How do I prove the continuity of this identity function? I'm self-learning topology and the notes asked me to verify the following.
id$:\mathbb{R}_{\text{Sogenfrey}}\rightarrow \mathbb{R}_{\text{usual}}$, Show id is continuous
Here id is the identity function id$(x)=x$ and $\mathbb{R}_{\text{Sogenfrey}}$ is the topology generated by $\{[a,b)\subseteq\mathbb{R}:a<b\}$
However, instead of proving it is continuous, I proved it is discontinuous. So I figured out I may have misunderstood something
Using the definition of continuity
I need to show every $u\in\mathbb{R_{usual}}\Rightarrow \mathrm{id}^{-1}(u)\in\mathbb{R_{Sogenfrey}}$
Let's say $u=\bigcup_{i\in I}(a_{i},b_{i})$, since $\mathrm{id}^{-1}(u)=u$
$(1)$ How is $\bigcup_{i\in I}(a_{i},b_{i})\in\mathbb{R_{Sogenfrey}}$?
The very next exercise in the notes is to prove that $\mathrm{id}:(X,\mathcal{T_{1}})\rightarrow(X,\mathcal{T_{2}})$ is continuous $\iff \mathcal{T_{1}}\ \ \mathrm{refines} \ \ \mathcal{T_{2}}$ and I managed to prove this fairly easily. If I'm not mistaken $\mathcal{T_{1}}\ \ \mathrm{refines} \ \ \mathcal{T_{2}} \ \ \ if \ \ \  \mathcal{T_{2}}\subseteq \mathcal{T_{1}}$
$(2)$ Now how does $\mathbb{R_{usual}} \subseteq \mathbb{R_{Sogenfery}}$? if I take for example $(a,b)\in \mathbb{R_{usual}}$ which is not in $\mathbb{R_{Sogenfery}}$
 A: Yes, to prove that this identity is continuous it suffices to show that $\mathbb{R_{usual}} \subseteq \mathbb{R_{Sorgenfrey}}$, which we prove if we demonstrate that each element $(a,b)$ of the basis of euclidean topology belongs to Sorgenfrey topology. This is because of the equality:
$$(a,b) = \sum_{n=1}^\infty\left[a+\frac\varepsilon n,b\right)$$ for a fixed $\varepsilon<b-a$. This equality shows that each set $(a,b)$ is a sum of open sets in Sorgenfrey topology, therefore it's open.
A: $• \space f:(X, \tau) \to (Y\tau') $ is continuous
$• \space f^{-1}(U) \in\tau$ for all $U\in\tau'$
$• \space f^{-1}(B)\in\tau$ for all $B\in{\mathscr{B}}_{Y}$ where ${\mathscr{B}}_{Y}$ is a basis of $(Y, \tau_Y) $
$• \space f^{-1}(B) =\bigcup_{i\in I}V_i$ where $V_i$'s are basic open set in $(X, \tau) $ and $B$ is a basic open set in $(Y, \tau') $.
All are equivalent.

Id$:\mathbb{R}_{\text{Sogenfrey}}\rightarrow \mathbb{R}_{\text{usual}}$
$B=(a, b) $ is a basic open set in $\mathbb{R}_{\text{usual}}$.
$\text{Id}^{-1}(B) =(a, b)=\bigcup_{n\in\mathbb{N}}[a+\frac{1}{n},b)$
