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I'm tried to prove that $\mathbb{R}_{S}$ = (Sorgenfrey line) is a Baire Space. I find that my prove is correct, but i'm not sure.

$\{U_n; n \in \mathbb{N}\}$ are a collection of open and dense sets in $\mathbb{R}_{S}$. If we can prove that $[a,b) \cap \left(\cap_{n \in \mathbb{N}} U_n \right) \neq \emptyset$ then $\left(\cap_{n \in \mathbb{N}} U_n \right)$ is dense and $\mathbb{R}_{S}$ is Baire.

Consider $x_1 \in [a,b) \cap U_1$, then exist $y_1$ such that $[x_1,y_1] \subset [a,b) \cap U_1$.

Consider $x_2 \in [x_1,y_1) \cap U_2$, then exist $y_2$ such that $[x_2,y_2] \subset [x_1,y_1) \cap U_2$.

Inductively we can construct a family of subsets $[x_{n+1}, y_{n+1}] \subset [x_{n}, y_{n}) \cap U_n$

How $[x_{n+1}, y_{n+1}] \subset [x_{n}, y_{n}]$ then exist $x \in \cap_{n \in \mathbb{N}} [x_{n}, y_{n}]$. Furthermore $x \in \cap_{n \in \mathbb{N}} [x_{n}, y_{n}] \subset \cap_{n \in \mathbb{N}} U_n$. Then $x \in [a,b) \cap \left(\cap_{n \in \mathbb{N}} U_n \right)$, therefore $\left(\cap_{n \in \mathbb{N}} U_n \right)$ is dense.

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I'm sure it's familiar enough to those deep in topology, but a definition of the Sorgenfrey line would really help make this question self-contained. – Steven Stadnicki Aug 26 '13 at 21:04
You are right. The Sorgenfrey line is $\mathbb{R}$ with the topology generated by the following set $\mathcal{B} = \{[a,b); a<b, a,b \in \mathbb{R}\}$. – jon jones Aug 26 '13 at 21:11
Looks good to me too. – Jonathan Y. Aug 26 '13 at 21:13

You have to be a little more careful: when you have $x_{n+1}\in[x_n,y_n)$, you need to specify that you can choose $y_{n+1}>x_{n+1}$ so that $[x_{n+1},y_{n+1}]\subseteq[x_n,y_n)$, because you need $[x_{n+1},y_{n+1})$ to be non-empty. Otherwise, though, the argument is fine.

An alternative is to note that a subset of $\Bbb R$ is dense in the Sorgenfrey topology if and only if it’s dense in the Euclidean topology. Then for each $n\in\Bbb N$ let $V_n$ be the Euclidean interior of $U_n$, and show that $V_n$ is a dense open set in the Euclidean topology on $\Bbb R$. Now you can simply appeal to the fact that $\Bbb R$ in its usual topology is a Baire space: $\bigcap_{n\in\Bbb N}V_n$ is dense in $\Bbb R$ in its usual topology and therefore in the Sorgenfrey topology, and of course so is the larger set $\bigcap_{n\in\Bbb N}U_n$.

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