How is this a basis for the ordinal space topology? I encountered this example in a book. Let $\Gamma$ be any ordinal and consider $[0,\Gamma]$ (or equivalently $\Gamma + 1$). In this space, consider the topology generated by sets of the form $(\alpha , \Gamma]$ and $[0 , \beta)$ where $\alpha$ and $\beta$ are ordinals not greater than $\Gamma$. The task is to show that sets of the form $]\alpha,\beta]$ form a basis of this topology. But I cannot understand why.
For instance, consider any set of the form $[0 , \beta)$. Any set of the form $]\alpha,\beta]$ does not contain the ordinal $0$, and hence $[0 , \beta)$ cannot be written as a union of these sets.
Am I wrong somewhere or what I am saying is indeed true?
PS: Here the interval notations have the usual meaning.
 A: You are correct, though the example can be repaired fairly easily.
The topology generated by the sets of the form $(\alpha,\Gamma]$ and $[0,\beta)$ is the smallest topology $\tau$ containing all of those sets. Plainly $\tau$ must contain all of the open intervals
$$[0,\beta)\cap(\alpha,\Gamma]=(\alpha,\beta)$$
for $\alpha<\beta\le\Gamma$ as well as the open nbhds $[0,\beta)$ of $0$ and the open nbhds $(\alpha,\Gamma]$ of $\Gamma$. And we can recognize
$$\{(\alpha,\beta):0\le\alpha<\beta\le\Gamma\}\cup\{[0,\beta):0<\beta\le\Gamma\}\cup\{(\alpha,\Gamma]:0\le\alpha<\Gamma\}$$
as a base for the usual order topology on $[0,\Gamma]$, so $\tau$ is that topology.
And $\{(\alpha,\beta]:0\le\alpha<\beta\le\Gamma\}$ is almost a base for $\tau$. It certainly contains all of the sets $(\alpha,\Gamma]$. It also contains all of the sets $(\alpha,\beta)$, since $(\alpha,\beta)=(\alpha,\beta+1]$. But as you point out, it does not contain any nbhd of $0$. One has to include the sets of the form $(\leftarrow,\alpha]=[0,\alpha]$ for $0\le\alpha\le\Gamma$ in order to have a base for $\tau$ consisting of right-closed intervals.
