# Clarification of the proof that if $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$

Claim: If $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$.

Proof: Suppose that $f$ is not uniformly continuous on $[a,b]$, then there exists $\epsilon > 0$ such that for each $\delta > 0$ there must exist $x,y \in [a,b]$ such that $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$ .

Thus for each $n \in \Bbb N$ there exists $x_n,y_n$ such that $|x_n-y_n| < \frac {1} {n}$, and By B-W, $x_n$ has a subsequence $x_{n_k}$ with limit $x_o$ that belongs to $[a,b]$.

Now here is where I am confused, the author next states that "Clearly we also have $x_0$ is the limit of the sub-sequence $y_{n_k}$". Are $x_n$ and $y_n$ not potentially different sequences? Why would the same indexes chosen to form a sub-sequence give the same convergent value.

• @G.Sassatelli thanks edited. – IntegrateThis Aug 6 '16 at 22:01
• @G.Sassatelli I don't understand your possible duplicate reference, are these not two sub-sequences of potentially different sequences $x_n$ and $y_n$ ?? – IntegrateThis Aug 6 '16 at 22:03
• Good point. Vote retracted. @jamesdickens – user228113 Aug 6 '16 at 22:05

For every $c>0$, there exists $k_0$ such that $k>k_0$ implies that $|x_0-x_{n_k}|<c/2$, we choose $k'_0>k_0$ such that ${1\over n_{k'_0}}<c/2$. Let $k>k'_0$, $|x_0-y_{n_k}|\leq |x_0-x_{n_k}|+|x_{n_k}-y_{n_k}|\leq |x_0-x_{n_k}|+{1\over n_k}<c/2+c/2<c.$
$\lim_{k\to \infty}|x_{n_k}-y_{n_k}| = \lim_{k\to \infty} \frac {1}{n_k} = 0$ thus lim $x_{n_k} =$lim $y_{n_k}$ ? But does this not suppose that lim$y_{n_k}$ exists.