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A and B are subsets of a topological space. How do i prove that if $A \cap B = \emptyset$ and A is an open set, then $\overline A \cap Int(\overline B) = \emptyset $?

Here's my part of proof:

$A \cap B = \emptyset \rightarrow A \cap \overline B = A \cap (B \cup \partial B) = (A \cap B) \cup (A \cap \partial B) = \emptyset \cup (A \cap \partial B) = A \cap \partial B$

$Int (A \cap \overline B) = Int (A \cap \partial B) = Int A \cap Int (\partial B) =$ (since A is open) $= A \cap \emptyset = \emptyset$

$Int (A \cap \overline B) = Int A \cap Int (\overline B) = A \cap Int (\overline B)$

$A \cap Int(\overline B) = \emptyset$

$\overline A \cap Int(\overline B) = (Int (A) \cup \partial A) \cap Int(\overline B) = (A \cup \partial A) \cap Int(\overline B) = (A \cap Int(\overline B)) \cup (\partial A \cap Int(\overline B)) = \emptyset \cup (\partial A \cap Int(\overline B)) = \partial A \cap Int(\overline B)$

Arctic Char
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Anton
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  • Hint: if $x\in\overline A\cap\mathop{\rm Int}(\overline B)$, then every neighborhood of $x$ contains a point in $A$, and there exists a neighborhood of $x$ contained within $\overline B$. Can you arrive at a contradiction using the two given assumptions? – Greg Martin Dec 11 '21 at 00:38

1 Answers1

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Since $A$ is open, $A^c$ is closed. Since $B$ is disjoint from $A$, $B\subset A^c$ and thus $\overline B \subset A^c$. Thus $A\cap \overline B = \emptyset$.

On the other hand, since $\operatorname{int} (\overline B) \subset \overline B$ by definition and is open, $A\cap \overline B = \emptyset$ implies that $A \subset [\operatorname{int} (\overline B)]^c$, which is closed. Thus $\overline A \subset [\operatorname{int} (\overline B)]^c$ by definition of closure. This is the same as $\overline A \cap \operatorname{int} (\overline B) = \emptyset$.

DanielWainfleet
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Arctic Char
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