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Given a topological space ($X$, $\mathcal{T}$) and the following defintions

  • $\partial A$ := { $x$ | every neighbourhood of $x$ contains points from $A$ and from $X \setminus A$ }
  • $\overline{A}$ := $A \cup \partial A$
  • $A^{\circ}$ := $A \setminus \partial A$.

Now i want to prove that $$ \overline{A \cap B} \subseteq \overline{A} \cap \overline{B} $$ Proof: With set theory $$ \overline{A} \cap \overline{B} = (A \cup \partial A) \cap (B \cup \partial B) = (A \cap B) \cup (\partial A \cap B) \cup (A \cap \partial B) \cup (\partial A \cap \partial B) $$ and $\overline{A \cap B} = (A \cap B) \cup \partial (A \cap B)$. So if $x \in \overline{A \cap B}$. Now i distinguish two cases.

(i) $x \in (A \cap B)$, then its obvious that $x \in \overline{A} \cap \overline{B}$ with the equations from above

(ii) $x \in \partial (A \cap B)$, here i have no idea how to proceed, because i have no idea how to decompose the expression $\partial (A \cap B)$, do you have any suggestions for me?

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your second line is a hint (the first dotted line) – Berci Oct 13 '12 at 15:13
up vote 6 down vote accepted

Remember that if $X\subseteq Y$ then $\overline{X}\subseteq \overline{Y}$.

As $A\cap B\subseteq A$ and $A\cap B\subseteq B$ then $\overline{A\cap B}\subseteq \overline{A}$ and $\overline{A\cap B}\subseteq \overline{B}$ so

$$\overline{A\cap B}=(\overline{A\cap B})\cap(\overline{A\cap B})\subseteq \overline{A}\cap \overline{B}$$

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+1. This is the most direct approach. Straight to the heart of the matter. – Did Oct 13 '12 at 18:09
Not exactly, I venture to say, since Andres uses properties which you are required to prove from the definitions in OP's question. – Mad Hatter Oct 13 '12 at 18:34
yes, but this property $X \subseteq Y \Rightarrow \overline{X} \subseteq \overline{Y}$ was easy to proof for me from my definitions, so this is okay. – Stefan Oct 14 '12 at 22:57

You can use the following characterization of the boundary $\partial A$ of a set $A$ in a topological space $X$:

A point $x \in X$ is in $\partial A$ if and only if for every neighborhood $U$ of $x$ we have $U \cap A \neq \emptyset$ and $U \cap A^c \neq \emptyset$.

There is also an analogous definition for the closure:

A point $x\in X$ is in the closure $\overline A$ of $A$ if and only if every neighborhood $U$ of $x$ satisfies $U \cap A \neq \emptyset$.

Now apply this to $A \cap B$.

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But this was essentially the definition of the boundary given in the question? – joriki Oct 13 '12 at 15:15
@joriki: Ah yes, I noticed this after I posted the answer. Still, the definition of the closure should be helpful, assuming it's allowed to use it. – fbg Oct 13 '12 at 15:22

Let $x \in \partial (A \cap B)$.

Let $U$ be any neighborhood of $x$. Then there exists some $y \in U \cap (A \cap B)$ and some $z \in U \backslash (A \cap B)$.

Then $y$ in $U \cap A$ and in $U \cap B$, and $z$ is either in $U \backslash A$ or in $U \backslash B$.

Now, the problem to face is that $z$ could be in $U \backslash A$ but when you change $U$ to $U'$ you get a $z'$ in $U \backslash B$. Anyhow, looking at $U \cap U'$ should help you fix this issue....

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The answer to the case you ask about is hidden in the definition of $\partial A$ (which is included in your question). Since it is formulated by means of neighbourhoods, you are rather supposed to use them.

Let $\mathscr{N}_x$ be the set of all neighbourhoods of $x$.

Notice, that the following (standard) characterization of $\overline{A}$ is a consequence of your definitions: \begin{equation}\tag{$\star$} x\in\overline{A}\quad\text{iff}\quad(\forall U\in\mathscr{N}_x)(N\cap A\neq\emptyset). \end{equation}

And now use ($\star$) to prove the remaining part.

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