Prove that if $B$ is open then $B\cap \bar{A}\subset\overline{A\cap B}$

Question

How can I prove that if $A,B\subset \mathbb{R}^n$ and $B$ is open then $B\cap \bar{A}\subset\overline{A\cap B}$ using balls?

I've seen this proof:

Let $x\in {\overline A}\cap B$. Since $B$ is open, we can chose $r>0$ such that $B_{r}(x)\subset B$. For some $0 < s < r$ we still have $B_s(x)\cap B\neq\emptyset$ and since $x\in\overline A$, then $B_s(x)\cap A\neq\emptyset$. Thus $B_s(x)\cap(A\cap B)\neq\emptyset$.

We have just shown $x\in \overline{A \cap B}.$

but it seems wrong since $s$ isn't arbitrary. The following is my attempt at a proof but I suppose it's also wrong since I don't use the fact that $B$ is open. Why is it wrong?

Let $x\in B\cap\overline{A}$. Then since $x\in\overline{A}$, we have that $\forall r>0 \ B(x;r)\cap A\neq\varnothing$. But $x\in B$ also so $B(x;r)\cap A\cap B\neq \varnothing$. Since $r$ was arbitrary, $x\in\overline{A\cap B}$.

Answer 1

The proof linked is correct. You can replace 'For some $s$...' with 'for all $s$...' and the logic goes through, so $s$ can indeed be taken to be arbitrary.

The incorrect assumption in your proof is that, if $B_r(x) \cap A \neq \emptyset$ and $B_r(x) \cap B \neq \emptyset$, then $B_r(x) \cap A \cap B \neq \emptyset$. If $B$ is not open, then these two nonempty sets can be disjoint, as illustrated by the case where $n = 1$, $B = [-1, 0]$, $A = (0, 1)$, $x = 0$, and $r = \frac{1}{2}$ (drawing this might be helpful).

Answer 2

I think you would need to develop it more. For example, point out that if B is an open set then there is an open ball that contains $x$ and is contained in $B$.

My suggestion:

Let $x \in \overline{A} \cap B$. How $x \in \overline{A}$, then for every $\varepsilon > 0$, we have that

$$B(x; \varepsilon) \cap A \ne \emptyset.$$

How $B$ is open in $\mathbb{R}^n$, then exists $r > 0$ such that

$$x \in B(x;r) \subset B.$$

Then $x \in \overline{A \cap B}$ because for every $\varepsilon' > 0$ we have that

$$B(x;\varepsilon') \cap A \ne \emptyset,$$

and

$$B(x; \varepsilon') \cap B \ne \emptyset.$$

Just note that if $\varepsilon' > r$, then

$$B(x;r) \subset B(x; \varepsilon') \Rightarrow B(x; \varepsilon') \cap B \ne \emptyset.$$

On the other hand, if $\varepsilon < r$ then

$$x \in B(x;\varepsilon') \cap B$$.