Wanted to check if this proof is valid (not completely sure about the end):
We claim that $$\bar A = A \cup A'$$
Proof:
Take some a $\in A \cup A'.$ If $a \in A$ then $a \in \bar A$. If $a \in A'$ then consider a sequence $(a_n) \rightarrow a$, where $(a_n) \in A$ $\forall n \in \mathbb N $. Then since $\bar A$ is closed it contains all its limit points and so $a \in \bar A$. Thus $\bar A \supset A \cup A'$.
Now take some $x \in \bar A$. Obviously if $x \in A \subset \bar A$ then $ x \in A \cup A'$. Consider some $y \in \bar A \setminus A$. We wish to show that $y \in A'$.
Assume otherwise, that is that $y \not\in A'$. Then $y$ is not a limit point of $A$ and so if we remove $y$ from $\bar A$ it is still closed. But this contradicts the minimality of $\bar A$ so $y$ is a limit point of $A$ and we have equality.
How do I rigorously show the contradiction here (or is it okay)?
Best Answer
How do you conclude that removing a non-limit point from $\overline A$ leaves you with a closed set, without using the theorem that you are proving?
Better argue explicitly that there is an open ball $B$ around $y$ that avoids $A$, so $\overline A\setminus B $ is the intersection of two closed sets ($\overline A$ and the complement of the open set $B$) and therefore still a closed set (that contains $A$).