A is a open set $\iff (\forall X) (A \cap \overline{X} \subset \overline{A \cap X} )$

$A \subset \mathbb{R}^n$ is open $\iff A \cap \overline{X} \subset \overline{A \cap X}$

The $\overline{X}$ represents the closure of a set.

If I assume that $A$ is open we have many cases for the left side $A \cap \overline{X}$.

1. $A \cap \overline{X} = \emptyset$
2. $A \cap \overline{X} = A$ since $A \subset \overline{X}$
3. $A \cap \overline{X} = \overline{X}$ since $\overline{X}\subset A$
4. $A \cap \overline{X} = C$ another set, not open or closed.

In the first case it states true even if $A$ is a closed set. Since $A$ and $\overline{X}$ are disjoint the $\cap$ is a empty set and we don't even need to evaluate the right hand since $\emptyset \subset Z$ where $Z$ is ANY set even $\emptyset$. And it holds true in the first case.

The second case, since $A \subset \overline{X}$ the result is the smaller set $A$. If $A$ is open this is open. If $A$ is closed this is closed. Lets assume $A$ is open (by hypothesis) therefore in the right hand we have $A \subset X = A$ (open) and the closure of $A$ is $\overline{A} = \partial A \cup \mathring{A}$ that is closed.

As left hand $A$ is open we have $A = \mathring{A}$ and the right hand $A$ is closed and we have $A = \overline{A}$

Since $\mathring{A} \subset A \subset \overline{A}$ the statement stands as true in the second case.

...

Starting this way is the best way to do? should I split in 4 cases? I think the proof will become too big and this is not the right approach.

(PS: I want ask a deep sorry, since yesterday I was VERY tired and had no clues how to follow up. Today morning this startup seems almost as easy as write a letter to Santa)

• the convention for sets is capital letters. – TKM Nov 3 '14 at 3:24
• Assume A is open, then take any point of $A\cap\bar{ X}$ and show it must be contained in $\overline{A\cap X}$. – TKM Nov 3 '14 at 3:27
• thank you. I will try that. sorry for the bad defined question at first. i'm sending from cell phone – Lin Nov 3 '14 at 3:32
• The left to right implication is true. But the other direction is clearly false. Let, for example, $A$ be a closed subset of $X$. – T.J. Gaffney Nov 3 '14 at 3:47
• You need a $\forall X$ for it to be true. $$A \text{ is open} \iff \bigl(\forall X\bigr)\bigl( A \cap \overline{X} = \overline{A\cap X}\bigr).$$ – Daniel Fischer Nov 3 '14 at 13:06

$\implies$ Suppose $A \subset \mathbb{R}^n$ is open, $x\in A\cap \overline{X}$, we need to show $x\in\overline{A \cap X}$. Let $G\subset \mathbb{R}^n$ open, then $A\cap G$ is also open due to openness of $A$. Since $x\in \overline{X}$ , $A\cap G\cap X\neq \emptyset$. Hence $\emptyset\neq G\cap (A\cap X)$, by definition, $x\in\overline{A \cap X}$.
$\impliedby$ We proved by contradiction. Suppose $\forall X$, $A\cap \overline{X}\subset \overline{A \cap X}$. Suppose $A$ is not open, then $A^c$ is not closed. Let $X=A^c$, then $A\cap \overline{X}\neq \emptyset$, i.e $\exists x\in A\cap \overline{X}$. By assumption, $x \in A\cap \overline{X}\subset \overline{A \cap X}$. But $A\cap X= \emptyset$. Hence $\overline{A \cap X} =\emptyset$ We get a contradiction.