# Set inclusion in topological space

1. Aug 13, 2014

### mahler1

The problem statement, all variables and given/known data.

Let $X$ be a topological space and let $A,B \subset X$. Then

(1) $A \cap \overline{B} \subset \overline{A \cap B}$ when $A$ is open

(2) $\overline{A} \setminus \overline{B} \subset \overline {A \setminus B}$.

The attempt at a solution.

In (1), using the properties (which I've already proved) that $\overline{A \cap B} \subset \overline {A} \cap \overline{B}$ and $\overline{\overline{A}}=\overline{A}$, I have $\overline{A \cap \overline{B}} \subset \overline{A} \cap \overline{\overline{B}}=\overline{A} \cap \overline {B}$. From here I couldn't arrive to anything and I don't know how to use the fact that $A$ is open.

In (2), I've tried to express $\overline{A} \setminus \overline{B}=\overline{A} \cap {\overline {B}}^c$. I want to show that the subset is contained in $\overline{A \setminus B}$, in order to do this I should show that for every $x \in \overline{A} \cap {\overline {B}}^c$ and for every open set $U$ with $x \in U$, I have $U \cap (\overline{A \setminus B}) \neq \emptyset.$ So, if $x \in \overline {A} \cap {\overline B}^c$, then for every $U$ open with $x \in U$, $U \cap A \neq \emptyset$ and there is an open set $V$ with $x \in V \subset \overline {B}^c$. I didn't know how to continue from that point.

Any suggestions would be appreciated.

2. Aug 13, 2014

### Fredrik

Staff Emeritus
Are you familiar with nets? For all $E\subset X$, we have $x\in\overline E$ if and only if x is the limit of a net in E. Also, every open neighborhood of x contains a point from E.

This is an example of how a convergent net can be found: For all x in X, the set of all open neighborhoods of x is a directed set. If you choose one point from each of those sets, you end up with a net that converges to x.

One idea for problem 1 is to let $x\in A\cap\overline B$ be arbitrary, and then use that x is an interior point of A to find a net in $A\cap B$ that converges to x. This approach seems to work, but I haven't made 100% sure.

I don't have time to think about problem 2 right now. Need to get some sleep.

3. Aug 14, 2014

### gopher_p

For (1), $A$ open is a necessary condition. So you need to use it somewhere. I would recommend working from "first principles" - i.e. the definitions of closure and intersection - rather than trying to invoke other theorems or De Morgan's laws. Though depending on your definition of closure, you may need to use some elementary results.

For (2), you might find (1) at least partly useful. Remember, $\overline B^c$ is an open set. I haven't worked it all the way through, but it could turn out to be a follow-your-nose kind of deal after using that.

4. Aug 14, 2014

### mahler1

You're right about using (1) for (2): Since ${\overline{B}}^c$ is open, then $\overline{A} \cap {\overline{B}}^c \subset \overline {A \cap B^c}=\overline {A \setminus B}$.

I think I could solve (1). Suppose $x \in A \cap \overline{B}$. I want to show that for all $U$ open such that $x \in U$, we have $U \cap (A \cap B) \neq \emptyset$. So let $x \in U$ open, since $x \in A$, $U \cap A \neq \emptyset$. Now, $A$ is also open, so $U \cap A$ is open and as $x \in \overline{B}$, then $\emptyset \neq (U \cap A) \cap B=U \cap (A \cap B)$, which is what I wanted to prove.

Thanks for the help!