Is Sorgenfrey topology generated by interval $[a,b)$ with a being irrational, $b$ being rational

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So the aim is to show that for any open set of Sorgenfrey topology, such open set can be expressed as any union of $[a,b)$, given a being irrational and b rational.

How do we show this? or disprove that $\{[a,b)| \text{ a irrational, b rational}\}$ is not a basis for Sorgenfrey Topology.

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$[0,1)$ is Sorgenfrey open, but not equal to a union of intervals $[a_i, b_i)$, $i \in I$ where all $a_i$ are irrational. If $0 \in [a_i, b_i)$ then $a_i < 0$ (we have $a_i \le 0$ and $a_i \neq 0$) and we'd have points outside $[0,1)$ in the union..

So the mentioned collection is not a base for the Sorgenfrey line.