order topology and subspace topology

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??

Answer 1

Let's call $\mathscr{T}$ the subspace topology over $X$ and $\mathscr{T}_\leq$ the order topology over $X$ derived from the usual order. As you said, $[0,1)$ is open in $X$ because $(-\frac{1}{2},1)$ is open in ${\rm I\!R}$, and $(-\frac{1}{2},1)\cap X = [0,1)$, but there is no point $x\in X$ such that $(x,\rightarrow) = [0,1)$, hence $0\not\in [0,1)^{\circ}$, therefore $[0,1)\not\in\mathscr{T}_\leq$.