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It actually says:

a) If $A\cup B=\emptyset$ and both are open (or both closed) then they are separated.

b)If $A$ and $B$ are separated then:

i) If $A\cup B$ is open $\Rightarrow A$ and $B$ are open.

ii) If $A\cup B$ is closed $\Rightarrow A$ and $B$ are closed.

José Osorio
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  • For a) do you mean to say $A\cap B=\emptyset$ instead? Because as it stands currently, $A\cup B=\emptyset$ implies that $A=B=\emptyset$. – T. Eskin Mar 07 '16 at 18:14

1 Answers1

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Remember that two sets are separated if neither has a point in the other's closure. So for $(a)$ we have:

  • Suppose $A\cap B=\emptyset$, $A, B$ are open. Assume they're not separated - WLOG, we have $x\in A\cap \overline{B}$. Since $A$ is open, there's some small $\epsilon$ such that $B_\epsilon(x)\subseteq A$ - but since $x\in \overline{B}$, what can you say about $B_\epsilon(x)\cap B$? Do you see why this means $A\cap B\not=\emptyset$?

  • Suppose $A\cap B=\emptyset$, $A, B$ are closed. Again, assume $x\in A\cap \overline{B}$ for a contradiction. Do you see why this means $x\in A\cap B$?

For $(b)$, we'll use similar arguments. First suppose $A\cup B$ is closed and $A$ is not closed. Let $x\in \overline{A}\setminus A$. Since $A\cup B$ is closed, this means $x\in B$ - contradicting the assumption that $A$ and $B$ are separated. Similarly, if $A\cup B$ is open and $A$ is not open, we can find some $x\in A$ with an open ball $B_\epsilon(x)$ disjoint from $A\setminus\{x\}$. Do you see how to use the assumption that $A\cup B$ is open to get a contradiction, now?

Noah Schweber
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