3
$\begingroup$

Given a uniformly continuous function $f(x)$ on the real numbers $\Bbb R$, then by the definition of uniform continuity this means: for any $\epsilon>0$ there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. My question is: Given $\epsilon_{n} =\frac{1}{3^{n}}$ (for example), and $y_{n}=x_{n}+w_{n}$ so that $|x_{n} -y_{n}|=|w_{n}|$, then we know that there is $\delta_{n}>0$, how can we find the corresponding $\delta_{n}$? I don't know if my question makes sense.

  • 3
    @Monica: First: you presented a *very* general question, with an arbitrary $f$. My comment is simply that there is no way to give an algorithm in such generality (specific functions *may* have algorithms that tell you how to find $\delta_n$). Second: $\delta_n$ is completely independent of $x_n$, $y_n$, and $w_n$; in the uniformly continuous case, $\delta$ depends *only* on $\epsilon$, so the points are utterly irrelevant. In fact, there is absolutely no point in mentioning them at all, except to clutter up the question.2011-02-28

0 Answers 0