1
$\begingroup$

Suppose that $f:X \rightarrow \mathbb{R}$ is uniformly continuous, where $X$ is a compact metric space. By definition of uniform continuity, there exists $\delta$, independent of point chosen such that $d_X(x,y)<\delta \Rightarrow |f(x)-f(y)|<\epsilon$.

Question: Suppose that $x,y \in X$ such that $d_X(x,y) > \delta$. I want to show that $|f(x)-f(y)|< \epsilon^*$, where $\epsilon^*$ depends on $\epsilon$. Can I perform the followings?

Since $y \notin B(x,\delta)$, choose $x_1 \in B(x,\delta)$ and consider $B(x_1,\delta)$. If $y \in B(x_1,\delta)$, then we have $|f(x)-f(y)|<2 \epsilon$. Otherwise, choose $x_2 \in B(x_1, \delta)$ and consider $B(x_2, \delta)$. If $y \in B(x_2, \delta)$, then we have $|f(x)-f(y)|<3 \epsilon$. Generally, if $y \in B(x_n,\delta)$, then $|f(x)-f(y)|<(n+1) \epsilon$.

I am wondering whether this process terminates for some $n_0 \in \mathbb{N}$. I think the answer is yes since $X$ is compact.

$\endgroup$
  • $\begingroup$ Where does it say that $X$ is compact? $\endgroup$ – Tryss Jul 26 '15 at 11:17
  • $\begingroup$ Edited. Thanks. $\endgroup$ – Idonknow Jul 26 '15 at 11:25
  • 1
    $\begingroup$ What about $\epsilon^*$ as a function of $\epsilon$? Do you allow it to be unbounded as $\epsilon \to 0$? $\endgroup$ – Siminore Jul 26 '15 at 11:28
  • $\begingroup$ No. $\epsilon^*$ is bounded $\endgroup$ – Idonknow Jul 26 '15 at 12:22
1
$\begingroup$

If $X $ is compact, there exists $k>0$ with $d (x,y)<k $ for all $x,y $. So you can take $\epsilon^*=k $, or you can choose any $n $ such that $n \epsilon >k $.

$\endgroup$

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.