For $A \subseteq \Bbb R$, the derived set of $A$, denoted by $A'$, is the set of all limit points of $A$.
Theorem: For $A \subseteq \Bbb R$, $A'$ is closed.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Assume that $a \in \Bbb R$ is a limit point of $A'$ . Then for all $\dfrac{\delta}{2}>0$ , there exists $y \in A'$ such that $y \neq a$ and $|y-a|<\dfrac{\delta}{2}$. Let $\delta' = \dfrac{|y-a|}{2}>0$.
$y \in A'$ $\implies$ there exists $x \in A$ such that $x \neq y$ and $|x-y|< \delta'$.
It follows that $|x-y| < \delta' = \dfrac{|y-a|}{2} < |y-a|<\dfrac{\delta}{2}$ and $|y-a| – |x-y| > |x-y|>0$.
As a result,
-
$|x-a| = |(x-y)+(y-a)| \le |x-y| +|y-a|<\dfrac{\delta}{2}+\dfrac{\delta}{2}<\delta$.
-
$|x-a| = |(x-y)+(y-a)| \ge ||y-a| -|x-y||> |x-y| >0$. Thus $x \neq a$.
Hence, for all $\delta >0$, there exists $x \in A$ such that $x \neq a$ and $|x-a| < \delta$. So $a$ is a limit point of $A$ and thus $a\in A'$. Finally, $A'$ is closed.
Best Answer
Another approach is to show the complement is open. Let $x\in (A')^c.$ Then, $x$ is not a limit point of $A$, which means there is an open neighborhood $x\in U(x)$, such that either $U(x)\cap A=\emptyset$ or $U(x)\cap A=\{x\}$. In either case, we have $x\in (A')^c.$
Now, if $y\in U(x)$, then there is an open neighborhood $y\in U(y)\subseteq U(x)$, so $U(y)\cap A=\emptyset,$ from which it follows that $y\in (A')^c$ and therefore that $U(x)\subseteq (A')^c$ so $(A)^c$ is open.