Suppose $A \subset \mathbb{R}$ be an uncountable set and let $B \subset \mathbb{R}$ be a set defined by $s \in B$ if and only if the sets {$x \in \mathbb{R}$: $x \in A$ and $x \gt s$} and {$x \in \mathbb{R}$: $x \in A$ and $x \lt s$} are uncountable.
How to prove that B is non-empty and open?
My trying was to prove that the interior set of $A$ is in the $B$ so $B$ is not empty. But how to prove that the uncountable set must have an interior point?
I am not sure how to show $B$ is open.
Thank you.
Let $A^+$ be the set of $x\in \Bbb R$ such that $A\cap (x,\infty)$ is uncountable. Let $A^-$ be the set of $x\in \Bbb R$ such that $(-\infty,x)\cap A$ is uncountable. Of course $B=A^+\cap A^-.$
$(1).$ Suppose $x\in B.$ Then $(x,\infty)\subset A^-.$ If there did not exist $y>x$ such that $(x,y)\subset A^+$ then for each $n\in \Bbb N$ there would exist $y_n\in (x,x+1/n)\setminus A^+.$ But then $A\cap (x,\infty)=\cup_{n\in \Bbb N}(A\cap (y_n,\infty))$ is a countable union of countable sets and hence is countable, contrary to $x\in A^+.$
So for $x\in B$ there exists $y>x$ such that $(x,y)\subset B.$
Similarly if $x\in B$ there exists $y'<x$ such that $(y',x)\subset B.$
So $B$ is open.
$(2).\;$ $A^-$ and $A^+$ are not empty, otherwise either $A=\cup_{n\in \Bbb Z}(A\cap (-\infty,n))$ is countable or $A=\cup_{n\in \Bbb Z}(A\cap (n,\infty))$ is countable.
$(2a).$ If $A^-=\Bbb R$ then $B=A^-\cap A^+=\Bbb R \cap A^+=A^+\ne \emptyset.$
$(2b).$ If $A^-\ne \Bbb R$ let $P=\Bbb R\setminus A^-.$ Let $x_0=\sup P,$ which exists in $\Bbb R.$
Otherwise for each $n\in \Bbb Z$ there would exist $y_n\in [n,\infty)\cap P,$ implying $n\in P.$ But then $A=\cup_{n\in \Bbb Z})(A\cap (-\infty,n))$ would be countable.
Observe that $x_0\in P,$ that is, $$(-\infty,x_0)\cap A \text { is countable.}$$ Because $x_0=\sup P,$ so for each $n\in \Bbb N$ there exists $y_n\in (x_0-1/n,x_0]\cap P,$ that is, such that $A\cap (-\infty, y_n)$ is countable. So $(-\infty,x_0)\cap A=\cup_{n\in \Bbb N}((-\infty,y_n)\cap A)$ is countable.
Observe that by definition of $P$ and of $x_0$ we have $$(x_0,\infty)\subset A^-.$$ Now $(x_0,\infty)$ cannot be disjoint from $A^+, $ otherwise $ (x_0,\infty)\cap A=\cup_{n\in \Bbb N}( (x_0+1/n,\infty)\cap A)$ would be countable, BUT then $A\subset [ (-\infty,x_0)\cap A]\cup \{x_0\}\cup [ (x_0,\infty)\cap A]$ would imply that $A$ is countable.
So there exists $y\in (x_0,\infty)\cap A^+\subset A^-\cap A^+=B.$