Order topology and subspace topology in $\mathbb R$

Question

let $X=(-\infty, -1) \cup [0,\infty)$, subspace of $\mathbb{R}$. Then is it different from the order topology? Say $(-1/2,1) \cap X =[0,1)$ is open in $X$, but not open in the order topology??

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I once questioned it... but still confused

so, [0,1) is not basis for order topology on X?( not open in order topology on X?)

Answer 1

Yes $[0,1)$ is not a basis for the order topology because a basis $\mathcal{B}$ for the order topology should contain all sets of the following types:

1) All open intervals $(a,b)$

2) Intervals of the form $[a_1,b)$

3) Intervals of the form $(a,b_0]$.

Then only we call the set a basis.