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Let the closure of a set $A$ be $\bar A$. On Page 62, Introduction to Boolean Algebras,Steven Givant,Paul Halmos(2000), an exercise goes like,

Show that $P \cap \bar Q \subseteq \overline{(P \cap Q)}$, whenever $P$ is open.

I felt muddled in face of this sort of exercises. Is there some way to deal with these problems and be assured about the result?

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up vote 2 down vote accepted

Suppose $x \in P \cap \overline{Q}$. To show that $x \in \overline{ P \cap Q }$ we will show that every open neighbourhood of $x$ meets $P \cap Q$.

If $U$ is any open neighbourhood of $x$, then as $x \in P$ it follows that $U \cap P$ is also an open neighbourhood of $x$. As $x \in \overline{Q}$, then $U \cap P$ meets $Q$, or, $U \cap ( P \cap Q ) = ( U \cap P ) \cap Q \neq \emptyset$. Therefore $x \in \overline{ P \cap Q }$.

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Let $x$ in $P\cap\bar Q$. Since $x$ is in $\bar Q$, there exists a sequence $(x_n)_n$ in $Q$ such that $x_n\to x$. Since $x$ is in $P$ and $P$ is open, $x_n$ is in $P$ for every $n$ large enough, say, $n\geqslant n_0$. Hence, for every $n\geqslant n_0$, $x_n$ is in $P\cap Q$. Thus, $x$ is in $\overline{P\cap Q}$ as limit of $(x_n)_{n\geqslant n_0}$.

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This only works if the space in question is Fréchet-Urysohn. – arjafi Jan 8 '13 at 9:37
What if at point $x$, there is not a countable basis? – Metta World Peace Jan 8 '13 at 9:39
@ArthurFischer: Is it possible to adapt this sequence argument to a net one? – Metta World Peace Jan 8 '13 at 9:42
@MettaWorldPeace: Probably so, but there is no need. See Arthur's answer for an alternative. – Cameron Buie Jan 8 '13 at 9:44
@MettaWorldPeace: Yes, of course. If $\{ x_\sigma \}_{\sigma \in \Sigma}$ is a net in $Q$ converging to $x$, then there is a $\sigma^\prime \in \Sigma$ such that $x_\sigma \in P$ for all $\sigma \geq \sigma^\prime$. Take the subnet determined by the directed set $\Sigma^\prime = \{ \sigma \in \Sigma : \sigma \geq \sigma^\prime \}$. This is a net in $P \cap Q$ which converges to $x$. – arjafi Jan 8 '13 at 9:47

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