$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)

  • $\begingroup$ the convention for sets is capital letters. $\endgroup$ – TKM Nov 3 '14 at 3:24
  • $\begingroup$ Assume A is open, then take any point of $A\cap\bar{ X}$ and show it must be contained in $\overline{A\cap X}$. $\endgroup$ – TKM Nov 3 '14 at 3:27
  • $\begingroup$ thank you. I will try that. sorry for the bad defined question at first. i'm sending from cell phone $\endgroup$ – Lin Nov 3 '14 at 3:32
  • 1
    $\begingroup$ The left to right implication is true. But the other direction is clearly false. Let, for example, $A$ be a closed subset of $X$. $\endgroup$ – T.J. Gaffney Nov 3 '14 at 3:47
  • $\begingroup$ You need a $\forall X$ for it to be true. $$A \text{ is open} \iff \bigl(\forall X\bigr)\bigl( A \cap \overline{X} = \overline{A\cap X}\bigr).$$ $\endgroup$ – Daniel Fischer Nov 3 '14 at 13:06

You don't need to separate into several cases.

$\implies$ 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.


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