Let $\mathcal{B}=\{[a,b):a\in \Bbb R, b\in \Bbb Q,a<b\}$
$\mathcal{B}$ is the basis for a topology $\tau_{1}$ on $\Bbb R$.
The space $(\Bbb R,\tau_1)$ is called the Sorgenfrey line
Prove If $\tau$ is the Euclidean topology on $\Bbb R$ then $\tau_1 \supsetneq \tau$.
Strategy
If I can show the union of open sets of the Sorgenfrey line is inside $\tau$
If I can show the intersections of the Sorgenfrey line is inside $\tau$
Then I think I have proved it
The empty set and $X$ are obvious.
Are my assumptions correct
If not what do l have to do?
Thanks
$[0,1) \in \mathcal{B}$ so $[0,1) \in \tau_1$, but $[0,1) \notin \tau$ as $0$ is not an interior point of $[0,1)$ in the Euccidean topology (any neighbourhood of $0$ contains some open interval $(-\varepsilon, \varepsilon)$, but such sets cannot be contained in $[0,1)$ as they always contain negative reals.)
Every basic open set $(a,b)$ ($a < b$) of $\tau_2$ is in $\tau_1$, as $$(a,b) = \bigcup \{[c,d)\mid c \in (a,b), d \in \Bbb Q, c < d < b\}$$ which is a union of sets from $\mathcal{B}$ so lies in $\tau_2$.
As all sets in $\tau$ are unions of these intervals, all Euclidean open sets are in $\tau_1$ and you'rre done showing the inclusion.