# Show that $A \cap B$ is open in $\Bbb R$ if and only if there exists $X \subseteq \Bbb R$ such that $A \cap B = A \cap X$ as follows.

Let $$A,B \subseteq \Bbb R$$ with $$A$$ is open in $$\Bbb R$$. Show that $$A \cap B$$ is open in $$\Bbb R$$ if and only if there exists $$X \subseteq \Bbb R$$ with $$X=\bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$ such that $$A \cap B = A \cap X$$. (Here, $$\Bbb R^+$$ denote the set of all positive real numbers and $$B(x,r)$$ denote the neighborhood of $$x$$ with radius $$r$$.)

Attempt: Suppose that there exists $$X \subseteq \Bbb R$$ with $$X=\bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$ such that $$A \cap B = A \cap X$$. We will show that $$A \cap B$$ is open in $$\Bbb R$$; that is, for all $$y \in A \cap B$$, there exists $$t \in \Bbb R^+$$ such that $$B(y,t) \subseteq A \cap B$$. To this end, let $$y \in A \cap B$$. By hypothesis, we have $$y \in A \cap X$$ which means $$y \in A$$ and $$y \in X$$. By definition of $$X$$, there exists $$x_0 \in X$$ and $$r_0 \in \Bbb R^+$$ such that $$y \in B(x_0,r_0)$$. This means that there exists $$r_1=\min\{\frac{|y-(x_0-r_0)|}{2}, \frac{|y-(x_0+r_0)|}{2}\} \in \Bbb R^+$$ such that $$B(y,r_1) \subseteq B(x_0,r_0) \subseteq X$$. Now, since $$A$$ is open in $$\Bbb R$$, there exists $$r_2 \in \Bbb R^+$$ such that $$B(y,r_2) \subseteq A$$. Define $$t=\min\{r_1,r_2\} \in \Bbb R^+$$, then we have $$B(y,t) \subseteq A \cap X$$. Hence, for arbitrary $$y \in A \cap B$$, we can find $$t \in \Bbb R^+$$ such that $$B(y,t) \subseteq A \cap X = A \cap B$$. Thus, $$A \cap B$$ is open in $$\Bbb R$$.

Conversely, suppose that $$A \cap B$$ is open in $$\Bbb R$$. We will show that there exists $$X \subseteq \Bbb R$$ with $$X= \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$ such that $$A \cap B = A \cap X$$. Let's define $$X = \bigcup_{s \in A \cap B \\ t \in \Bbb R^+} B(s,t).$$ First, we must show that $$X = \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$. Let $$y \in X$$. By the definition of $$X$$, there exists $$s_1 \in A \cap B$$ and $$t_1 \in \Bbb R^+$$ such that $$y \in B(s_1,t_1)$$. If we define $$r=\min\{\frac{|y-(s_1-t_1)|}{2}, \frac{|y-(s_1+t_1)|}{2}\} \in \Bbb R^+,$$ then $$B(y,r) \subseteq \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\},$$ so that $$y \in \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$. Hence, $$X \subseteq \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$.

Conversely, let $$y \in \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$. Then there exists $$x_0 \in X$$ and $$r_0 \in \Bbb R^+$$ such that $$y \in B(x_0,r_0)$$. Since $$x_0 \in X$$ and $$A \cap B$$ is open in $$\Bbb R$$, there exists $$s_2 \in A \cap B$$ and $$r_2 \in \Bbb R^+$$ such that $$x_0 \in B(s_2, r_2) \subseteq A \cap B$$. Hence, there exists $$x_0 \in A \cap B$$ and $$r_0 \in \Bbb R^+$$ such that $$y \in B(x_0,r_0)$$. By the definition of $$X$$, we have $$y \in X$$. Hence, $$X \supseteq \bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}.$$

Therefore, $$X =\bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}.$$

Next, by construction, it's clear that $$X \subseteq \Bbb R$$. Now, we will show that $$A \cap B = A \cap X$$.

$$(\subseteq)$$ Let $$y \in A \cap B$$. Since $$A \cap B$$ is open in $$\Bbb R$$, there exists $$r \in \Bbb R^+$$ such that $$B(y,r) \subseteq A \cap B$$. By the definition of $$X$$, we have $$y \in B(y,r) \subseteq X$$. Hence, $$y \in A \cap X$$. Since $$y$$ was arbitrarily given, we can conclude that $$A \cap B \subseteq A \cap X$$.

$$(\supseteq)$$ Let $$y \in A \cap X$$. Since $$y \in X$$ and $$A \cap B$$ is open in $$\Bbb R$$, there exists $$s_y \in A \cap B$$ and $$r_y \in \Bbb R^+$$ for which $$y \in B(s_y,r_y) \subseteq A \cap B$$. Hence, $$y \in A \cap B$$. Since $$y$$ was arbitrarily given, we can conclude that $$A \cap X \subseteq A \cap B$$.

Therefore, $$A \cap B = A \cap X$$. In all, if $$A \cap B$$ is open in $$\Bbb R$$, then there exists $$X \subseteq \Bbb R$$ with $$X=\bigcup \{B(x,r):x \in X \text{ and } r \in \Bbb R^+\}$$ such that $$A \cap B = A \cap X$$.

Does my approach above work? Thanks in advanced.

Your current proof seems OK to me (though lengthy!) Here's a simpler way to prove the backward $$(\impliedby)$$ implication:
Let $$A,B \subseteq \Bbb R$$ with $$A$$ open in $$\Bbb R$$. Suppose there exists an $$X\subseteq\Bbb R$$ satisfying the all of the above conditions. Then $$X$$ must be open in $$\Bbb R$$ since the sets $$B(x,r)$$ are (by definition) open in $$\Bbb R$$ and an arbitrary union of open sets is always open. Next, observe that since $$A$$ is open in $$\Bbb R$$, the set $$A\cap X$$ must also be open in $$\Bbb R$$ because a finite intersection of open sets is always open. By hypothesis, since $$A\cap B=A\cap X$$, we must have that $$A\cap B$$ is open in $$\Bbb R$$.
• Thanks. What about the proof ofthe forward $(\implies)$ implication? Commented Jun 25, 2023 at 5:18