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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.

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1 Answer 1

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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 $\Bbb R$.

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  • $\begingroup$ Thanks. What about the proof ofthe forward $(\implies)$ implication? $\endgroup$
    – math404
    Commented Jun 25, 2023 at 5:18
  • $\begingroup$ I'm still trying to see if there's a simpler way to prove the forward (right) implication. If I have something, I'll update my answer. $\endgroup$ Commented Jun 25, 2023 at 5:19
  • $\begingroup$ OK. But, what I want to know is about whether my recent attempt correct, not the other way. $\endgroup$
    – math404
    Commented Jun 25, 2023 at 5:26
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    $\begingroup$ @math404 : As I have stated at the very beginning of my answer, your current proof does seem OK to me. $\endgroup$ Commented Jun 25, 2023 at 5:27

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