Let $\{0,1\}\times\mathbb{N}$ be equipped with the lexicographical order. Let $\tau_{LO}$ be the topology regarding the lexicographical order on $\{0,1\}\times\mathbb{N}$. Show that $\tau_{LO}$ is not discret.
To show that $\tau_{LO}$ is not discret it is enough to give $U\in\mathcal{P}(\{0,1\}\times\mathbb{N})$ with $U\notin\tau_{LO}$.
The lexicographical order $<$ is defined as follows:
$u=(u_1,u_2)<(v_1,v_2)=v\iff\begin{cases}u_1<v_1~\text{or}\\u_1=v_1~\text{and} ~u_2<v_2\end{cases}$
A base of the order topology is given by:
$\mathcal{O}=\{(u,v)|u,v\in X, u<v\}\cup\{(-\infty, u), (u,\infty)|u\in X\}\cup\{X\}$.
So there is a set, which can not be written as union of sets in $\mathcal{O}$.
Such a set is $\{(1,0)\}\in\mathcal{P}(\{0,1\}\times\mathbb{N})$.
So an interval $(u,v)$ with $u,v\in\{0,1\}\times\mathbb{N}$ which contains the tupel $(1,0)$ [not the interval] always has other elements and is never a singleton.
Suppose there are such $u,v$. Then $u<(1,0)<v$.
Hence $u_1<1$ or $u_1=1$ and $u_2<0$. Obviously the seconds case does not hold. So $u_1=0$. The successor of $(1,0)$ is obviously $v=(1,1)$.
So we have $(0,u_2)<(1,0)<(1,1)$ and how ever you choose $u_2$ there will always be infinite many other tupel 'between' $(0,u_2)$ and $(1,0)$.
In other words: A set containing $(1,0)$ can never be open, because $(1,0)$ has no unique predecessor.
And of course this shows explicit that $\{(1,0)\}$ is not open. So $\tau_{LO}$ is not the discret topology.
Is this correct?
Thanks in advance.