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

$f:[a,b] \rightarrow R$ 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.

• Why would $(y_{n_k})_k$ be convergent "since the domain is bounded"? Bounded sequences do not converge in general, that's the point of Bolzano's theorem: bounded sequences have a subsequence which is convergent. Apr 25, 2018 at 0:41
• $y_n$ is bounded on $[a,b]$, which means $y_{n_k}$ is convergent by Bolzano's theorem. Is it wrong? Apr 25, 2018 at 0:46