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.

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?

share|improve this question
add comment

2 Answers

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 $x \in \overline{ P \cap Q }$.

share|improve this answer
add comment

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 $(x_n)_{n\geqslant n_0}$.

share|improve this answer
This only works if the space in question is Fréchet-Urysohn. –  Arthur Fischer 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 $x$. –  Arthur Fischer Jan 8 '13 at 9:47
show 1 more comment

Your Answer


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.