Show that the lexicographic order topology for $\mathbb{N}\times \mathbb{N}$ is not the discrete My question is:
Show that the lexicographic order topology for $\mathbb{N}\times \mathbb{N}$ is not the discrete
I have been thinking on the fact that on the discrete topology all singleton sets are open. If  can I find one singleton not open the proof is done?
For example $\{(0,0)\}$ is not open because it has not a predecessor
is it right? I don't know how to write it properly
 A: For order topologies we also include as (basic) open sets all intervals of the form $$( \leftarrow , a ) = \{ x \in X : x < a \}$$ where $a \in X$. So in the lexicographic order order on $\mathbb{N} \times \mathbb{N}$ we have $\{ \langle 0 , 0 \rangle \} = ( \leftarrow , \langle 0 , 1 \rangle )$.  (We similarly include all intervals of the form $( a , \rightarrow ) = \{ x \in X : a < x \}$ as (basic) open sets.)
It is relatively easy to prove that if $<$ is a strict linear order on a set $X$, then $x \in X$ is isolated in the order topology iff 


*

*$x$ has an immediate predeccessor, or is the minimum element; and

*$x$ has an immediate successor, or is the maximum element.


Can you think of points in $\mathbb{N} \times \mathbb{N}$ (points similar to the one you chose, in fact) where one of these two fails?
A: HINT: In two steps,


*

*Show that $\{x\}$ is open if and only if $x$ is a successor of some $y$ and the predecessor of some $z$; or $x$ is the minimal element and it has a successor.

*Show that for $n>0$, the point $\langle n,0\rangle$ is not a successor in the lexicographic order, and not the minimal element.
$\implies$ Conclude that $\Bbb{N\times N}$ is not dicrete.
A: Definitely not. First, being closed is not the same as not open. In fact, if every singleton set is open, then every set is both open and closed.
Also, $\{(0,0)\}$ is open, because it's the set of points less than $(1,0)$.
Have you tried looking at $\{(0,1)\}$? Is it open?
