This is part of Exercise 9 in Section 2.2 of Topology and Groupoids, by Brown.
For each $x \in \mathbb R, N \subseteq \mathbb R$ is a neighborhood of $x$ if and only if there are real numbers $x^{\prime}, x^{\prime \prime}$ such that $x \in [ x^{\prime}, x^{\prime \prime}) \subseteq N$. This is called the half-open topology on $\mathbb R$.
I am supposed to show that this topological space is separable, and that if $\hat A$ is the set of limit points of a set $A \subseteq \mathbb R$, then $A \setminus \hat A$ is countable.
My attempt:
The space is separable because the subset $\mathbb Q \subseteq \mathbb R$ is countable, and every neighborhood of every point must intersect $\mathbb Q$.
Showing that $A \setminus \hat A$ is countable seems harder. I thought about proving it by contradiction. Assume it were uncountable. So there are uncountably many $x \in A$ that have at least one neighborhood that contains no other points of $A$. So there are uncountably many intervals $[a, b)$ that contain only one point of $A$.
I was going to say that none of the intervals can intersect, but I'm not sure that's true.
I'm stuck. Any help is appreciated.
Edit:
If there are uncountably many such points, does that mean they have to form at least a finite interval? Then there couldn't be uncountably many intervals $[a, b)$ each only containing one of the points within a finite interval, right?
Best Answer
You're right about the countable dense subset. (every set $[a,b)$ contains an open interval $(a,b)$ which contains a rational etc.)
If $a \in A\setminus \hat{A}$, then there is a basic subset $[a,f(a))$ with $f(a) \in \Bbb Q$ such that $[a,f(a)) \cap A = \{a\}$ (this is what not being a limit point entails, plus we use the density of $\Bbb Q$). Suppose that we have $a_1 < a_2$ in $ A\setminus \hat{A}$ and $f(a_1) = f(a_2)$. But then $a_2 \in [a_1, f(a_1)$ and this contradicts how $f(a_1)$ was chosen. So $f: A\setminus \hat{A} \to \Bbb Q$ is injective and so the domain of $f$ is at most countable.