(professor hints)

A Road Map to Glory

  • Write Down the negation of the definition of an accumulation point.
  • Prove that there exists a positive real number $\delta$ for which $$(x_0-\delta, x_0+\delta) \cap D=\{x_0\}$$
  • Prove that the only number $x$ satisfying $x\in D$ and $ |x-x_0| < \delta$ is $x=x_0$.
  • Prove that for such an $x$, $|f(x)-f(x_0)|<\epsilon$ for every positive number $\epsilon$

I have trouble starting from the second bullet point. After that I wouldn't know how to connect it with the third. I wanted to ask for help regarding these two bullets. I understand that due to the negation of accumulation point there exists a finite neighborhood of $x_0$.Im not sure how this connects to the third bullet. I understand that it the $\delta$-neighborhood of $x_0$, however how is that neighborhood finite.

  • $\begingroup$ How do you define “accumulation point”? $\endgroup$ – José Carlos Santos Oct 24 at 23:35
  • 1
    $\begingroup$ Let S be a set of real numbers. A real number B is an accumulation point of S iff every neighborhood of B contains infinitely many points of S. $\endgroup$ – Ever Olivares Oct 24 at 23:37

If $x_0$ is not an accumulation point of $D$ the there exists $r>0$ such that $(x_0-r,x_0+r)$ contains at most finitely many points of $D$ other than $x_0$. Suppose these points (if any) are $x_1,x_2,..,x_n$. Let $\delta =\min \{r,|x-x_0|,...,|x-x_n|\}$. Verify that $(x_0-\delta,x_0+\delta) \cap D =\{x_0\}$.

For the last step just note that $|f(x)-f(x_0)|= |f(x_0)-f(x_0)|=0$.

  • $\begingroup$ Ok, after a bit of studying your response. Wouldn't it be better to let $\delta >0$ from the beginning, thus not having to define $\delta$ by a min. I had a question regarding the question. Is it assumed that $ x \in (x_0-\delta,x_0+\delta)$, because if so I can simply derive that $|x-x_0| <\delta$. Given $D=\{x_0\}$ thus the only inputs would be $x_0$. Sorry Im still trying to grasp the concept completely for this proof. Thank you for responding however! $\endgroup$ – Ever Olivares Oct 28 at 1:02
  • $\begingroup$ @EverOlivares Assuming that $x_0$ is not an accumulation point only lets you conclude that some interval around it contains finitely many points of $D$. But what you need is an interval which contains no point other the $x_0$. For this it is necessary to define $\delta$ the way I have done. $\endgroup$ – Kabo Murphy Oct 28 at 5:13

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