# order topology and subspace topology

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??

• As $X$ is not convex, you'd expect that the subspace topology is different from the order topology. I mean, $-1/2$, and $1$ are in $X$ but $(-1/2,1)$ is not contained in it. – David Molano Jun 25 '15 at 16:59
• Also take a look at how I edited your question to make the $\LaTeX$ compile. Looks like you were on the right track. – graydad Jun 25 '15 at 17:00
• @LeviathanTheEsper [0,1) is not basis for order topology on X? – solafide Jun 25 '15 at 17:28
• Only that set wouldn't be a basis, would it? I mean, if it were, it would hold that for each $x\in X$, there is at least one basis element $B$ containing $x$. Take $x=-5$, for example, there is not a $B$ in $\{[0,1)\}$ to which $-5$ belongs. – David Molano Jun 25 '15 at 20:07

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 $[0,1)\not\in\mathscr{T}_\leq$.