Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone give me a hint/explain how to show this inclusion? $A,B$ are subsets of a topological space. If $x \in A$, showing that $x \in \overline{A - B}$ is obvious, but I'm not sure how to show that if $x$ is in the boundary of $A$, then $x \in \overline{A - B}$

By $\overline A$, I mean closure of $A$.

share|improve this question
    
what does the overlined A mean? the closure? –  Jorge Fernández Sep 13 '12 at 23:49
    
Yes, I meant closure. –  JohnDoe Sep 13 '12 at 23:51
3  
Try using the characterization that $x\in\overline A$ if and only if every open set containing $x$ has nonempty intersection with $A$: Note that if $x\in \overline A\setminus\overline B$, then there is an open set $O$ containing $x$ with $O\cap B=\emptyset$. Now show that if $U$ is open and contains $x$, then it contains a point of $A\setminus B$. –  David Mitra Sep 14 '12 at 0:02

5 Answers 5

The closure of a set $X$ is equal to the union of $X$ and its boundary $\partial X$ (those points $b$ in the topological space such that every open set containing $b$ contains at least one point in $X$ and one not in $X$). Let $x \in \overline{A} - \overline{B}$. There are two cases to consider:

  1. $x \in A$ (and $x \notin \overline{B}$)
  2. $x \notin A$ and $x \in \partial A$ (and $x \notin \overline{B}$)

You mentioned that case 1 is obvious, so I'll explain case 2. To show that $x \in \overline{A - B}$, assume that $x \notin A - B$ (since then it would also be in $\overline{A-B}$). We'll show that this implies $x \in\partial(A - B)$. To that end, let $U$ be any open set containing $x$. We must show that $U \cap (A-B) \neq \emptyset$. Since $x \notin \overline{B}$, there is an open set $V$ containing $x$ such that $V \cap B = \emptyset$. Now $V \cap U \neq \emptyset$ is an open set containing $x$, so using the fact that $x \in \partial A$, there exists $y \in (V\cap U) \cap A$. This puts $y \in U \cap (A-B)$. Thus, by definition, $x \in \overline{A - B}$, as required.

Hope this helps!

share|improve this answer

Let $x\in\overline A - \overline B$. You have to show $x \in \overline{A - B}$. For that let $U$ be an arbitrary neighborhood of $x$. Now you have to show $U\cap (A-B)\neq \emptyset$.

Since $x\notin\overline B$ the set $X- B$ is a neighbourhood of $x$ (where $X$ is the topological space).

Then $U\cap(X- B)$ is a neighbourhood of $x$, too.

Since $x\in\overline A$ you have $A\cap (U\cap(X-B))\neq\emptyset$. But $A\cap (U\cap(X- B))=U\cap (A-B)$.

share|improve this answer

Suppose that $x\in\operatorname{cl}A\setminus\operatorname{cl}B$; then $x\notin\operatorname{cl}B$, so $x$ has an open neighborhood $U$ such that $U\cap B=\varnothing$. Suppose, to get a contradiction, that $x\notin\operatorname{cl}(A\setminus B)$; then $x$ also has an open nbhd $V$ such that $V\cap(A\setminus B)=\varnothing$. Let $W=U\cap V$; then $W\cap\Big(B\cup(A\setminus B)\Big)=\varnothing$. But $B\cup(A\setminus B)=A\cup B$, so $W\cap(A\cup B)=\varnothing$. In particular, $W\cap A=\varnothing$, contradicting the hypothesis that $x\in\operatorname{cl}A$. Thus, $x\in\operatorname{cl}(A\setminus B)$, and therefore $\operatorname{cl}A\setminus\operatorname{cl}B\subseteq\operatorname{cl}(A\setminus B)$.

share|improve this answer

Let's show that $$ \overline{A} - \overline{B} \subset \overline{A - \overline{B}}. $$ This implies the desired inclusion because $\overline{A - \overline{B}} \subset \overline{A - B}$. (We are just saying that it is sufficient to prove for $B$ closed!)

Let $G = \left(\overline{B}\right)^c$. Then, we want to show that $$ \overline{A} \cap G \subset \overline{A \cap G}. $$ But this is clear because for $x \in \overline{A} \cap G$, if $V$ is a neighborhood of $x$, then $V \cap G$ is also a neighborhood of $x$, and therefore intersects $A$, since $x \in \overline{A}$. That is, $V \cap G \cap A \neq \emptyset$. So, every neighborhood of $x$ intersects $A \cap G$, and therefore, $x \in \overline{A \cap G}$.

share|improve this answer
    
If fact we can say that $\overline{\overline A\cap G}=\overline{A\cap G}$, see this question: Show that $\overline{U\cap \overline{A}}=\overline{U\cap A}.$ –  Martin Sleziak Sep 14 '12 at 4:06
    
@MartinSleziak: You are right. I resisted the temptation to add this information to the answer... But I think it is very good to have it on the comments. –  André Caldas Sep 14 '12 at 23:05

Another possible proof using characterization of closure with sequences.

Let $l\in \overline{A} - \overline{B}$. As $l\notin\overline{B}$, there exist an open set $U$ such that $l\in U$ and $U\cap B = \emptyset$. As $l\in\overline{A}$, there exist a convergent sequence $(x_n)$ of $A$ that tends to $l$. Let $N\in\mathbb{N}$ such that for all $n\geq N$, we have $x_n\in U$. We define a new sequence $(y_n)$ by $y_n = x_{N+n}$ for all $n\geq 0$. Clearly, $(y_n)$ is a convergent sequence of $A-B$ that tends to $l$ so $l\in\overline{A-B}$.

share|improve this answer
1  
You should replace sequences by nets here, which will also lead to the goal. –  Flanders Sep 14 '12 at 16:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.