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 $x_0$.

Question

(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 $\epsilon$

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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.

Answer 1

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 $|f(x)-f(x_0)|= |f(x_0)-f(x_0)|=0$.