0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @G.Sassatelli thanks edited. $\endgroup$ – IntegrateThis Aug 6 '16 at 22:01
  • $\begingroup$ @G.Sassatelli I don't understand your possible duplicate reference, are these not two sub-sequences of potentially different sequences $x_n$ and $y_n$ ?? $\endgroup$ – IntegrateThis Aug 6 '16 at 22:03
  • 1
    $\begingroup$ Good point. Vote retracted. @jamesdickens $\endgroup$ – user228113 Aug 6 '16 at 22:05
1
$\begingroup$

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 $k>k'_0$, $|x_0-y_{n_k}|\leq |x_0-x_{n_k}|+|x_{n_k}-y_{n_k}|\leq |x_0-x_{n_k}|+{1\over n_k}<c/2+c/2<c.$

$\endgroup$
  • $\begingroup$ This is a great answer, but why instinctually should I know this to be true. $\endgroup$ – IntegrateThis Aug 6 '16 at 22:04
  • 3
    $\begingroup$ Instinctively? One (sub)sequence converges, and the other is tailgating it (the difference between the two sequence goes to zero). If a car is chasing another one, getting closer and closer, and the first ends up in Toronto, then the second car also is in Toronto. $\endgroup$ – Clement C. Aug 6 '16 at 22:32
0
$\begingroup$

Is it true perhaps that

$\lim_{k\to \infty}|x_{n_k}-y_{n_k}| = \lim_{k\to \infty} \frac {1}{n_k} = 0$ thus lim $ x_{n_k} = $lim $ y_{n_k}$ ? But does this not suppose that lim$ y_{n_k}$ exists.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.