow R$ is continuous, then $f$ is uniformly continuous.

Proof from my lecture:

Suppose $f$ is not uniformly continuous. Then, $\exists \varepsilon$ such that $\forall \delta$, $\exists x,y \in [a,b]$ with $|x-y|<\delta$, and $|f(x)-f(y)|<\varepsilon$.

Let $\delta = 1/n$ for $n\in Z^+$, and $x_n,y_n \in [a,b]$. Then, $|x_n-y_n|<1/n.$ By Bolzano theorem, there exists a subsequence $x_{n_k}$ such that $\lim_{k \to \infty} x_{n_k} = x$. In addition,$y_{n_k}$ has a subsequence $y_{n_{k_l}}$ such that $\lim_{l \to \infty} y_{n_{k_l}}=y.$

Then, $|x_{n_{k_l}}-y_{n_{k_l}}|<1/n_{k_l}\le 1/l$.So if we take a limit from $l$ to $\infty$, we get $x=y$.

My question is why do we use $y_{n_{k_l}}$ instead of $y_{n_{k}}$? Since the domain is bounded, $y_{n_k}$ is also convergent subsequence, but my lecturer uses another subsequence of a subsequence and I feel it makes the proof more complicated. But, I guess there must be a reason for this. Could you tell me what am I missing here?

Thank you in advance.

Answer 1

Based on the comments below your question, it looks like it boils down to a confusion about what Bozano's theorem guarantees.

Every real bounded sequence $(a_n)_n$ has a convergent subsequence $(a_{n_k})_k$.

This is not saying that if I get two bounded sequences $(a_n)_n$ and $(b_n)_n$, by applying Bolzano to each of them separately I will get the same sequence of indices $n_1 < \dots < n_k<\dots$ such that $(a_{n_k})_k$ and $(b_{n_k})_k$ are both convergent. The subsequence of indices may not be the same, and that is why your teacher applies Bolzano's theorem to a subsequence of $(y_n)_n$ instead of directly to $(y_n)_n$. (But you seem to be confused by the notation $(n_k)_k$ -- it may be worth going and see what it actually means, i.e., the definition of a subsequence.)

An example: the sequences $(a_n)_n$ and $(b_n)_n$ defined by $a_n = (-1)^n$ and $b_{2n}=b_{2n+1}=(-1)^n$ are both bounded. If you apply Boznao's theorem to the first, you may get $(n_k)_k = (2k)_k$ (even indices), as $(a_{2n})_n$ is a constant and hence convergent subsequence of $(a_n)_n$.

But $(b_{2n})_n$ is not a convergent subsequence of $(b_n)_n$. Appplying Bolzano to $(b_n)_n$ may have given you another sequence of indices, e.g. $(n_k)_k = (4k)_k$ (as $(b_{4n})_n$ is a convergent subsequence of $(b_n)_n$).

Answer 2

You want to get two sequences $x_n,y_n$ with the properties $|x_n-y_n|<1/n,|f(x_n)-f(y_n)|>\varepsilon,\lim_{n \to \infty} x_n=x,\lim_{n \to \infty} y_n=y$. From this, you could deduce $x=y$, which would allow you to derive a contradiction.

Why don't you have this immediately? Because $x_n$ and $y_n$ don't have to converge. You can use sequential compactness to obtain subsequences $x_{f(n)},y_{g(n)}$ that converge, where $f,g$ are strictly increasing functions from $\mathbb{N}$ to $\mathbb{N}$. That's all well and good, but $f(n) \neq g(n)$ in general so you don't know $|x_{f(n)}-y_{g(n)}|$ is small anymore. What you want to do instead is to pick the same subsequence of indices for both of them. Proceeding this way, you're stuck: there is no guarantee that $f(n)$ and $g(n)$ have any terms in common, so there is no way to extract a common subsequence of both of them to finish the problem.

The workaround is to first find $f(n)$ with $x_{f(n)}$ converging, and then look at the sequence $y_{f(n)}$. You can then find a subsequence $y_{h(f(n))}$ that converges. Since $x_{f(n)}$ converges, so does $x_{h(f(n))}$, so now $x_{h(f(n))}$ and $y_{h(f(n))}$ have the properties we wanted in the first paragraph.