# Let $f: D\rightarrow \mathbb{R}$ and assume that $x_0 \in D$ is not an accumulation point of $D$. Prove that $f$ is continuous at $x_0$.

(professor hints)

• 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.

• How do you define “accumulation point”? Commented Oct 24, 2019 at 23:35
• 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. Commented Oct 24, 2019 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 $$|f(x)-f(x_0)|= |f(x_0)-f(x_0)|=0$$.
• 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! Commented Oct 28, 2019 at 1:02
• @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. Commented Oct 28, 2019 at 5:13