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Q. $B$ is an open set in $X$ and $A$ is everywhere dense in $X$. Show that $B\subseteq \operatorname{cl}(A\cap B)$.

My approach: Should be suffice to say that if $(A\cap B)$ is dense in $B$ then the proof is easily done. Although I am not sure how that can be done.

Jochen
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Sepia Glen
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    Let $x\in B$. Since $A$ is dense in $X$, there is a sequence $(x_n)\in A^{\mathbb N}$ that converges to $x$. Since $B$ is open, there is $N\in\mathbb N$ s.t. $x_n\in B$ for all $n\geq N$. Therefore, $(x_n)_{n\geq N}\in (A\cap B)^{\mathbb N}$, and thus $x\in Cl(A\cap B)$. The claim follows. – Surb Sep 27 '21 at 10:39

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Let $b \in B$ be arbitrary and let $U$ be an open neighbourhood of $b$ in $X$. Then $U \cap B$ is also an open neighbourhood of $b$ so by density of $A$ in $X$ we get $$U \cap (A \cap B) = A \cap (U \cap B)\ne \emptyset.$$ Therefore, every open neighbourhood of $b$ intersects $A \cap B$ so we conclude $b \in \overline{A \cap B}$. Since $b \in B$ was arbitrary, it follows $B \subseteq \overline{A \cap B}$.

mechanodroid
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