Prove that $\bar A = A' \cup A$

Question

Let $A \in R^n$: I need to prove that: $$\bar A = A' \cup A$$

So I Said if $x \in \bar A$, then $x$ is in: $$A \cup \bar A\backslash A$$ in the first case I'm done, but how could I prove the second case? (if ($x \in \bar A \backslash A$) )

Edit: Someone suggested a solution in the comments but I didn't understand at all why this is true: $$A∩U=A∩U∖{x}≠∅$$

Thanks Ishan, could someone give me a hint at least on how to move on? — Daniel98, Jun 04 '20 at 20:56
See http://mathonline.wikidot.com/the-closure-of-a-set-equals-the-union-of-the-set-and-its-acc — Paweł Czyż, Jun 04 '20 at 20:57
@Daniel98 the last paragraph of my answer proves your second case. Feel free to ask if anything confuses you. (You can skip the first couple of paragraphs.) — Syed Bukhari, Jun 04 '20 at 21:36

Syed Bukhari · Answer 1 · 2020-06-05T00:24:35.247

In one direction, if $x \in A \cup A^{'}$ then either $x \in A$ or $x \in A^{'}$ .

. If $x \in A \implies x \in \overline{A} $ (trivially).

. If $x \in A^{'}$, by the definition of a limit point, every neighborhood of $x$ intersects $A$ in a point different from $x$. Because their intersections are non empty, $x \in \overline{A}$ by definition.

In the other direction, if $ x \in \overline{A}$ then either $x \in A$ or $x \notin A $ (more specifically $x\in \overline{A} \backslash A$). Now,

. Suppose $ x \in A $. Then certainly $ x \in A \cup A^{'}$

. Now suppose $x \notin A. $ We are given that $ x \in \overline{A}.$ By the definition of a closure of a set, all neighborhoods of $x$ must intersect $A$. Now we have that these intersections are non empty and $x \notin A,$ so both of these imply that these neighborhoods MUST intersect $A$ at some other (different) point(s) of $A$, which is exactly the definition of a limit point. Therefore $x \in A{'}$ and finally $ \implies x \in A \cup A^{'}.$

Prove that $\bar A = A' \cup A$

1 Answers1