Give two subsets that the closure ( A $\cap$ B) = $\emptyset$ and Cl(A) $\cap$ Cl(B) = $\mathbb{R}$

Question

Let U be the topology on $\mathbb{R}$ such that U = {U $\subset$ $\mathbb{R}$ $\vert$ if x $\in$ U, then there exists a,b $\in$ $\mathbb{R}$ such that x $\in$ (a,b) $\subset$ U}

Let L be the topology on $\mathbb{R}$ such that L = {V $\subset$ $\mathbb{R}$ $\vert$ if x $\in$ V, then there exists a,b $\in$ $\mathbb{R}$ such that x $\in$ [a,b) $\subset$ V}

Give an example of two subsets A and B of $\mathbb{R}$$_{U}$ such that Cl(A $\cap$ B) = $\emptyset$ and Cl(A) $\cap$ Cl(B) = $\mathbb{R}$. Does your example work in $\mathbb{R}$$_{L}$

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I came up with: Let A = (-$\infty$, 0) and B = (0,$\infty$). Then Cl(A $\cap$ B) = $\emptyset$ because their intersection is empty, hence the closure of the empty set is empty. But, I wasn't entirely sure if I had the right assumption that Cl(A) = $\mathbb{R}$ and Cl(B) = $\mathbb{R}$ because I can't construct a 'smaller' U-closed set around either intervals but then I considered if ($\infty$,0] and [0, $\infty$) were U-closed, so I checked $\mathbb{R}$$\setminus$($\infty$,0] = (0, $\infty$) and $\mathbb{R}$$\setminus$[0, $\infty$) = (-$\infty$,0) and they were both U-open so-- I'm back where I started. I 'think' because they're 'infinite' that maybe their closure are $\mathbb{R}$ but can someone provide or clarify an example for this?

Thank you!

Answer 1

$(-\infty;0]$ and $[0;\infty)$ are indeed closed, so $Cl((-\infty;0)) = (-\infty;0]$ and your idea will not work, because then $Cl((-\infty;0)) \cap Cl((0;\infty)) = \{0\}$.

Consider this : $Cl(A \cap B) = \emptyset$ implies that $A \cap B = \emptyset$ because $A \cap B \subset Cl(A \cap B)$. Also, $Cl(A) \cap Cl(B) = \mathbb{R}$ implies that $Cl(A) = Cl(B) = \mathbb{R}$.

So what you need to find is two disjoint dense subsets of $\mathbb{R}$. The simplest example I have would be $\mathbb{Q}$ and $\mathbb{R} \backslash \mathbb{Q}$.

Answer 2

Take $A=\mathbb{Q}$, the rational numbers in $\mathbb{R}$, and $B = \mathbb{P}$, the irrational numbers.

Then $A \cap B = \emptyset$ so $\operatorname{Cl}(A \cap B) = \emptyset$ too, while $\operatorname{Cl}(\mathbb{Q}) = \operatorname{Cl}(\mathbb{P}) = \mathbb{R}$, so $\operatorname{Cl}(\mathbb{Q}) \cap \operatorname{Cl}(\mathbb{P}) = \mathbb{R}$, as required.

This example works in both the usual topology $U$ as well as in the lower limit topology $L$.