1
$\begingroup$

Bit stuck on this one so would appreciate any help.

Let $f : (0, 1)\to\mathbb R$ be uniformly continuous and $(x_n)$ and $(y_n)$ be sequences in $(0, 1)$ where $$\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = 0.$$

Show that $$\lim_{n\to\infty} f(x_n) − f(y_n) = 0.$$

Think I will have to use the triangle inequality but struggling to see where.

$\endgroup$
2
$\begingroup$

$f$ is uniformly continuous implies that for every $c>0$ there exists $d>0$ such that $|x-y|<c$ implies that $|f(x)-f(y)|<d$, there exists $N$ such that $n>N$ implies that $|x_n|<c/2$ and $|y_n|<c/2$, we deduce that $|x_n-y_n|<|x_n|+|y_n|<c$ it implies that $|f(x_n)-f(y_n)|<d$.

$\endgroup$
0
$\begingroup$

$|f(x)-f(y)|<\epsilon$, $x,y\in(0,1)$, $|x-y|<\delta$.

Now we know that $\lim_{n}(x_{n}-y_{n})=0$, so for some $N$, $|x_{n}-y_{n}|<\delta$, $n\geq N$.

Then $|f(x_{n})-f(y_{n})|<\epsilon$ for all such $n$.

$\endgroup$

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.