Prove that $U$ is a topology on $\Bbb{R}$ I was going through some exercises about topology, and I got stock on one of them.
If $X=\mathbb{R}$ and $u \in U$ if and only if $u$ is a subset of $\mathbb{R}$ and $\forall s\in u \exists t>s$ such that $[s,t) \subseteq u$, where $[s,t)=\{x \in \mathbb{R};s\leq x<t\}$.
Now the book says that I need to proof that $U$ is a topology of $X$, but I don't know what to do with this one. Do I have to prove that $[s,t)$ is a topology? Is that the definition of a half open topology? Can I use it? 
 A: So $\mathcal{T}$ is defined by
$$\mathcal{T}=\{O \subseteq \Bbb R: \forall s \in O: \exists t> s: [s,t) \subseteq O\}$$
We can check directly that this defines a topology on $\Bbb R$:
For $O=\emptyset$, the condition is satisfied voidly (a universal statement over the  empty set is always true), so $\emptyset \in \mathcal{T}$.
For $O=\Bbb R$ we can just take $t=s+1$ for any $s \in O$ and the condition holds, so $\Bbb R \in \mathcal{T}$.
If $O_1, O_2 \in \mathcal{T}$ then let $s \in O_1 \cap O_2$ be arbitrary. Then $s \in O_1$, so the condition on open sets gives us $t_1 > s$ such that $[s,t_1) \subseteq O_1$. Also, $s \in O_2$ so the condition on open sets gives us $t_2 > s$ such that $[s,t_1) \subseteq O_2$. But then $t=\min(t_1,t_2)>s$ too and $[s,t) \subseteq [s,t_1) \cap [s,t_2) \subseteq O_1 \cap O_2$, and as $s$ was arbitrary, $O_1 \cap O_2 \in \mathcal{T}$, by definition. This shows that we have closedness under finite intersections.
If $O_i, i \in I$ is a family of sets from $\mathcal{T}$, then let $s \in \bigcup_{i \in I} O_i$ be arbitrary. Then for some $j \in I$, $s \in O_j$ by the definition of a union. Because $O_j \in \mathcal{T}$ we have $t > s$ such that $[s,t) \subseteq O_j \subseteq \bigcup_{i \in I} O_i$ and so $\bigcup_{i \in I} O_i \in \mathcal{T}$, and we have closedness under arbitrary unions.
We could also have shown that $\mathcal{B}=\{[s,t): s, t \in \Bbb R: s < t\}$ fulfills the axioms for a base for a topology (easy as the union is $\Bbb R$ and $\mathcal{B}$ is closed under finite intersections; facts that were implicitly used in the above proof). The resulting topology is called the Sorgenfrey line or (sometimes) the lower limit topology on $\Bbb R$.
