0
$\begingroup$

In general topology the following result holds:

In a topological space $X$, $\overline{A}-\overline{B} \subset \overline{A-B}$ for any subsets $A$, $B$ of $X$.

I wonder if the result can be generalized, so I put $A=(A-B) \cup (A \cap B)$ and $B=(B-A) \cup (A \cap B)$ instead of $A$ and $B$ in $\overline{A}-\overline{B}$ (This method comes from the book General Topology by John Kelley). After some computations I get the following identity:

$$\overline{A} - \overline{B} = \overline{A-B} - \overline{B}$$

Does it have a simpler form? Why have I never seen this result on any general topology books?

  • 0
    Could you show your computations? – 2017-02-07
  • 0
    That you never see a lot of such formulae, is because they're not very interesting. Have you looked in Sierpinski's (old!) book? He has lots of these fomulae for closures. – 2017-02-07
  • 0
    That (old!) book sounds interesting. I'm checking it out right now XD. – 2017-02-07

2 Answers 2

0

Following the approach you propose, I get:

$$ \begin{align} \overline{A} - \overline{B} &= \overline{(A-B) \cup (A \cap B)} - \overline{(B-A) \cup (A \cap B)} \\ &= \Big(\overline{A-B} \cup \overline{A \cap B}\Big) - \Big(\overline{B-A} \cup \overline{A \cap B}\Big) \\ &= \overline{A-B} - \Big(\overline{B-A} \cup \overline{A \cap B}\Big) \\ &= \overline{A-B} - \overline{B} \enspace, \end{align}$$ which doesn't equal $\overline{A-B} - \overline{B-A}$. A simple counterexample is provided by two intersecting open discs in $\mathbb{R}^2$.

On the other hand, we can prove that

$$\overline{A} - \overline{B} = \overline{A-B} - \Big(\overline{A} \cap \overline{B}\Big) \enspace. $$

Since $\overline{A} \cap \overline{B} \subseteq \overline{B}$,

$$ \overline{A} - \overline{B} = \overline{A-B} - \overline{B} \subseteq \overline{A-B} - \Big(\overline{A} \cap \overline{B}\Big) \enspace. $$

In the other direction, we have

$$ \begin{align} \overline{A} - \overline{B} &= \overline{A-B} - \Big(\overline{B-A} \cup \overline{A \cap B}\Big) \\ &= \Big(\overline{A-B} - \overline{B-A}\Big) \cap \Big(\overline{A-B} - \overline{A \cap B}\Big)\\ &\supseteq \overline{A-B} - \Big(\overline{A} \cap \overline{B}\Big) \enspace. \end{align}$$ Since $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$, it's clear that

$$\overline{A-B} - \Big(\overline{A} \cap \overline{B}\Big) \subseteq \overline{A-B} - \overline{A \cap B} \enspace.$$

To complete the proof, it remains to show that $\overline{A-B} - \Big(\overline{A} \cap \overline{B}\Big) \subseteq \overline{A-B} - \overline{B-A}$. This can be done by noting that

$$ \overline{A-B} - \overline{B-A} = \overline{A-B} - \Big(\overline{A-B} \cap \overline{B-A}\Big) $$

and $$ \overline{A-B} \cap \overline{B-A} \subseteq \overline{A} \cap \overline{B} \enspace. $$ I have to agree with @Henno that it's not a very interesting result, but perhaps it's a good homework problem.

  • 0
    I got stuck when I'm trying to show $\overline{A-B} \cap \overline{B-A}$ is empty. Could you add more details about how you get the last step? – 2017-02-07
  • 0
    @user413924 You got stuck because that last step was false. I've removed it. – 2017-02-07
  • 0
    $\overline{A- B} \cap \overline{B-A}$ could be the whole space. Take two disjoint dense subsets. – 2017-02-07
0

The formula $\overline{A} - \overline{B}= \overline{A-B}- \overline{B}$ is quite correct, as showing two inclusions will show: if $x$ lies in the left hand side, pick a neighbourhood $V$ of it witnessing that $x \notin \overline{B}$, i.e. $V \cap B = \emptyset$. Then for any open neighbourhood $O$ of $x$, the smaller neighbourhood $O\cap V$ intersects $A$ and misses $B$, so the original $O$ intersects $A-B$, showing $ x \in \overline{A-B}$ and as $x \notin \overline{B}$ still holds, $x $ is in the right hand side. The other inclusion is obvious from $A-B \subseteq A$, so $\overline{A-B} \subseteq \overline{A}$.