Sorgenfrey line is hereditarily separable

Question

How to prove that the Sorgenfrey line is hereditarily separable?

Answer 1

This has a proof that for any ordered space separable implies hereditarily separable. And the Sorgenfrey line is a subspace of a separable ordered space (e.g. the double arrow).

Answer 2

Hint: Let $S$ be a subset of the Sorgenfrey line. From each interval of the form $[q,r)$ in the original line, where $q,r \in \mathbb{Q}$, pick one point from $S$ if possible. Then characterize the points in $S$ that are not limits of the points you just chose.

Answer 3

Let $A \subset ℝ$ be arbitrary. If $A$ is countable, then $A$ itself is dense in itself and the proof is done. Assume that $|A| > |ℕ|$.

We will construct countable subsets $D_1, D_2$ of $A$ and show that their sum $D$ is countable and dense in $A$ (i.e. $D \cap U \neq ∅$ for all $\emptyset \neq U \in \tau_S|A$). This will prove, that the Sorgenfrey line is hereditarily separable.

Let $\mathcal{B} = \{ [p, q) \cap A: p,q \in \mathbb{Q}, p < q \} \setminus ∅$. And let $D_1$ be a set that contains one element from each $B \in \mathcal{B}$ (say "hello" to the Axiom of choice). Note that $D_1$ is an infinite countable set.

Remark. In the Euclidean topology $D_1$ is dense. However, in the Sorgenfrey line, it is not necessarily dense. Consider $A = [0, π]$. The singleton $\{π\} = [π, 4) \cap A$ is open in $\tau_S|A$, but it might be the case that we had bad luck when using the axiom of choice and $D_1 = [0, π] \cap \mathbb{Q}$, so $D_1 \cap \{π\} = ∅$, hence $D_1$ is not dense in $A$. We define $D_2$ to deal with such cases.

Let $ D_2 = \{A \cap \sup B : B \in \mathcal{B} \} \setminus ∅$ and $D = D_1 \cup D_2$, which is countable as a sum of countable sets. We show that $D$ is dense in $A$.

Take arbitrary nonempty basic open set $[x, y) \cap A$ in $A$. It suffices to show, that $([x,y) \cap A) \cap D \neq ∅$. There are two cases:

Case I. $[x,y) \cap A = \{ r \}$. Take $p, q \in \mathbb{Q}$ such that $p \leq x \leq r < q < y$. We have \begin{align} \sup ([p,q) \cap A) &= \sup (\overbrace{([x,q) \cap A)}^{\ni r} \cup (\overbrace{[q, y) \cap A}^{= ∅})) \\ &= \sup ([x,y) \cap A) \\ &= \sup \{ r \} \\ &= \{r\}. \end{align} By construction $r \in D_2 \subset D$.

Case II. $\{r, s\} \in [x,y) \cap A$. Without loss of generality assume $r < s$. Take $p, q \in \mathbb{Q}$ such that $x < p < s < q < y$. Since $[p,q) \cap A$ is not empty ($s$ is there!), by construction, there exists $d \in D_1 \subset D$ such that $$ d \in [p, q) \cap A \subset [x, y) \cap A. $$ □